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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: TrueTears on December 26, 2008, 02:16:52 am

Title: TrueTears question thread
Post by: TrueTears on December 26, 2008, 02:16:52 am: I think i might be posting too much threads for just 1 question, so i'll make a separate thread XD (please excuse my noob questions haha). Here goes...

1. Simplify the following expression, in an exact form.

$sin( tan^{-1}(-2)) = \frac{2\sqrt{5}}{5}$

Let $tan^{-1}(-2) = x$

$\implies tan(x)=-2$

$\implies \frac{1}{sin^2(x)}=1 + \frac{1}{(-2)^2}$

$\implies sin^2(x) = 4/5$

$\implies sin(x) = \frac{2\sqrt{5}}{5}$

however the answers in my book say its $-\frac{2\sqrt{5}}{5}$ . How do i get the negative?

2. Let $sec \beta = b$ where $\beta \in [ \frac{\pi}{2},\pi]$ . Find, in terms of $\beta$ , 2 values of x in the range $[-\pi,\pi]$ which satisfies each of these equations:

a) $sec x = -b$

b) $cosec x = b$

Let $tan \gamma = c$ where $\gamma \in[ \pi,\frac{3\pi}{2}]$ . find in terms of $\gamma$ , 2 values of x in the range of $[0,2\pi]$ which satisfies

a) $cot x = c$

3. Given that $tan \alpha = p$ , where $\alpha$ is an acute angle, find each of the following in terms of $p$ .
a) $tan( \frac{3\pi}{2} + \alpha)$
Title: Re: TrueTears question thread
Post by: vce08 on December 26, 2008, 02:19:40 am: You get the neg when u square root the sin^2(x) thingy.
Title: Re: TrueTears question thread
Post by: Damo17 on December 26, 2008, 08:01:08 am: Quote from: TrueTears on December 26, 2008, 02:16:52 am

$\implies sin^2(x) = 4/5$

$\implies sin(x) = \frac{2\sqrt{5}}{5}$

however the answers in my book say its $-\frac{2\sqrt{5}}{5}$ . How do i get the negative?

$sin^2(x) = 4/5$
$\implies sin(x) =\pm \sqrt\frac{4}{5}$
$= \pm \frac{2\sqrt{5}}{5}$
Title: Re: TrueTears question thread
Post by: TrueTears on December 26, 2008, 01:27:05 pm: yeah but why is the positive not one of the answers?
Title: Re: TrueTears question thread
Post by: Matt The Rat on December 26, 2008, 01:29:49 pm: It'd be where the $tan^{-1}(-2)$ lies
Title: Re: TrueTears question thread
Post by: TrueTears on December 26, 2008, 01:43:24 pm: but isn't $tan$ negative in the 2nd or 4th quadrant, so $tan(x)=-2$ could be either the 2nd or 4th quadrant. $sin$ is also positive in 2nd quadrant and negative in 4th quadrant, so shouldn't both positive and negative answers be included?
Title: Re: TrueTears question thread
Post by: Matt The Rat on December 26, 2008, 01:59:55 pm: Remember where $tan^{-1}(x)$ is defined as a function ( $Range: (\frac{-\pi}{2},\frac{\pi}{2})$ ) and as that leaves only the fourth quadrant as possible then it must be a negative answer.
Title: Re: TrueTears question thread
Post by: TrueTears on December 26, 2008, 02:01:02 pm: ahhh thanks Matt The Rat :D
Title: Re: TrueTears question thread
Post by: Mao on December 26, 2008, 02:03:52 pm: quite right. but remember, the range of arctan is $\left( -\frac{\pi}{2},\frac{\pi}{2}\right)$ , so technically, you don't actually have the second quadrant to work with.

remember that arcsin and arctan only spits out values in 4th and 1st Q, and arccos only gives 1st and 2nd Q
Title: Re: TrueTears question thread
Post by: Mao on December 26, 2008, 02:09:02 pm: the rest of the questions aren't hard to do, but very hard to explain without a drawing instrument, and i do not have my tablet nearby...
but the easiest way to do it is to draw graphs, then set up hypothetical points and see what works :)
Title: Re: TrueTears question thread
Post by: TrueTears on December 27, 2008, 03:49:14 pm: Simplify $(cis(\frac{\pi}{2} - \alpha))^7$

$\implies cis( \frac{7\pi}{2} - 7\alpha)$

$\implies \frac{cis(\frac{7\pi}{2})}{cis(7\alpha)}$

$\implies \frac{cis(\frac{-\pi}{2})}{cis(7\alpha)}$

$\implies cis( \frac{-\pi}{2} - 7\alpha)$

however the answers in my book says its meant to be $cis( \frac{3\pi}{2} - 7\alpha)$

why is it $\frac{3\pi}{2}$ and not $\frac{-\pi}{2}$ ? $\frac{3\pi}{2}$ is not allowed in the interval of $(-\pi,\pi]$ for Argument, isn't it?
Title: Re: TrueTears question thread
Post by: TrueTears on December 27, 2008, 03:57:16 pm: And also...

Show that $cos(\alpha) - sin(\alpha) i = cis(-\alpha)$

thanks
Title: Re: TrueTears question thread
Post by: Damo17 on December 27, 2008, 04:01:47 pm: That interval would be true if the answer $\frac{-\pi}{2}$ was the actual Argument. However when you have the $-7\alpha$ in there as part of the Argument then most likely what they are doing is making it neater by making the $\frac{-\pi}{2}$ into a positive angle $\frac{3\pi}{2}$ . If you substituted a value for $\alpha$ then the answer you come up with would have to be within the interval of $(-\pi,\pi]$ .
Title: Re: TrueTears question thread
Post by: TrueTears on December 27, 2008, 04:07:03 pm: ah i see, so how do i know when to leave the answers as they are or to change them to be within the interval of $(-\pi,\pi]$

eg, $cis(-2\pi+4\alpha)$ was simplified to $cis(4\alpha)$

however, $cis(-3\pi+6\alpha)$ was left as $cis(-3\pi+6\alpha)$ (according to book's answers)

so, how do i know when to keep simplifying or stop?
Title: Re: TrueTears question thread
Post by: Damo17 on December 27, 2008, 04:48:04 pm: Quote from: TrueTears on December 27, 2008, 03:57:16 pm
And also...

Show that $cos(\alpha) - sin(\alpha) i = cis(-\alpha)$

thanks

Maybe use:

$\frac{1}{cis\theta}= cis(-\theta)$

$\frac{1}{cos\theta+sin\theta i}=cis(-\theta)$

$cos\theta-sin\theta i=cis(-\theta)$

Or

If $z=r cis(\theta)$ then $\tilde{z}= r cis(-\theta)$

in this case:
$z= cis(\theta)$
$= cos(\theta)+sin(\theta)i$

Therefore $cos(\theta)-sin(\theta)i=cis(-\theta)$
Title: Re: TrueTears question thread
Post by: TrueTears on December 27, 2008, 04:52:50 pm: ahh thanks again Damo17.
Title: Re: TrueTears question thread
Post by: TrueTears on December 27, 2008, 04:57:30 pm: Quote from: TrueTears on December 27, 2008, 04:07:03 pm
ah i see, so how do i know when to leave the answers as they are or to change them to be within the interval of $(-\pi,\pi]$

eg, $cis(-2\pi+4\alpha)$ was simplified to $cis(4\alpha)$

however, $cis(-3\pi+6\alpha)$ was left as $cis(-3\pi+6\alpha)$ (according to book's answers)

so, how do i know when to keep simplifying or stop?

yeah so is there a generalization which i can follow?
Title: Re: TrueTears question thread
Post by: Damo17 on December 27, 2008, 05:11:29 pm: Quote from: TrueTears on December 27, 2008, 04:57:30 pm
Quote from: TrueTears on December 27, 2008, 04:07:03 pm
ah i see, so how do i know when to leave the answers as they are or to change them to be within the interval of $(-\pi,\pi]$

eg, $cis(-2\pi+4\alpha)$ was simplified to $cis(4\alpha)$

however, $cis(-3\pi+6\alpha)$ was left as $cis(-3\pi+6\alpha)$ (according to book's answers)

so, how do i know when to keep simplifying or stop?

yeah so is there a generalization which i can follow?

The $-2\pi$ does not change the point of $z$ . Adding or subtracting $2\pi$ from any angle will yield the same base angle/ same point on the graph. That is why that have taken it out.

As to the $-3\pi$ , that does change the point on the graph so they have kept it in.

Unfortunately I am limited in my knowledge of this so if you still don't understand i'm sure someone else could explain it much better.
Title: Re: TrueTears question thread
Post by: TrueTears on December 27, 2008, 05:15:12 pm: ah i seem to understand now, but isn't $-3\pi$ the same as $-\pi$ ? so why can't $cis(-3\pi+6\alpha)$ be simplified to $cis(-\pi+6\alpha)$ ?
Title: Re: TrueTears question thread
Post by: Damo17 on December 27, 2008, 05:19:12 pm: Quote from: TrueTears on December 27, 2008, 05:15:12 pm
ah i seem to understand now, but isn't $-3\pi$ the same as $-\pi$ ? so why can't $cis(-3\pi+6\alpha)$ be simplified to $cis(-\pi+6\alpha)$ ?

I thought that myself and to be honest I have no idea why.
Title: Re: TrueTears question thread
Post by: TrueTears on December 27, 2008, 06:57:28 pm: Quote from: TrueTears on December 27, 2008, 05:15:12 pm
ah i seem to understand now, but isn't $-3\pi$ the same as $-\pi$ ? so why can't $cis(-3\pi+6\alpha)$ be simplified to $cis(-\pi+6\alpha)$ ?
hmmm, could anyone else see if they can find a reason why ? :D

in the meanwhile....

Factorise the first expression into linear factors over C, given that the second expression is one of the linear factors.

$z^3 + (1-i)z^2 + (1-i)z - i$ , $z-i$
Title: Re: TrueTears question thread
Post by: Matt The Rat on December 27, 2008, 07:11:30 pm: The expressions are equal. $cis(-3\pi+6\alpha) = cis(-\pi+6\alpha)$

For the factorisation,

$(z-i)(Az^2 + Bz + C) = z^3 + (1-i)z^2 + (1-i)z - i$

is the way I'd approach it. Then just equate co-efficients and then factor the quadratic bracket again to have it as its linear factors.
Title: Re: TrueTears question thread
Post by: TrueTears on December 27, 2008, 07:27:43 pm: ahhh, thanks very much matt the rat
Title: Re: TrueTears question thread
Post by: Damo17 on December 27, 2008, 07:29:28 pm: Quote from: Matt The Rat on December 27, 2008, 07:11:30 pm
The expressions are equal. $cis(-3\pi+6\alpha) = cis(-\pi+6\alpha)$

I did know that, but I am not sure why they would leave the $-3\pi$ there instead of putting it to just $-\pi$ .
Title: Re: TrueTears question thread
Post by: shinny on December 27, 2008, 08:12:37 pm: Was the value or a possible range of alpha given? The value inside the cis should be within the principal domain, so that may cause it to be left as -3pi. If it isn't, then without having an idea of what alpha is, there is no way to determine what value should be used to put it within the principal domain and anything is acceptable really.
Title: Re: TrueTears question thread
Post by: TrueTears on December 27, 2008, 09:08:25 pm: no range was given it was just like that, so im guessing both answers are correct.

Also, this question:

If $z = 1 + i$ is a zero of the polynomial $z^3+az^2+bz+10-6i$ , find the values of $a$ and $b$ given that they are real.

Well this is a fairly easy question, since the coefficients are real, so 2 of the roots would be $1 + i$ and $1 - i$ . (conjugate pairs.)

So u sub both those in the former equation and solve the 2 simultaneous equations, however i keep getting $a = 3-3i$ and $b = -8+6i$ , even after solving on calc. But, the question says they are real numbers. Can anyone see what's wrong?
Title: Re: TrueTears question thread
Post by: shinny on December 27, 2008, 09:21:36 pm: The conjugate root theorem ONLY applies if all coefficients are real, which isn't the case here. I might have a go at it later in latex if I get the time.
Title: Re: TrueTears question thread
Post by: TrueTears on December 27, 2008, 09:26:47 pm: but isn't all the coefficients of $z$ all real?
Title: Re: TrueTears question thread
Post by: ell on December 27, 2008, 09:26:55 pm: Imagine it like this: $z^3+az^2+bz^{1}+(10-6i)z^{0}$
Title: Re: TrueTears question thread
Post by: TrueTears on December 27, 2008, 09:29:58 pm: Quote from: ell on December 27, 2008, 09:26:55 pm
Imagine it like this: $z^3+az^2+bz^{1}+(10-6i)z^{0}$
ahhhh i see, now i get it thanks ell and shinny
Title: Re: TrueTears question thread
Post by: Damo17 on December 27, 2008, 09:50:32 pm: Quote from: TrueTears on December 27, 2008, 09:08:25 pm
no range was given it was just like that, so im guessing both answers are correct.

Also, this question:

If $z = 1 + i$ is a zero of the polynomial $z^3+az^2+bz+10-6i$ , find the values of $a$ and $b$ given that they are real.

Well this is a fairly easy question, since the coefficients are real, so 2 of the roots would be $1 + i$ and $1 - i$ . (conjugate pairs.)

So u sub both those in the former equation and solve the 2 simultaneous equations, however i keep getting $a = 3-3i$ and $b = -8+6i$ , even after solving on calc. But, the question says they are real numbers. Can anyone see what's wrong?

Sub in $z=1+i$ .

$z^3+az^2+bz+10-6i$

$= (1+i)^3+a(1+i)^2+b(1+i)+10-6i$

$=-2+2i+a2i+b+ib+10-6i$

$-2+b+10=0$

      $b=-8$

and $2i+a2i+ib-6i=0$
$2i+a2i-8i-6i=0$
         $a2i=14i$
      $a=6$
Title: Re: TrueTears question thread
Post by: TrueTears on December 27, 2008, 11:23:00 pm: ah yeah, i just got that answer as well xD

and also, this question:

Find the solutions of the equation $z^4-2z^2+4=0$ in polar form.

just not sure how to factorise it...
Title: Re: TrueTears question thread
Post by: humph on December 27, 2008, 11:26:39 pm: Make the substitution $w = z^2$ first, and solve the corresponding quadratic. Then substitute back in $z = \pm \sqrt{w}$ for your four solutions.
Title: Re: TrueTears question thread
Post by: TrueTears on December 27, 2008, 11:32:54 pm: ahh yes thanks humph i got it.
Title: Re: TrueTears question thread
Post by: TrueTears on December 28, 2008, 01:06:37 am: another 2 Q's

1. Prove that for any complex number $z$ , $3|z-1|^2 = |z+1|^2$ if and only if
$|z-2|^2=3$
Title: Re: TrueTears question thread
Post by: Mao on December 28, 2008, 01:51:14 am: 1. let z=x+yi

$3|z-1|^2 = |z+1|^2$

$\implies 3\times |(x-1)+yi|^2 = |(x+1)+yi|^2$

$\implies 3 \times \left( (x-1)^2 + y^2) = (x+1)^2 + y^2$

$\implies 3x^2 - 6x + 3 + 3y^2 = x^2 + 2x + 1 + y^2$

$\implies 2x^2 - 8x +2 + 2y^2 = 0$

$\implies x^2 - 4x +1 + y^2 = 0$

$\implies (x-2)^2 -3 + y^2 = 0$

$\implies |(x-2)+yi|^2 = 3$

$\implies |z-2|^2 = 3$
Title: Re: TrueTears question thread
Post by: Mao on December 28, 2008, 01:57:48 am: 2.

$x^2-y^2=y$

$\implies x^2 -y^2 -y -\frac{1}{4}=-\frac{1}{4}$

$\implies x^2 -\left(y+\frac{1}{2}\right)^2 = -\frac{1}{4}$

$\implies 4 \left(y+\frac{1}{2}\right)^2 - 4x^2 = 1$
Title: Re: TrueTears question thread
Post by: TrueTears on December 28, 2008, 02:01:09 am: thanks mao for that proof, lol i dono why i posted that other Q up there, i got it straight after i posted LOL, probs too late in the night o.O
Title: Re: TrueTears question thread
Post by: TrueTears on December 28, 2008, 01:30:00 pm: 1. On the Argand plane sketch $|\frac{z-1-i}{z}|=1$

how do i simplify $|\frac{z-1-i}{z}|=1$ so i can sketch it?

2. If the real part of $\frac{z-1}{z+1}$ is zero, find the locus of points representing z in the complex plane.

Thanks
Title: Re: TrueTears question thread
Post by: humph on December 28, 2008, 01:51:45 pm: Quote from: TrueTears on December 28, 2008, 01:30:00 pm
1. On the Argand plane sketch $|\frac{z-1-i}{z}|=1$

how do i simplify $|\frac{z-1-i}{z}|=1$ so i can sketch it?
As you know that $z \neq 0$ , you can multiply both sides by $|z|$ . So you have that
$|z-1-i| = |z|$
$\Longrightarrow (x-1)^2 + (y-1)^2 = x^2 + y^2$
$\Longrightarrow x^2 - 2x + 1 + y^2 - 2y + 1 = x^2 + y^2$
$\Longrightarrow y = 1 - x$

Quote from: TrueTears on December 28, 2008, 01:30:00 pm
2. If the real part of $\frac{z-1}{z+1}$ is zero, find the locus of points representing z in the complex plane.

Thanks
$\frac{z-1}{z+1} = \frac{(x-1)+iy}{(x+1)+iy}$
$\qquad = \frac{((x-1)+iy)((x+1)-iy)}{((x+1)+iy)((x+1)-iy)}$
$\qquad = \frac{x^2 - 1 + 2iy + y^2}{(x+1)^2 + y^2}$
$\qquad = \frac{x^2 - 1 + y^2}{(x+1)^2 + y^2} + i\frac{2y}{(x+1)^2 + y^2}$
Now the real part of this is zero, which is the first term. This means that the numerator of the first term is equal to zero; that is,
$x^2 - 1 + y^2 = 0$
$\Longrightarrow x^2 + y^2 = 1$
So the locus of points representing z for which the real part of $\frac{z-1}{z+1}$ is zero is merely the unit circle in the complex plane.

Interestingly enough, a question very similar to this was on my complex analysis exam last semester! (Mind you, it was the very first question, a "warm-up question" if you will.)
Title: Re: TrueTears question thread
Post by: TrueTears on December 28, 2008, 01:57:00 pm: ahhh thanks humph, so wat happens to the $i\frac{2y}{(x+1)^2 + y^2}$ ?
Title: Re: TrueTears question thread
Post by: humph on December 28, 2008, 02:08:17 pm: Oh yeah, I forgot. You can't have that $z=-1$ as otherwise $\frac{z-1}{z+1}$ is undefined. So the answer actually should be $\{z \in \mathbb{C} : |z|=1, \; z \neq -1 \}$ .

The $i\frac{2y}{(x+1)^2 + y^2}$ is just the imaginary part of $\frac{z-1}{z+1}$ ; it's unimportant in this question.
Title: Re: TrueTears question thread
Post by: TrueTears on December 28, 2008, 02:12:37 pm: ahhh, thanks that clears it all up xD
Title: Re: TrueTears question thread
Post by: TrueTears on December 28, 2008, 05:42:31 pm: and...

If $Arg z = \frac{\pi}{4}$ and $Arg(z-3) = \frac{\pi}{2}$ , find $Arg(z-6i})$ .
Title: Re: TrueTears question thread
Post by: vce08 on December 28, 2008, 06:03:23 pm: Quote from: TrueTears on December 28, 2008, 05:42:31 pm
and...

If $Arg z = \frac{\pi}{4}$ and $Arg(z-3) = \frac{\pi}{2}$ , find $Arg(z-6i})$ .

Is the answer - pi/4?
Title: Re: TrueTears question thread
Post by: TrueTears on December 28, 2008, 06:07:46 pm: yeah lol
Title: Re: TrueTears question thread
Post by: ed_saifa on December 28, 2008, 06:20:58 pm: Quote from: TrueTears on December 28, 2008, 05:42:31 pm
and...

If $Arg z = \frac{\pi}{4}$ and $Arg(z-3) = \frac{\pi}{2}$ , find $Arg(z-6i})$ .
http://vcenotes.com/forum/index.php/topic,2787.msg34557.html#msg34557
Title: Re: TrueTears question thread
Post by: TrueTears on December 28, 2008, 07:47:31 pm: o thanks very much
Title: Re: TrueTears question thread
Post by: TrueTears on December 28, 2008, 09:32:19 pm: and just another Q, i got $\sqrt{\frac{1+cos(\alpha)}{1-cos(\alpha)}}$

How can i simplify that further?

My book does...

$\sqrt{(\frac{1+cos(\alpha)}{1-cos(\alpha)}) (\frac{1+cos(\alpha)}{1+cos(\alpha)})}$

why do u multiply by $\frac{1+cos(\alpha)}{1+cos(\alpha)}$ ? its not rationalising or anything is it?

Why can't u multiply it by $\frac{1-cos(\alpha)}{1-cos(\alpha)}$ ?
Title: Re: TrueTears question thread
Post by: vce08 on December 28, 2008, 09:35:31 pm: cos the base becomes 1 - (cos(x))^2 which is equal to (sin(x))^2 when u use the book thingy
Title: Re: TrueTears question thread
Post by: humph on December 28, 2008, 09:59:26 pm: Quote from: TrueTears on December 28, 2008, 09:32:19 pm
and just another Q, i got $\sqrt{\frac{1+cos(\alpha)}{1-cos(\alpha)}}$

How can i simplify that further?

My book does...

$\sqrt{(\frac{1+cos(\alpha)}{1-cos(\alpha)}) (\frac{1+cos(\alpha)}{1+cos(\alpha)})}$

why do u multiply by $\frac{1+cos(\alpha)}{1+cos(\alpha)}$ ? its not rationalising or anything is it?

Why can't u multiply it by $\frac{1-cos(\alpha)}{1-cos(\alpha)}$ ?

$\sqrt{\frac{1+\cos(\alpha)}{1-\cos(\alpha)}} = \sqrt{\left(\frac{1+\cos(\alpha)}{1-\cos(\alpha)}\right) \left(\frac{1+\cos(\alpha)}{1+\cos(\alpha)}\right)}$
$= \sqrt{\frac{\left(1+\cos(\alpha)\right)^2}{1-\cos^2(\alpha)}}$
$= \sqrt{\frac{\left(1+\cos(\alpha)\right)^2}{\sin^2(\alpha)}}$
$= \frac{1+\cos(\alpha)}{\sin(\alpha)}$

$\sqrt{\frac{1+\cos(\alpha)}{1-\cos(\alpha)}} = \sqrt{\left(\frac{1+\cos(\alpha)}{1-\cos(\alpha)}\right) \left(\frac{1-\cos(\alpha)}{1-\cos(\alpha)}\right)}$
$= \sqrt{\frac{1-\cos^2(\alpha)}{\left(1-\cos(\alpha)\right)^2}}$
$= \sqrt{\frac{\sin^2(\alpha)}{\left(1-\cos(\alpha)\right)^2}}$
$= \frac{\sin(\alpha)}{1-\cos(\alpha)}$

So there are two ways to simplify it. The first way is "nicer" because the denominator is simpler. But it doesn't make that much of a difference...
Title: Re: TrueTears question thread
Post by: TrueTears on December 28, 2008, 10:09:21 pm: ahh thanks humph
Title: Re: TrueTears question thread
Post by: TrueTears on December 29, 2008, 07:04:15 pm: could someone please list the newton notation way to differentiate $sin^{-1}(x)$ , $cos^{-1}(x)$ and $tan^{-1}(x)$ . Like for product rule in newton notation it's just $f(x)g'(x) + g(x)f'(x)$ . My book doesn't make it very clear how to differentiate the inverse of sin, cos, tan. yeah so pretty much could someone please teach me how to differentiate those 3 inverses. lol
Title: Re: TrueTears question thread
Post by: Matt The Rat on December 29, 2008, 07:22:19 pm: There's a couple of different ways to do it.

For $sin^{-1}(x) , cos^{-1}(x) and tan^{-1}(x)$ the derivatives are given on the formulae sheet.

When they become slightly more complex (ie, $sin^{-1}(\frac{3}{x}))$ then either than chain rule or implicit diff could be used

Chain Rule: Make the substitution $u = \frac{3}{x}$ and then $y = sin^{-1}(u)$

Implicit Diff: $y = sin^{-1}(\frac{3}{x}) => sin(y) = \frac{3}{x}$ Use that info to make a right angled triangle (with $x$ as the hypotenuse, $3$ as the opposite side and $\sqrt{x^2-9}$ and $y$ as the appropriately placed angle) as you'll need to work out $cos(y)$ for the derivative.

I haven't finished the examples completely but just shown the groundwork of how to approach the problem.
Title: Re: TrueTears question thread
Post by: TrueTears on December 29, 2008, 07:30:59 pm: yeah, im just beginning the differentiations for inverse trig functions, i think i get how to do the basic differentiations now, but theres one question in the an example i dont quite get.

Differentiate with respect to x
$sin^{-1}(x^2-1)$

Let $y = sin^{-1}(x^2-1)$ and $u = x^2-1$

so $\frac{dy}{dx} = \frac{1}{\sqrt{1-u^2}}$ x $2x$

$\implies \frac{2x}{\sqrt{1-(x^2-1)^2}}$

after simplification it becomes $\frac{2x}{\sqrt{x^2}\sqrt{2-x^2}}$

then this is where i don't understand. My book says $\frac{dy}{dx} = \frac{2}{\sqrt{2-x^2}}$ if $0 < x< \sqrt{2}$

and $\frac{dy}{dx} = \frac{-2}{\sqrt{2-x^2}}$ if $-\sqrt{2} < x < 0$ .

How do u get 2 answers for $\frac{dy}{dx}$ ? And how do u get the restrictions for $x$ , ie $0 < x< \sqrt{2}$ and $-\sqrt{2} < x < 0$ .
Title: Re: TrueTears question thread
Post by: Matt The Rat on December 29, 2008, 07:42:40 pm: $\frac{2x}{\sqrt{x^2}\sqrt{2-x^2}}$

$=> \frac{2x}{x\sqrt{2-x^2}}$

$=> \frac{2}{\sqrt{2-x^2}}$

$= \frac{dy}{dx}$

The domain of the derivative is to do with where the $sin^{-1}$ function is defined.

As its domain initially was within $[-1,1]$ then

$-1 < x^2 - 1 < 1$

$\therefore -\sqrt{2} < x < \sqrt{2}$

The terminals are not included as you would be dividing by 0 in the derivative.
Title: Re: TrueTears question thread
Post by: TrueTears on December 29, 2008, 07:46:36 pm: ahh yes i think i understand, but wat about the 0, how do u know to split it at $0 < x < \sqrt{2}$ and $-\sqrt{2} < x < 0$

and how come $\frac{dy}{dx} = \frac{-2}{\sqrt{2-x^2}}$ if $-\sqrt{2} < x < 0$ . How do u get the negative?
Title: Re: TrueTears question thread
Post by: humph on December 29, 2008, 09:28:38 pm: Should be
$\frac{dy}{dx} = \frac{2x}{\sqrt{x^2}\sqrt{2-x^2}} = \frac{2x}{|x|\sqrt{2-x^2}}$
Now remember that
$\frac{x}{|x|} = \begin{cases} 1 & \text{if $x > 0$}\\ -1 & \text{if $x < 0$}\\ \text{undefined} & \text{if $x=0$} \end{cases}$
Title: Re: TrueTears question thread
Post by: Ahmad on December 29, 2008, 10:12:15 pm: Quote from: humph on December 28, 2008, 09:59:26 pm
$\frac{\sin(\alpha)}{1-\cos(\alpha)}$

This is probably perfectly acceptable, but it might be pursued further:

$\frac{\sin(\alpha)}{1-\cos(\alpha)} = \frac{2\sin(\alpha/2) \cos(\alpha/2)}{1-(1 - 2\sin^2(\alpha/2))} = \cot(\alpha/2)$
Title: Re: TrueTears question thread
Post by: TrueTears on December 30, 2008, 04:36:04 pm: ah thanks.

also,

1. show that the parabola with equation $ax^2+bx+c$ has no points of inflexion.

2. Determine the points of inflexion for the following function:

$(x^2-1)(x^2+1) = y$

$\implies \frac{dy}{dx} = 4x^3$

$\implies \frac{d^2y}{dx^2} = 12x^2$

so point of inflexion occurs when $\frac{d^2y}{dx^2} = 0$ $\implies x = 0$ . But this is not a point of inflexion, why is that?
Title: Re: TrueTears question thread
Post by: /0 on December 30, 2008, 04:55:59 pm: Quote from: TrueTears on December 30, 2008, 04:36:04 pm
ah thanks.

also, show that the parabola with equation $ax^2+bx+c$ has no points of inflexion.

$\frac{dy}{dx} = 2ax+b$

$\frac{d^2y}{dx^2} = 2a$

Assuming $a \neq 0$ , $\frac{d^2y}{dx^2} \neq 0$ , so the graph will always be either concave up or concave down.
Title: Re: TrueTears question thread
Post by: humph on December 30, 2008, 09:28:44 pm: Quote from: TrueTears on December 30, 2008, 04:36:04 pm
ah thanks.

also,

1. show that the parabola with equation $ax^2+bx+c$ has no points of inflexion.

2. Determine the points of inflexion for the following function:

$(x^2-1)(x^2+1) = y$

$\implies \frac{dy}{dx} = 4x^3$

$\implies \frac{d^2y}{dx^2} = 12x^2$

so point of inflexion occurs when $\frac{d^2y}{dx^2} = 0$ $\implies x = 0$ . But this is not a point of inflexion, why is that?
If $x$ is a point of inflection of $f$ , then $f''(x)=0$ . Unfortunately, the converse is not true: $f''(x)=0$ does not necessarily imply that $x$ is a point of inflection of $f$ . That is, $f''(x)=0$ is necessary but not sufficient for $x$ to be a point of inflection of $f$ .
Title: Re: TrueTears question thread
Post by: TrueTears on December 30, 2008, 10:07:18 pm: ah i see, so what are the conditions for point of inflexion?

And also this question:

The equation of a curve C is $x^3+xy+2y^3 = k$ , where $k$ is a constant

a) find $\frac{dy}{dx}$ $\implies$ $\frac{dy}{dx} = \frac{-3x^2-y}{x+6y^2}$

C does have a tangent parallel to the y axis

b) show that the y coordinate at the point of contact satisfies $216y^6+4y^3+k = 0$

how do u do part b)?

many thanks
Title: Re: TrueTears question thread
Post by: humph on December 31, 2008, 01:45:05 am: Quote from: TrueTears on December 30, 2008, 10:07:18 pm
ah i see, so what are the conditions for point of inflexion?
I think it involves calculating $f''$ at points near and either side of the point you're interested in. Can't remember the exact details though, but it should be in a textbook.

Quote from: TrueTears on December 30, 2008, 10:07:18 pm
And also this question:

The equation of a curve C is $x^3+xy+2y^3 = k$ , where $k$ is a constant

a) find $\frac{dy}{dx}$ $\implies$ $\frac{dy}{dx} = \frac{-3x^2-y}{x+6y^2}$

C does have a tangent parallel to the y axis

b) show that the y coordinate at the point of contact satisfies $216y^6+4y^3+k = 0$

how do u do part b)?

many thanks
" $C$ does have a tangent parallel to the $y$ -axis" means that the gradient is infinite at this point (this is quite dodgy mathematically but hopefully you understand what I mean geometrically - a straight vertical line is kinda like a linear function with infinite gradient).
Anyway, this will obviously happen when the denominator of $f'$ is equal to zero. You found this to be $x+6y^2$ , so we have that $x+6y^2=0 \implies x = -6y^2$ .
So what we have is that when the tangent of the curve is parallel to the $y$ -axis, and so the gradient is "infinite", we must have that the point $(x,y)$ at which this occurs must satisfy the relation $x = -6y^2$ . In particular, we can substitute this value of $x$ into the equation of the curve (as this point lies on the curve) in order to find that $216y^6+4y^3+k = 0$ .
Title: Re: TrueTears question thread
Post by: TrueTears on December 31, 2008, 01:50:17 am: ah yessss thanks so much , that was wat i was confused about, coz a vertically has undefined gradient lol
Title: Re: TrueTears question thread
Post by: TrueTears on December 31, 2008, 02:47:39 pm: yeah and just back to the point of inflexion stuff, my book says "A point of inflexion will occur at $x = \alpha$ if $f''(x) = 0$ , and $f''(\alpha + \beta)$ and $f''(\alpha - \beta)$ have different signs."

So to work out the points of inflexion for $f(x) = sin(x)$ for $x \in [0,2\pi]$

$f''(x) = -sin(x)$

$\implies -sin(x) = 0$

$\implies x = 0 , \pi , 2\pi$

so $f''(0+1) = -0.841$ ... and $f''(0-1) = 0.841$ ... So there will be a point of inflexion at $x=0$ since both signs are different.

But this is where i don't get, if you do the same to $\pi, 2\pi$ you get:

$f''(\pi+1) = 0.841$ ... and $f''(\pi-1) = -0.841$ ... So both signs are different, so there should be a point of inflexion at $x = \pi$ . However my book's answers says there isn't. The same goes for $2\pi$ . Can someone clarify this?
Title: Re: TrueTears question thread
Post by: Matt The Rat on December 31, 2008, 03:10:02 pm: For Spec, I don't think you need to be too worried expect about points of inflexion. Just remember $f"(x)=0$ and you should be right. If not, blame Coblin; it's what he told me. But the way the book describes it is correct.
Title: Re: TrueTears question thread
Post by: /0 on December 31, 2008, 04:36:32 pm: I think $x=0,\pi,2\pi$ should all be points of inflection. An easy way to visualise it is as if you are driving a car on the curve. Every time you are not 'turning' left or right, but driving straight (momentarily) you will get a point of inflection.
Title: Re: TrueTears question thread
Post by: TrueTears on December 31, 2008, 06:42:01 pm: Quote from: /0 on December 31, 2008, 04:36:32 pm
I think $x=0,\pi,2\pi$ should all be points of inflection. An easy way to visualise it is as if you are driving a car on the curve. Every time you are not 'turning' left or right, but driving straight (momentarily) you will get a point of inflection.

ah, kk so i guess the book's answer is wrong o.o

And also some other questions

1. Consider $f(x) = tan^{-1} + tan^{-1}(\frac{1}{x})$ , $x$ does not equal 0

If x > 0, find $f(x)$ .

My book does $f(x) = tan^{-1}(x) + \frac{\pi}{2} - tan^{-1}(x)$ . How do u get the $\frac{\pi}{2}$ ?

2. $A = \frac{\sqrt{4+p^2}}{2} + \frac{p^2}{2\sqrt{4+p^2}} -1$ . Show that $A \ge 0$ for all $p$

3. Co-ordinate Z is described by $x = 3( cos(\theta) - sin(\theta) )$ and $y = 2( sin(\theta) + cos(\theta) )$ , find the locus of Z as $\theta$ varies.

Many thanks again, i really appreciate all your helps XD
Title: Re: TrueTears question thread
Post by: /0 on January 01, 2009, 02:29:51 am: 1.
The derivative of the entire expression is 0, so the graph is constant.

$f(1) = \frac{\pi}{2}$

Alternatively, derive from a triangle... or use the fact that $\tan{f(x)}= undefined$ to deduce that $f(x) = \frac{\pi}{2}$ .

2.

$A = \frac{p^2+2}{\sqrt{p^2+4}}-1$

$\Rightarrow \frac{dA}{dp} = \frac{p(p^2+6)}{(p^2+4)^\frac{3}{2}}$

If the graph has non-negative gradient, then:

$\frac{dA}{dp} \geq 0 \Rightarrow p(p^2+6) \geq 0$

Since $p(p^2+6)$ is the product of two numbers, and since $p^2+6$ is always greater than 0, it follows that $p \geq 0$ .

So the graph is rising for $p \geq 0$ .

When $p=0$ , $A = 0$

The gradient of the graph is negative for $x<0$ , but the lowest point is A = 0, from which point it starts rising. $\therefore \forall p\in R,\ A \geq 0$ (for all p in R)

3.

$\frac{x}{3} = \cos{\theta}-\sin{\theta}$ , $\frac{y}{2} = \sin{\theta}+\cos{\theta}$

$\Rightarrow \left(\frac{x}{3}\right)^2+\left(\frac{y}{2}\right)^2=2$

$\frac{x^2}{9}+\frac{y^2}{4}=2$
Title: Re: TrueTears question thread
Post by: TrueTears on January 01, 2009, 08:46:45 pm: hm i get Q 2 and Q 3, but still cloudy on Q 1 o.O

and also just this Q

Find an anti derivative to $cot^2(x)$
Title: Re: TrueTears question thread
Post by: ell on January 01, 2009, 09:34:54 pm: Quote from: TrueTears on January 01, 2009, 08:46:45 pm
hm i get Q 2 and Q 3, but still cloudy on Q 1 o.O

and also just this Q

Find an anti derivative to $cot^2(x)$

$\int \cot^2(x)\ dx$

$=\int \tan^2\left(\frac{\pi}{2}-x\right)\ dx$

$=\int \sec^2\left(\frac{\pi}{2}-x\right)-1\ dx$

$=-\tan \left(\frac{\pi}{2}-x\right) - x +C$

$=-\cot (x) - x +C$
Title: Re: TrueTears question thread
Post by: TrueTears on January 02, 2009, 03:02:05 pm: thanks ell,

also

1. $\int \frac{e^{2x}}{1+e^x} dx$ just a bit stuck lol

2. $\int sin(3x)cos^5(3x) dx$ i did this question, but my answer looks totally different from book, wondering if someone can confirm if it is right or not.

so $\int sin(3x)cos^5(3x) dx$ = $\int cos^4(3x)cos(3x)sin(3x) dx$

$\implies \int (1-sin^2(3x))^2cos(3x)sin(3x) dx$

let $u = sin(3x)$

$\implies \frac{du}{dx} = 3cos(3x)$

$\implies du = 3cos(3x)dx$

$\implies \frac{1}{3} \int 3(1-sin^2(3x))^2cos(3x)sin(3x) dx$

$\implies \frac{1}{3} \int (1-u^2)^2u du$

$\implies \frac{1}{3} \int u^5 - 2u^3 + u du$

$\implies \frac{1}{3} (\frac{1}{6}u^6 - \frac{1}{2}u^4 + \frac{1}{2}u^2)$

$\implies \frac{1}{18}u^6 - \frac{1}{6}u^4 + \frac{1}{6}u^2$

sub $u = sin(3x)$ back in and we get $\frac{1}{18}sin(3x)^6 - \frac{1}{6}sin(3x)^4 + \frac{1}{6}sin(3x)^2 + c$

However answer in my book has $-\frac{1}{18}cos^6(3x) + c$

Is that a different form, or is my answer wrong?
Title: Re: TrueTears question thread
Post by: ed_saifa on January 02, 2009, 04:58:54 pm: $\int \frac{e^{2x}}{1+e^x} dx$

$\int \frac{e^{x}.e^{x}}{1+e^x} dx$

Let $u= 1 + e^x$

Rearranging gives $e^x= u-1$

$\frac{du}{dx}=e^x$

$dx=\frac{du}{e^x}$

$\int \frac{e^{2x}}{1+e^x} dx$ = $\int \frac{(u-1)^2}{u}.\frac{du}{u-1}$

= $\int \frac{u-1}{u}du$

= $\int 1-\frac{1}{u}du$

= $u- \ln|u| + C$

= $e^x + 1 -ln|e^x +1| + C$

For question 2 you are correct.
Title: Re: TrueTears question thread
Post by: TrueTears on January 02, 2009, 05:03:54 pm: ahhh thank you very much ed_saifa
Title: Re: TrueTears question thread
Post by: TrueTears on January 02, 2009, 06:50:50 pm: and
1. $\int sec^4(x) dx$
Title: Re: TrueTears question thread
Post by: ed_saifa on January 02, 2009, 07:21:12 pm: $\int sec^4(x)dx$

$\int sec^4(x)dx$ = $\int sec^2(x).sec^2(x)$

$sec^2(x)= 1 + tan^2(x)$

$\int sec^2(x)(1 + tan^2(x))dx$

Let $u=tan(x)$

$\frac{du}{dx}=sec^2(x)$

$dx=\frac{du}{sec^2(x)}$

$\int sec^2(x)(1 + tan^2(x))dx$ = $\int 1 + u^2 du$

= $u + \frac{u^3}{3} + C$

= $tan(x) + \frac{tan^3(x)}{3} + C$
Title: Re: TrueTears question thread
Post by: TrueTears on January 02, 2009, 07:33:33 pm: thanks again ed_saifa ^^, just realised it before u posted xD

just some other one ><

1. Let $U_{n} = \int^{\frac{\pi}{4}}}_{0} tan^n(x) dx$ where $n \in Z$ and $n>1$

Express $U_{n} + U_{n-2}$ in terms of $n$
Title: Re: TrueTears question thread
Post by: /0 on January 02, 2009, 07:46:22 pm: Quote from: TrueTears on January 02, 2009, 06:50:50 pm
and
1. $\int sec^4(x) dx$

2. Let $U_{n} = \int^{\frac{\pi}{4}}}_{0} tan^n(x) dx$ where $n \in Z$ and $n>1$

Express $U_{n} + U_{n-2}$ in terms of $n$

$U_n+U_{n-2} = \int_0^{\frac{\pi}{4}} \tan^n(x)+\tan^{n-2}(x)\,dx$

$=\int_0^\frac{\pi}{4} \tan^{n-2}(x)(1+\tan^2(x))\,dx$

$=\int_0^\frac{\pi}{4} \tan^{n-2}(x)\sec^2(x)\,dx$

Let $u=\tan(x)\Rightarrow \frac{du}{dx} = \sec^2(x)$

$dx = \frac{du}{\sec^2(x)}$

$\Rightarrow U_n+U_{n-2} = \int_0^{1} tan^{n-2}(x) du=\int_0^1 u^{n-2}du$

$\therefore U_n+U_{n-2} = \left[ \frac{u^{n-1}}{n-1}\right]_0^1 = \frac{1}{n-1}$
Title: Re: TrueTears question thread
Post by: TrueTears on January 02, 2009, 07:52:19 pm: thanks /0 and also

Let $\alpha = \frac{\pi}{2} - \theta$ . Show that $\int^{\frac{\pi}{2}}}_{0} \frac{d\theta}{1+tan\theta} = \int^{\frac{\pi}{2}}}_{0} \frac{d\alpha}{1+cot\alpha}$

I keep getting $-\int^{\frac{\pi}{2}}}_{0} \frac{d\alpha}{1+cot\alpha}$ for Q2, this is my working:

$\alpha = \frac{\pi}{2} - \theta$

$\implies \theta = \frac{\pi}{2} - \alpha$

$\implies \frac{d\alpha}{d\theta} = - 1$

$\implies d\alpha = - d\theta$

$\implies -\int^{\frac{\pi}{2}}}_{0} \frac{d\alpha}{1+tan(\frac{\pi}{2} - \alpha)} = -\int^{\frac{\pi}{2}}}_{0} \frac{d\alpha}{1+cot\alpha}$

Can anyone check if anything is wrong?
Title: Re: TrueTears question thread
Post by: /0 on January 02, 2009, 07:56:49 pm: When you change $d\theta$ to $d\alpha$ you also have to change the terminals on the integral to:

$-\int_{\frac{\pi}{2}-0}^{\frac{\pi}{2}-\frac{\pi}{2}} \frac{d\alpha}{1+\cot{\alpha}}$

This will cancel out the negative sign.

(since $\alpha = \frac{\pi}{2} - \theta$ )
Title: Re: TrueTears question thread
Post by: TrueTears on January 02, 2009, 08:11:48 pm: how u get the $-\int_{\frac{\pi}{2}-0}^{\frac{\pi}{2}-\frac{\pi}{2}}$ ?
Title: Re: TrueTears question thread
Post by: /0 on January 02, 2009, 08:19:26 pm: The $\frac{\pi}{2}$ and $0$ terminals are valid only with respect to $\theta$ . If you transform $\theta$ into another variable, you also have to transform the terminals to fit the new variable. Transforming the terminals according to $\alpha = \frac{\pi}{2} - \theta$ assign new terminals to fit the new variable and ensures the integral is still being taken over the same area. Page 270 has some information on it.
Title: Re: TrueTears question thread
Post by: TrueTears on January 02, 2009, 08:39:01 pm: thanks /0. xD
Title: Re: TrueTears question thread
Post by: TrueTears on January 02, 2009, 08:45:58 pm: If $F'(x) = f(x)$ , then an antiderivative of $3f(3-2x)$ is?
Title: Re: TrueTears question thread
Post by: /0 on January 02, 2009, 08:54:48 pm: $\int 3f(3-2x)dx = 3 \int f(3-2x)dx$

$u = 3-2x \Rightarrow \frac{du}{dx} = -2$

$\Rightarrow 3 \int f(3-2x)dx = 3 \int f(u)\frac{du}{-2}=\frac{-3}{2} \int f(u)du = \frac{-3}{2}F(u)+C=-\frac{3}{2}F(3-2x)+C$

$\therefore$ An antiderivative is $-\frac{3}{2}F(3-2x)$
Title: Re: TrueTears question thread
Post by: TrueTears on January 02, 2009, 08:58:06 pm: yeap thx. Hey can u take a look at Q 27 b) and Q 28 c) on pg 281, /0 xD
Title: Re: TrueTears question thread
Post by: brendan on January 02, 2009, 09:00:03 pm: Quote from: ed_saifa on January 02, 2009, 07:21:12 pm
Let $u=tan(x)$

$\frac{du}{dx}=tan(x)$

thats not right...

$\frac{d(tan(x))}{dx}=sec^2(x)$
Title: Re: TrueTears question thread
Post by: /0 on January 02, 2009, 09:52:49 pm: Quote from: TrueTears on January 02, 2009, 08:58:06 pm
yeap thx. Hey can u take a look at Q 27 b) and Q 28 c) on pg 281, /0 xD

lol

27b)

You have $U_n+U_{n-2}$ . If you find $U_2=\int_0^\frac{\pi}{4} \tan^2(x)dx$ then you can find $U_4$ , then $U_6$ .

28c)

$\int_0^\frac{\pi}{2} \frac{d\theta}{1+\tan{\theta}}+\int_0^\frac{\pi}{2} \frac{d\theta}{1+\cot{\theta}}=\int_0^\frac{\pi}{2} \left(\frac{1}{1+\tan\theta}+\frac{1}{1+\cot\theta}\right)d\theta=\int_0^\frac{\pi}{2}d\theta=\frac{\pi}{2}$

Let $\int_0^\frac{\pi}{2} \frac{d\theta}{1+\tan\theta} = \int_0^\frac{\pi}{2} \frac{d\phi}{1+\cot\phi}=x$

If $\int_0^\frac{\pi}{2} \frac{d\phi}{1+\cot\phi}=x$ ,

we can say... for instance, that $\int_0^\frac{\pi}{2} \frac{da}{1+\cot{a}}=x$ or $\int_0^\frac{\pi}{2} \frac{dp}{1+\cot{p}}=x$ , simply by changing variables.

Hence it follows that $\int_0^\frac{\pi}{2} \frac{d\theta}{1+\cot{\theta}}=x$ .

Therefore, $\int_0^\frac{\pi}{2} \frac{d\theta}{1+\tan{\theta}}=\int_0^\frac{\pi}{2} \frac{d\theta}{1+\cot{\theta}}$

So... $\int_0^\frac{\pi}{2} \frac{d\theta}{1+\tan{\theta}}=\int_0^\frac{\pi}{2} \frac{d\theta}{1+\cot{\theta}}= \frac{\pi}{4}$
Title: Re: TrueTears question thread
Post by: ed_saifa on January 02, 2009, 10:30:45 pm: Quote from: Brendan on January 02, 2009, 09:00:03 pm
Quote from: ed_saifa on January 02, 2009, 07:21:12 pm
Let $u=tan(x)$

$\frac{du}{dx}=tan(x)$

thats not right...

$\frac{d(tan(x))}{dx}=sec^2(x)$
Whoops, I'll change it.
Title: Re: TrueTears question thread
Post by: TrueTears on January 02, 2009, 10:58:51 pm: thx /0 XD

and just this Q im half way through but im stuck...

$\int \frac{x}{x^2+2x+3} dx$
Title: Re: TrueTears question thread
Post by: /0 on January 02, 2009, 11:36:00 pm: Quote from: TrueTears on January 02, 2009, 10:58:51 pm
thx /0 XD

and just this Q im half way through but im stuck...

$\int \frac{x}{x^2+2x+3} dx$

$\int \frac{x}{x^2+2x+3}dx = \int \frac{x}{(x+1)^2+2}dx$

Let $u=x+1 \Rightarrow \frac{du}{dx}=1$

$\Rightarrow \int \frac{u-1}{u^2+2}du$

$=\int \frac{u}{u^2+2}du - \int \frac{1}{u^2+2}du$

$=\frac{1}{2} \int \frac{2u}{u^2+2}du - \int \frac{1}{u^2+2}$

Since $\int \frac{f'(x)}{f(x)}dx = \log_e|f(x)|+C$

$\Rightarrow \frac{1}{2}\log_e|u^2+2|- \frac{1}{\sqrt{2}}\int \frac{\sqrt{2}}{u^2+(\sqrt{2})^2}$

$=\frac{1}{2}\log_e|u^2+2|-\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{u}{\sqrt{2}}\right)+C$

$=\frac{1}{2}\log_e|x^2+2x+3|-\frac{\sqrt{2}}{2}\tan^{-1} \left(\frac{\sqrt{2}(x+1)}{2}\right)+C$
Title: Re: TrueTears question thread
Post by: TrueTears on January 02, 2009, 11:37:57 pm: yeah i got to this step

= $\frac{1}{2} \int \frac{2u}{u^2+2}du - \int \frac{1}{u^2+2}$

$\Rightarrow \frac{1}{2}\log_e|u^2+2|- \frac{1}{\sqrt{2}}\int \frac{\sqrt{2}}{u^2+(\sqrt{2})^2}$

how u know to use log here, is it just trial and error?
Title: Re: TrueTears question thread
Post by: ell on January 02, 2009, 11:40:07 pm: Quote from: TrueTears on January 02, 2009, 11:37:57 pm
yeah i got to this step

= $\frac{1}{2} \int \frac{2u}{u^2+2}du - \int \frac{1}{u^2+2}$

$\Rightarrow \frac{1}{2}\log_e|u^2+2|- \frac{1}{\sqrt{2}}\int \frac{\sqrt{2}}{u^2+(\sqrt{2})^2}$

how u know to use log here, is it just trial and error?

When it's in the form $\frac{f'(x)}{f(x)}$

edit: oops just realised /0 already pointed this out. you can get that answer by substituting $v=u^2+2$
Title: Re: TrueTears question thread
Post by: vce08 on January 02, 2009, 11:42:37 pm: Quote from: TrueTears on January 02, 2009, 11:37:57 pm
yeah i got to this step

= $\frac{1}{2} \int \frac{2u}{u^2+2}du - \int \frac{1}{u^2+2}$

$\Rightarrow \frac{1}{2}\log_e|u^2+2|- \frac{1}{\sqrt{2}}\int \frac{\sqrt{2}}{u^2+(\sqrt{2})^2}$

how u know to use log here, is it just trial and error?

the other integral thingy is in the form of tan^-1 i think
Title: Re: TrueTears question thread
Post by: TrueTears on January 02, 2009, 11:43:47 pm: ahh i see, u can also use another substitution for u there and get same thing.
Title: Re: TrueTears question thread
Post by: TrueTears on January 03, 2009, 01:22:42 am: are we required to know how to sketch something like $y = \frac{1}{\sqrt{9-4x^2}}$ in the spesh course without a calculator?
Title: Re: TrueTears question thread
Post by: Matt The Rat on January 03, 2009, 01:49:30 am: You could be asked to. It's a reciprocal graph for a square root function graph and reciprocal graphs are part of the spec course. Unlikely to be asked, but could be.
Title: Re: TrueTears question thread
Post by: TrueTears on January 03, 2009, 07:23:24 pm: ah kk

Also just this stupid Q >.<

How do u solve the following simultaneous equation for x and y

$y = 9-x^2$ ......(1)

$y = \frac{1}{\sqrt{9-x^2}}$ ......(2)
Title: Re: TrueTears question thread
Post by: ell on January 03, 2009, 07:44:36 pm: Quote from: TrueTears on January 03, 2009, 07:23:24 pm
ah kk

Also just this stupid Q >.<

How do u solve the following simultaneous equation for x and y

$y = 9-x^2$ ......(1)

$y = \frac{1}{\sqrt{9-x^2}}$ ......(2)

From (1): $x^2=9-y$ (3)

From (2): $x^2=9-\frac{1}{y^2}$ (4)

(3) = (4):

$9-y=9-\frac{1}{y^2}$

$\implies \frac{1}{y^2}-y=0$

Multiply both sides by $y^2$ :

$1-y^3=0$

$\implies y=1$

Substitute $y=1$ into (1):

$\implies x=\pm2\sqrt{2}$ (Substituting into (2) also yields the same answer.)
Title: Re: TrueTears question thread
Post by: TrueTears on January 03, 2009, 07:49:00 pm: ah thank you so much
Title: Re: TrueTears question thread
Post by: TrueTears on January 04, 2009, 07:05:39 pm: just some questions:

1. Find the volume of the solid generated when the region enclosed by $y = \sqrt{3x+1}$ , $y = \sqrt{3x}$ , $y = 0$ and $x = 1$ is rotated about the x axis.

2. The region defined by the inequalities $y \ge x^2-2x+4$ and $y \le 4$ is rotated about the line $y = 4$ . Find the volume generated.

3. The region bounded by the parabola $y^2 = 4(1-x)$ and the y axis is rotated about the a) x axis, b) y axis
prove that the volumes of the solids formed are in the ratio 15:16

For this question, i can do part b), which is just u solve for x, and then u have the intergral: $\pi \int^{2}_{0} (1 - \frac{1}{4}y^2)^2 dy$ = $\frac{16\pi}{15}$

However, how do u do part a) ?

4. The region bounded by the graph $y = \frac{1}{\sqrt{x^2+9}}$ , the x axis, the y axis and the line $y = 4$ is rotated about the y axis. Find the volume of the solid formed.

Many thanks again XD
Title: Re: TrueTears question thread
Post by: ell on January 04, 2009, 07:39:12 pm: 1. Graph the functions first to visualise what volume you need and to determine what terminals your integrals will use. The top function is $\sqrt{3x+1}$ and the bottom function is $\sqrt{3x}$ so find the difference in volumes between the two functions.

$\begin{alignat}{2} \mbox{Volume} & = \pi\int_{-\frac{1}{3}}^{1}3x+1\ dx-\pi\int_{0}^{1}3x\ dx \\ & = \pi\left[\frac{3x^2}{2}+x\right]_{-\frac{1}{3}}^{1}-\pi\left[\frac{3x^2}{2}\right]_{0}^{1} \\ & = \pi\left[\frac{3}{2}+1-\frac{1}{6}+\frac{1}{3}\right]-\pi\left[\frac{3}{2}\right] \\ & = \frac{7\pi}{6}\ \mbox{cubic units} \\ \end{alignat} $
Title: Re: TrueTears question thread
Post by: TrueTears on January 05, 2009, 12:22:06 am: ah yes thanks ell, just new to this was kinda confused.

I get the all the other questions now cept for Q2.
2. The region defined by the inequalities $y \ge x^2-2x+4$ and $y \le 4$ is rotated about the line $y = 4$ . Find the volume generated. What do u do when it says rotated around y = 4? I've only met questions which says rotate around x/y axis.

Also just this new question:

if $\frac{dx}{dy}$ is inversely proportional to y, and y = 2 when x = 0 and y = 4 when x = 2. Find y when x = 3.
Title: Re: TrueTears question thread
Post by: /0 on January 05, 2009, 12:32:31 am: It is the same as rotating the region defined by $y \geq x^2-2x$ and $y \leq 0$ about the line $y=0$ . You just translated it down by 4 but the volume is the same.

$\frac{dx}{dy} = \frac{k}{y}\Rightarrow x = klog_e|y|+C$ , differentiating with respect to y. Then solve for k and C.
Title: Re: TrueTears question thread
Post by: TrueTears on January 05, 2009, 12:43:26 am: Quote from: /0 on January 05, 2009, 12:32:31 am
It is the same as rotating the region defined by $y \geq x^2-2x$ and $y \leq 0$ about the line $y=0$ . You just translated it down by 4 but the volume is the same.

$\frac{dx}{dy} = \frac{k}{y}\Rightarrow x = klog_e|y|+C$ , differentiating with respect to y. Then solve for k and C.

yeah i did for that $\frac{dx}{dy}$ question, keep getting wrong answer ><
Title: Re: TrueTears question thread
Post by: TrueTears on January 05, 2009, 12:45:33 am: actually nvm, i think i got it

What about this question

Find the constants a and b if $x = t(a cos2t + b sin2t)$ is a solution of the differential equation $\frac{d^2x}{dt^2} + 4x = 2 cos2t$ , i worked out $\frac{d^2x}{dt^2}$ and subbed it back in and everything, but the 4x cancels out the $\frac{d^2x}{dt^2}$ . What do i do?
Title: Re: TrueTears question thread
Post by: shinny on January 05, 2009, 01:06:32 am: Bit too late at night for calculus, but from what I can see at first glance, I think you forgot to product rule the x (don't forget the t outside the brackets) to get to the second derivative. If you don't do that, then yeh it'll cancel out I think.

Note: Only did a few mental calculations so there's a 50% chance this is wrong =P
Title: Re: TrueTears question thread
Post by: ell on January 05, 2009, 04:07:18 am: Quote from: TrueTears on January 05, 2009, 12:45:33 am
Find the constants a and b if $x = t(a cos2t + b sin2t)$ is a solution of the differential equation $\frac{d^2x}{dt^2} + 4x = 2 cos2t$ , i worked out $\frac{d^2x}{dt^2}$ and subbed it back in and everything, but the 4x cancels out the $\frac{d^2x}{dt^2}$ . What do i do?

DEs are always messy so I'll omit some of the working

$x=t(a\cos 2t+b\sin 2t)$

$\frac{dx}{dt}=a\cos 2t+b\sin 2t-2ta\sin 2t+2tb\cos 2t$

$\frac{d^2x}{dt^2}=(-4a-4bt)\sin 2t+(4b-4at)\cos 2t$

Substituting into given equation:

$(-4a-4bt)\sin 2t+(4b-4at)\cos 2t+4t(a\cos 2t+b\sin 2t)=2\cos 2t$

after some work:

$\implies -4a\sin 2t+4b\cos 2t =2\cos 2t$

Equate:
$-4a=0\implies a=0$ and $4b=2 \implies b=\frac{1}{2}$
Title: Re: TrueTears question thread
Post by: TrueTears on January 05, 2009, 09:03:41 pm: ah yes thanks ell, soz was doing this a bit late at night, i did what shinny said lol forgot the product rule grrrrrrr

Oh and also, When u are doing differential equations, how do u know when to sub let $A=e^{-2k}$ like this example:

Find the general solution of each of following differential equations

a) $\frac{dy}{dx} = 2y+1, y > \frac{-1}{2}$

so $\frac{dx}{dy} = \frac{1}{2y+1}$ and $x = \int \frac{1}{2y+1} dy$

then after some working u get $y = \frac{1}{2} ( e^{2x-2k} -1 )$ But then my book says Let $A = e^{-2k}$ and so the equation becomes $y = \frac{1}{2} ( Ae^{2x} -1 )$

just wondering what's the point of letting $A = e^{-2k}$ , and when do u know when to do it?

And just this question:
A tank initially contains 200L of pure water. A salt solution containing 5 kg of salt per litre is added at the rate of 10 litres per minute and the mixed solution is drained simultaneously at the rate of 12 litres per minute. there is m kg of salt in the tank after t minutes. Find an expression for $\frac{dm}{dt}.$
Title: Re: TrueTears question thread
Post by: kj_ on January 05, 2009, 09:34:56 pm: Quote from: TrueTears on January 05, 2009, 09:03:41 pm
just wondering what's the point of letting $A = e^{-2k}$ , and when do u know when to do it?
Notice that when you let $A = e^{-2k}$ , your equation appears a lot neater - nothing else as the answers are equivalent.

Quote from: TrueTears on January 05, 2009, 09:03:41 pm
And just this question:
A tank initially contains 200L of pure water. A salt solution containing 5 kg of salt per litre is added at the rate of 10 litres per minute and the mixed solution is drained simultaneously at the rate of 12 litres per minute. there is m kg of salt in the tank after t minutes. Find an expression for $\frac{dm}{dt}.$
The expression for $\frac{dm}{dt}.$ is defined by Inflow - Outflow.
Now, inflow is defined by input concentration by volume rate, and thus IN = 5kg/L x 10L/min = 50 kg/min.
Outflow is defined by output concentration by volume rate, and this is where it gets tricky.
You'll know that concentration is defined by $\frac {mass}{volume}$ .
Your mass is unknown and defined as $m$ .
Your volume is $200 - 2t$ . Why? You initially have 200L of water, and you're adding water at 10L/min. At the same time, water is flowing out at 12L/min, and therefore, each minute, you are losing 2 litres of water. This is a variable of time, and thus the expression stated above.
And finally, your outflow volume rate is 12L/min, and therefore:
OUT = $\frac{m}{200-2t}$ x $12L/min$
$=\frac{12m}{200-2t}$

Thus, your expression is $\frac{dm}{dt} = 50 - \frac{12m}{200-2t}$

EDIT: Note that in the specialist math syllabus, you're not required to solve these type of differential equations. Actually, you're not taught how to :buck2:
Title: Re: TrueTears question thread
Post by: TrueTears on January 05, 2009, 09:52:04 pm: ahh thank you soo much kj_, yeah it was that output part that confused me, just didn't quite recognize that overall its losing 2L per min xD. thanks again
Title: Re: TrueTears question thread
Post by: TrueTears on January 05, 2009, 10:27:23 pm: just 2 more new questions

1. The rate of decay of a substance is $km$ $(k \in R^+)$ where $m$ is the mass of substance remaining. Show that the half life, the time in which the amount of the original substance remaining is halved, is $\frac{1}{k} log_{e}(2)$

2. (http://img210.imageshack.us/img210/4716/cylinderlp7.jpg)
How do u do Q 3. Just a bit confused about part a)

Sorry about Large picture >< not sure how to make it smaller.

Many thanks
Title: Re: TrueTears question thread
Post by: Mao on January 06, 2009, 12:05:33 am: 1.

$\frac{dm}{dt}=-k\cdot m$

$\implies \frac{dt}{dm}=-\frac{1}{km}$

let $m_0$ be the amount of mass initially

$\implies t=-\frac{1}{k}\int_{m_0}^{m}m\; dm$

$\implies t=-\frac{1}{k}\log_e \left(\frac{m}{m_0}\right)$

when $m=\frac{1}{2}m_0$ (half decayed)

$\implies t=-\frac{1}{k}\log_e \frac{1}{2}=\frac{1}{k}\log_e (2)$
Title: Re: TrueTears question thread
Post by: Mao on January 06, 2009, 12:19:43 am: 2. (actually 3 :P )

this question isn't straight forward. the variable 'A' is the cross sectional area of cylinder, with the side as a chord on the circle. There is no simple formula for this, hence we construct a circle $a^2+b^2=4$ (radius 2 circle centered at origin), where a is the horizontal axis and b is the vertical axis. We hence have the upper half of the circle $b=\sqrt{4-a^2}$

imagine x starts from (-2,0) on our a axis, hence our a value would be $-2+x$ , and its height would be $2b = 2\sqrt{4-(-2+x)^2}=2\sqrt{4x-x^2}$

this is the length of the chord across the cylinder. Hence, $A=l\times w = 6\times 2\sqrt{4x-x^2}=12\sqrt{4x-x^2}$

$\frac{dx}{dt}=-\frac{0.025}{12}\sqrt{\frac{x}{4x-x^2}}=-\frac{1}{480\sqrt{4-x}}$

the rest should be easy enough =]
Title: Re: TrueTears question thread
Post by: TrueTears on January 06, 2009, 08:37:02 pm: thanks mao :D

Kinematics q

1. The position of an object travelling in a horizontal line is x metres from a point O on the line at t seconds. The position is described by $x = 3t - t^2 , t \ge 0$
Find the displacement of the particle in the 5th second.

so x(5) = -10. And displacement means 'the position of the object relative to the origin (O)'. But my book says its -6. How do u get that
Title: Re: TrueTears question thread
Post by: Flaming_Arrow on January 06, 2009, 09:00:42 pm: $x = 3t - t^2 , t \ge 0$

$\mbox {displacement of the particle in the 5th second} = x(5) - x(4)$

$x(5) = -10 \mbox {and} x(4) = -4$

$-10+4 = -6$

$\implies \boxed {{-6m}}$
Title: Re: TrueTears question thread
Post by: TrueTears on January 06, 2009, 09:03:55 pm: wats difference b/w 5th second and in 5 seconds? is 5th second just between 4-5 seconds? ( soz for stupid Q ><)
Title: Re: TrueTears question thread
Post by: Flaming_Arrow on January 06, 2009, 09:07:22 pm: when it says find the distance traveled in the 5th second of motion that means t5-t4, the distance traveled in the 5TH SECOND OF MOTION.
if it says find the distance traveled in the first 5 seconds, thats the total distance traveled which is t5-t0, which is just t5
Title: Re: TrueTears question thread
Post by: TrueTears on January 06, 2009, 09:08:13 pm: ahhh thanks.
Title: Re: TrueTears question thread
Post by: TrueTears on January 06, 2009, 10:13:35 pm: From a balloon ascending with a velocity of 10 m/s. A stone was dropped and reached the ground in 12 seconds. Given the gravitational acceleration is 9.8m/s. Find the height of the balloon when the stone was dropped.
Title: Re: TrueTears question thread
Post by: Mao on January 06, 2009, 10:18:29 pm: $s=ut+\frac{1}{2}at^2=-10\times 12 + 4.9\times 12^2$

btw, acceleration is in ms^-2 :)
Title: Re: TrueTears question thread
Post by: TrueTears on January 06, 2009, 10:23:46 pm: thanks mao
Title: Re: TrueTears question thread
Post by: TrueTears on January 07, 2009, 12:47:10 am: A particle travels in a straight line(going right) with constant velocity of 4 m/s for 12 seconds. It is then subjected to a constant acceleration in the opposite direction for 20 seconds which returns the particle to its original position. Find the acceleration of the particle.

This is my working:

Left - negative and Right - positive

so the distance travelled right is just (4)(12) since it has constant velocity, a = 0, so it travelled 48 m right.

so next we have the going back journey which is going left back to original position.

a = -a, x = -48 and t = 20 u = 0, since the particle is instantaneously at rest when its turning around

using the formula $x = ut + \frac{1}{2}at^2$

$\implies -48 = 0(20) + \frac{1}{2}(-a)(20)^2$

$\implies a = -0.24$ However the answer is -0.64 m/s^2. Can anyone see what i did wrong?
Title: Re: TrueTears question thread
Post by: /0 on January 07, 2009, 12:57:39 am: You still have to take into account the particle is moving 4m/s to the right when it receives the negative acceleration, so u = 4.
Title: Re: TrueTears question thread
Post by: TrueTears on January 07, 2009, 01:02:34 am: o i see. thanks again ST
Title: Re: TrueTears question thread
Post by: TrueTears on January 07, 2009, 04:49:01 pm: 1. The maximum rate at which a bus can accelerate or decelerate is 2 m/s^2. It has a max speed of 60 km/h. Find the shortest time the bus can take to travel between two bus stops 1km apart on a straight stretch of road.

2. Two cars A and B, each moving with constant acceleration, are travelling in the same direction along the parallel lanes of a divided road. When A passes B, the speeds are 64km/h and 48km/h respectively. Three minutes later, B passes A, travelling at 96km/h. Find the distance travelled by A and B this instant(since they first passed) and the speed of A.
Title: Re: TrueTears question thread
Post by: Mao on January 07, 2009, 05:01:19 pm: 1,
it takes $s=\frac{v^2}{2a}=\frac{(50/3)^2}{4}=\frac{625}{9}\; m$ to accelerate from 0 to 60 (and from 60 to 0)

this amounts to ~138.9 m in total, and takes $t=2\times \frac{(50/3)}{2}=16\frac{2}{3}$ seconds

The rest of the way, $861\frac{1}{9}$ m, takes $51\frac{2}{3}$ seconds

In total, shortest time would be $68\frac{1}{3}$ seconds.
Title: Re: TrueTears question thread
Post by: Mao on January 07, 2009, 05:08:30 pm: 2.

acceleration of B: $\frac{96-48}{3.6\times 3\times 60}=\frac{2}{27}$ m/s²

distance travelled by B: $s=\frac{48}{3.6}\times 180 + \frac{1}{2}\times \frac{2}{27}\times 180^2=3600\; m$

acceleration of A: $a=2\times \frac{s-ut}{t^2}=2\times \frac{3600-64/3.6\times 180}{180^2}=\frac{2}{81}$ m/s²

velocity of A: $v=u+at=\frac{64}{3.6}+\frac{2}{81}\times 180=\frac{200}{9}\; ms^{-1}=80\; kmh^{-1}$
Title: Re: TrueTears question thread
Post by: TrueTears on January 07, 2009, 06:57:47 pm: thanks again mao XD
Title: Re: TrueTears question thread
Post by: TrueTears on January 07, 2009, 07:05:58 pm: just another Q,

The acceleration $a$ $m/s^2$ of an object whose displacement is x metres from the origin is given by $a = -x, \sqrt{3} \le x \le \sqrt{3}$ . If the object starts from the origin with a velocity of $\sqrt{3}$ m/s, then its velocity $v$ when displacement is $x$ metres from the origin is given by:

This is a multiple choice question, and my working is:
a = -x and the given information is when $t = 0, x = 0, v = \sqrt{3}$
$a = \frac{d}{dx}(\frac{1}{2}v^2)$

$(\frac{1}{2}v^2) = \int -x dx$

$\implies (\frac{1}{2}v^2) = -\frac{1}{2}x^2 + c$

subbing in the formation gets $c = \frac{3}{2}$

so $v^2 = -x^2 + 3$

$v = + or - \sqrt{-x^2+3}$

But when x = 0 $v = \sqrt{3}$

so the negative answer is discarded and the answer should be $v = \sqrt{-x^2+3}$

But my books answer says its $v = + or - \sqrt{-x^2+3}$ . why is that?
Title: Re: TrueTears question thread
Post by: Flaming_Arrow on January 07, 2009, 07:33:56 pm: velocity can be negative $\mbox {velocity} = \frac {\mbox {displacement}}{\mbox{time}}$
Title: Re: TrueTears question thread
Post by: TrueTears on January 07, 2009, 07:51:56 pm: Quote from: Flaming_Arrow on January 07, 2009, 07:33:56 pm
velocity can be negative $\mbox {velocity} = \frac {\mbox {displacement}}{\mbox{time}}$

yeah i know but for that question if u sub in x = 0 v should = $\sqrt{3}$ (positive) since its given in the question. But if u have + or - out on the front, u cant have $-\sqrt{3}$ as an answer since when x= 0 v doesnt equal $-\sqrt{3}$ . So i dono why the answer has + or - as the answer
Title: Re: TrueTears question thread
Post by: /0 on January 08, 2009, 09:20:36 am: It only says the object 'starts' from the origin at $v=\sqrt{3}$ . It could return to the origin later with a speed of $v=-\sqrt{3}$

In response to your PM, when a rock is thrown down from a cliff, it will begin with a positive velocity vector downwards. Since gravity is also a positive vector of magnitude 10 downwards, it will speed up the rock.
If a rock is thrown up from a cliff, it will have a positive velocity upwards, and gravity vector can be turned so it has a negative vector upwards, so it will slow the rock down until it has a positive velocity vector downwards.
Title: Re: TrueTears question thread
Post by: TrueTears on January 08, 2009, 03:03:53 pm: thanks ST
Title: Re: TrueTears question thread
Post by: TrueTears on January 08, 2009, 03:57:02 pm: When u have something like

Find the cartesian equation which corresponds to the following vector equations and state the domain and range:

$r$ (t) = $\frac{1}{t}$ $i$ + $\frac{1}{t+1}$ $j$ , $t \ne 0, -1$

so Let (x,y) be any point on the cartesian equation

$x = \frac{1}{t}$

$y = \frac{1}{t+1}$

and after eliminating t we get $y = \frac{x}{x+1}$ as cartesian equation.

so range of $x = \frac{1}{t}$ is the domain of the cartesian equation and the range of $y = \frac{1}{t+1}$ is the range of the cartesian equation.

range of $x = \frac{1}{t} \implies R \diagdown {0}$ so domain of cartesian equation is $R \diagdown {0}$

and range of $y = \frac{1}{t+1} \implies R \diagdown {0}$ so range of cartesian equation is $R \diagdown {0}$

But the answer has $R \diagdown {0,-1}$ as domain and $R \diagdown {0,1}$ as range for the cartesian equation. Is this because $t \ne 0, -1$ , so you sub in -1 into the $x = \frac{1}{t}$ and 0 into the $y = \frac{1}{t+1}$ equation and u get the values which are not allowed? If so then there's another question but this doesn't work

Find the cartesian equation and state the domain and range of the following vector equation:

$r(t) = \frac{1}{t+4}i + (t^2+1)j$ $t \ne -4$

$x = \frac{1}{t+4}$

$y = (t^2+1)$

and the cartesian equation is $y = (\frac{1}{x}-4)^2 + 1$

So the domain is $R \diagdown {0}$ for the cartesian equation, but the range is $R^+$ , But shouldn't the range be $R^+ \diagdown {17}$ ? Since $t \ne -4$ and the range for the cartesian equation is deduced from the $y = (t^2+1)$ equation, and if u sub in t = -4 u get y = 17, but u cant have t = -4 so 17 should not be allowed as one of the y values. But why answer just has $R^+$ for the range, why is that?

(all those things after $\diagdown$ should have a { } around them, i dono why it doesn't appear ><)
Title: Re: TrueTears question thread
Post by: /0 on January 08, 2009, 04:17:08 pm: Quote from: TrueTears on January 08, 2009, 03:57:02 pm
But the answer has $R \diagdown {0,-1}$ as domain and $R \diagdown {0,1}$ as range for the cartesian equation. Is this because $t \ne 0, -1$ , so you sub in -1 into the $x = \frac{1}{t}$ and 0 into the $y = \frac{1}{t+1}$ equation and u get the values which are not allowed?

Yeah, true

The range for the second problem is $R^+$ because even though $t\neq -4$ , you can have $t=4$ .

Also, for braces, $\{ \}$ , type \{ and \}
Title: Re: TrueTears question thread
Post by: TrueTears on January 08, 2009, 04:18:16 pm: ahhh i see, thanks again
Title: Re: TrueTears question thread
Post by: TrueTears on January 08, 2009, 04:55:00 pm: Find a vector equation which corresponds to the following:

1 $y = 3-2x$

2 $x^2+y^2 =4$

My book doesn't have any info or steps on how to approach this, is there a systematic way of doing it?

like for 1. i know its just $r(t) = ti + (3-2t)j$ , but that is more or less a bit of trial and error.
Title: Re: TrueTears question thread
Post by: /0 on January 08, 2009, 05:49:45 pm: Check out Chapter 1.8, Parametric equations of circles, ellipses and hyperbolas.

$\frac{x^2}{4}+\frac{y^2}{4} = 1$

A substitution you could make is $\frac{x^2}{4}=\sin^2{t}$ , $\frac{y^2}{4}=\cos^2{t}$ , or $\frac{x^2}{4}=\cos^2{t}$ , $\frac{y^2}{4}=\sin^2{t}$

And possible solutions are:
$r(t) = \pm2\sin{t}\mathbf{i}\pm2\cos{t}\mathbf{j}$
$r(t)=\pm2\sin{t}\mathbf{i}\mp2\cos{t}\mathbf{j}$
$r(t) = \pm 2\cos{t}\mathbf{i} \pm 2\sin{t}\mathbf{j}$
$r(t)=\pm 2\cos{t} \mathbf{i} \mp 2\sin{t}\mathbf{j}$

For a linear equation such as $y=3-2x$ , let one of the variables equal the parameter, just like you use parameters simultaneous equations in methods. e.g. $y=t \Rightarrow x = \frac{3-t}{2}$ could be another solution.
Or, to get your solution, you let $x=t$ (and it's probably a better way of putting it)
Title: Re: TrueTears question thread
Post by: TrueTears on January 08, 2009, 06:41:02 pm: o yeah nice thx

And just this question

The path of a particle defined as a function of time, t, is given by the vector equation $r(t) = (1+t)i + (3t+2)j$ where $t \ge 0$ . Find:

a) the distance of the particle from the origin when t = 3
b) the times at which the distance of the particle from the origin is one unit.

For part a), i don't know why my answer is wrong, so all u do is sub in t = 3, $r(3) = 4i + 11j$
so $|r(3)| = \sqrt{4^2+11^2} = \sqrt{137}$ but my book answer says its $\sqrt{53}$ , how do u get that?

And also for part b), without even working my book's answer says $t = \frac{-2}{5}$ and $t = -1$ , but its already stated $t \ge 0$ , how can u get negative answers? and besides t is time, u can have negative time? i don't get this question at all ><
Title: Re: TrueTears question thread
Post by: TrueTears on January 08, 2009, 07:14:22 pm: and another Q

Express $r(t) = (e^t + e^{-t})i + (e^{t} - e^{-t})j, t \ge 0$ in cartesian form.
Title: Re: TrueTears question thread
Post by: Mao on January 08, 2009, 07:38:12 pm: did you mean, $r(t)=(e^t+e^{-t})i+(e^{t}-e^{-t})j$ ?

$(e^t+e^{-t})^2 - (e^{t}-e^{-t})^2 = 4$

$\implies x^2 - y^2 = 4$

looking at the expression of x and y in terms of t, we find that $x\ge 2$ (minimum value, t=0) and that $y\ge 0$ (minimum value, t=0)

$\implies y=\sqrt{x^2-4}, x\ge 2$ [only the positive stream is taken, as $y\ge 0$ ]

The following is NOT on the course, however it is still very useful

the hyperbolic functions:

$\sinh (x) = \frac{e^{x}-e^{-x}}{2}$ , $\cosh (x) = \frac{e^{x}+e^{-x}}{2}$
graph those two and learn what they look like

and this identity should be useful: $\cosh^2 (x) - \sinh^2 (x)=1$

also, $\frac{d}{dx}\sinh (x) = \cosh (x)$ , $\frac{d}{dx}\cosh (x) = \sinh (x)$

:)
Title: Re: TrueTears question thread
Post by: TrueTears on January 08, 2009, 08:01:42 pm: ahh clever mao, XD yeah soz typo btw lol
Title: Re: TrueTears question thread
Post by: TrueTears on January 08, 2009, 08:26:05 pm: Also just another small Q

Theres a vector function like $r(t) = -cos(2 \pi t)i + sin(2 \pi t)j$
And my book says the particle moves around the circle with a period of one unit.
How do u work that out? Is it just $\frac{2\pi}{2\pi} = 1$ ? ie $\frac{2\pi}{n}$ where n is whatever is in the $cos(2 \pi t)$ or $sin(2 \pi t)$ ?

But what about these functions $r(t) = cos^2(3 \pi t)i + 2cos^2(3 \pi t)j, t \ge 0$

and $r(t) = cos(2 \pi t)i + cos(4 \pi t)j$

In general, how do u work out the 'time' it takes to do one cycle of whatever vector function it is?

And also how would u get the cartesian equation of $r(t) = cos(2 \pi t)i + cos(4 \pi t)j$ ?

(answered by mao already XD)
Title: Re: TrueTears question thread
Post by: TrueTears on January 09, 2009, 12:16:54 am: If $r(t) = ati + \frac{a^2t^2}{4}j$ where $a \ge 0$ and $t \ge 0$
Find when the magnitude of the angle between $\dot r (t)$ and $\ddot r(t)$ is 45 degrees.

EDIT: forgot the angle between LOL
Title: Re: TrueTears question thread
Post by: Mao on January 09, 2009, 01:35:07 am: $\dot{r} (t)=a \textbf{i} + \frac{a^2t}{2}\textbf{j}$

$\ddot{r} (t)=0 \textbf{i} + \frac{a^2}{2}\textbf{j}$

$\cos \theta = \frac{1}{\sqrt{2}}=\frac{\dot{r}(t)\cdot \ddot{r}(t)}{\left|\dot{r}(t)\right|\cdot \left|\ddot{r}(t)\right|}=\dfrac{\frac{a^4t}{4}}{\sqrt{a^2+\frac{1}{4}a^4t^2}\times \frac{a^2}{2}}=\frac{at}{\sqrt{4+a^2t^2}}$

$\implies \frac{1}{\sqrt{2}}=\frac{at}{\sqrt{4+a^2t^2}}$

$\implies \sqrt{4+a^2t^2} = \sqrt{2}\cdot at$

$\implies 4+a^2t^2 = 2a^2t^2$

$\implies a^2t^2 = 4 \implies at=2$ (discarding negative as a>0 and t>0)

$\impleis t=\frac{2}{a}$

at this point, $\textbf{r}\left(\frac{2}{a}\right)=2\textbf{i}+\textbf{j}$
Title: Re: TrueTears question thread
Post by: TrueTears on January 09, 2009, 01:37:25 am: and...
The position r(t) of a projectile at time $t ( t \ge 0 )$ is given by $r(t) = 400ti + (300t - 4.9t^2)j$ , if the projectile is initially at ground level find the initial angle of projection from the horizontal
Title: Re: TrueTears question thread
Post by: Mao on January 09, 2009, 02:02:17 am: $\dot{r} (t)=400\textbf{i}+(300-9.8t)\textbf{j}$

$\dot{r} (0)=400\textbf{i}+300\textbf{j}$

$\cos \theta = \frac{\dot{r}(0)\cdot \textbf{i}}{|\dot{r}(0)|\cdot |\textbf{i}|}=\frac{4}{5}$

$\theta = \cos^{-1}\frac{4}{5}= 36.87^o$ i think...
Title: Re: TrueTears question thread
Post by: TrueTears on January 09, 2009, 01:46:01 pm: thanks mao, but how do u know to use the $\dot{r}(t)$ function?
Title: Re: TrueTears question thread
Post by: shinny on January 09, 2009, 01:48:35 pm: $\dot{r}(t)$ is simply the derivative of $r(t)$ , i.e. $r'(t)$ . Is that what you were trying to find?
Title: Re: TrueTears question thread
Post by: TrueTears on January 09, 2009, 01:56:19 pm: soz typo, i meant how do u know to use the $\dot{r} (t)$ function here to work out the angle
Title: Re: TrueTears question thread
Post by: TrueTears on January 09, 2009, 02:28:12 pm: and also:

The velocity vector of a particle at time t seconds is given by $v(t) = (2t+1)^2i + \frac{1}{\sqrt{2t+1}}j$ . Find the magnitude and direction of the acceleration after one second.

I've worked out everything except for the direction part, how do u work out the direction of the acceleration?
My working is as follows:

$v(t) = \dot{r} (t) = (2t+1)^2i + \frac{1}{\sqrt{2t+1}}j$

$\ddot{r} (t) = 4(2t+1)i - \frac{1}{(2t+1)^{\frac{3}{2}}}j$

$\ddot{r} (1) = 12i - \frac{\sqrt{3}}{9}j$

$|\ddot{r} (1) | = \frac{\sqrt{11667}}{9}$

Now im just stuck on how to work out the direction ><
Title: Re: TrueTears question thread
Post by: shinny on January 09, 2009, 02:31:10 pm: Ah sorry, the $\dot{r}(t)$ function is of course what we refer to as the particle's velocity. Velocity not only indicates the speed at which something is moving, but the DIRECTION it is moving. The angle we are trying to find is the angle between the ground and it's initial direction. Hence, just draw up a triangle and find cos theta using the resolutes of the direction.
Title: Re: TrueTears question thread
Post by: TrueTears on January 09, 2009, 02:34:17 pm: ahhhh that explains it all, thanks shinny.
Title: Re: TrueTears question thread
Post by: shinny on January 09, 2009, 02:34:31 pm: Quote from: TrueTears on January 09, 2009, 02:28:12 pm
and also:

The velocity vector of a particle at time t seconds is given by $v(t) = (2t+1)^2i + \frac{1}{\sqrt{2t+1}}j$ . Find the magnitude and direction of the acceleration after one second.

I've worked out everything except for the direction part, how do u work out the direction of the acceleration?
My working is as follows:

$v(t) = \dot{r} (t) = (2t+1)^2i + \frac{1}{\sqrt{2t+1}}j$

$\ddot{r} (t) = 4(2t+1)i - \frac{1}{(2t+1)^{\frac{3}{2}}}j$

$\ddot{r} (1) = 12i - \frac{\sqrt{3}}{9}j$

$|\ddot{r} (1) | = \frac{\sqrt{11667}}{9}$

Now im just stuck on how to work out the direction ><

The direction of the acceleration is simply $12i - \frac{\sqrt{3}}{9}j$ :) Could convert this to a unit vector, but I don't think its necessary?
Title: Re: TrueTears question thread
Post by: TrueTears on January 09, 2009, 02:41:18 pm: yeah.. that's what i thought as well, but the answer in the book i guess converted to unit vector, guess ill try that as well lol
Title: Re: TrueTears question thread
Post by: TrueTears on January 09, 2009, 06:30:42 pm: 1. A body moves horizontally along a straight line in a direction of $N\alpha^{\circ}W$ with a constant speed of 20m/s. If $i$ is a horizontal unit vector due east and $j$ is a horizontal unit vector due north and it $tan \alpha^{\circ} = \frac{4}{3}$ find the velocity of the body at time $t$ .
Title: Re: TrueTears question thread
Post by: /0 on January 09, 2009, 07:09:24 pm: I'm not sure about this, but since $\tan \alpha^{\circ} = \frac{4}{3}$ and $\alpha$ is in quadrant 2, if you draw a diagram of that triangle then you'll see that the velocity vector in that direction is $v(t) = k(-4\mathbf{i}+3\mathbf{j})$ for some constant k.

$|v(t)| = 20$

$\sqrt{(-4k)^2+(3k)^2} = 5k = 20$

$\therefore k = 4$

So the velocity at a time t is $v(t) = -16\mathbf{i}+12\mathbf{j}$
Title: Re: TrueTears question thread
Post by: TrueTears on January 10, 2009, 02:19:57 pm: thx /0, that is the correct answer,

Also just this question

(http://img519.imageshack.us/img519/2378/dynamicsie1.jpg)
question 9 part a) just a bit confused.
Title: Re: TrueTears question thread
Post by: humph on January 10, 2009, 03:01:29 pm: 9 a) is asking for the resultant force as a vector, 9 b) is asking for the magnitude of that vector and then the unit vector.

So you're supposed to weight each of the forces correctly and then add them together. They've given you the direction of each vector, but they're not normalised. So the first thing to do would be to consider the three vectors when normalised; they are $\frac{1}{5}(-4,3), \; \frac{1}{5}(3,4) , \; \frac{1}{2}(0,-2)$ . Then you multiply these normalised vectors by the magnitude of the force of each one; $\frac{12}{5}(-4,3), \; \frac{16}{5}(3,4) , \; \frac{15}{2}(0,-2)$ . Then you add the three together; $\frac{12}{5}(-4,3) + \frac{16}{5}(3,4) + \frac{15}{2}(0,-2) = (0,5)$ . So that's the resultant force. Then the magnitude is 5 and the direction is $(0,1)$ .
Title: Re: TrueTears question thread
Post by: TrueTears on January 10, 2009, 03:05:12 pm: i kinda get it, but what do u mean by "normalised"? As in the unit vector in each of those 3 forces directions?
Title: Re: TrueTears question thread
Post by: ell on January 10, 2009, 03:09:53 pm: Quote from: TrueTears on January 10, 2009, 03:05:12 pm
i kinda get it, but what do u mean by "normalised"?

To normalise a vector is to take the unit vector.
Title: Re: TrueTears question thread
Post by: TrueTears on January 10, 2009, 03:42:28 pm: and also Q 16 part a), Q 18 and Q 19. Just slightly confused ><""

(http://img301.imageshack.us/img301/5604/dynamics3cg2.jpg)
Title: Re: TrueTears question thread
Post by: /0 on January 10, 2009, 04:07:20 pm: I am a noob in dynamics but whatever...

16a)

Let OX lie on the x-axis. The resultant force vector can be found by vector addition. Let the vector of 2N lie so that's its tail coincides with the head of the 3N vector. The resultant vector is the third side of the triangle.
The angle between the 2N and 3N vectors is $180-50 = 130^{\circ}$

So the resultant vector has magnitude $\sqrt{3^2+2^2 - 2(2)(3)\cos{130^{\circ}}}$ by the cosine rule.

From the vector triangle you can then use the cosine rule again to find the other angles, and hence the angle of the resultant vector with respect to the OX vector.

17. Same principle; draw a vector triangle

18. Draw a diagram and split the vectors into horizontal and vertical components, with the 10N vector on the x-axis. Since the resultant vector acts only along the 10N vector, the vertical vectors must cancel out, i.e.

$8 \sin{60^{\circ}} = P$

$\therefore P = 4\sqrt{3}N$ .

Since P does not have a horizontal component, it does not contribute to the force along the 10N vector direction, so the resultant force along the 10N vector direction is:

$8\cos{60^{\circ}}+10 = 14N$
Title: Re: TrueTears question thread
Post by: TrueTears on January 10, 2009, 04:51:35 pm: yeah thanks i get all of that now
Title: Re: TrueTears question thread
Post by: TrueTears on January 10, 2009, 07:49:01 pm: Also this question

1. In a lift that is accelerating upwards at 2m/s^2 a spring balance shows the apparent weight of the body to be 2.6 kg wt. What would be the reading if the lift were at rest?

(and also btw what does kg wt mean?)
Title: Re: TrueTears question thread
Post by: Mao on January 10, 2009, 09:57:15 pm: 1 kg wt is '1 kilogram weight', i.e. 1g N

When accelerating upwards, the lift exerts a force upwards on the weight, i.e. the weight exerts an opposite force down towards the lift, in the same direction as gravity. Hence, the normal reaction force would be
$(2+g)\times m=2.6g\; kg\; wt\implies m\approx \frac{2.6\times 9.8}{11.8}\approx 0.216 \; kg$

hence when not accelerating, the mass would exert 0.216 kg wt of force on the spring balance.
Title: Re: TrueTears question thread
Post by: TrueTears on January 11, 2009, 12:48:48 am: thanks mao
Title: Re: TrueTears question thread
Post by: TrueTears on January 11, 2009, 12:50:17 am: and also just a tiny bit stuck on Q 12, i think my 2nd equation is missing something...

(http://img79.imageshack.us/img79/6376/dynamics4vy1.jpg)
Title: Re: TrueTears question thread
Post by: Mao on January 11, 2009, 12:20:52 pm: 8kg mass: $T_2=8a$

12kg mass: $T_1-T_2 = 12a$

5kg mass: $5g - T_1 = 5a$
Title: Re: TrueTears question thread
Post by: TrueTears on January 11, 2009, 04:41:57 pm: thanks mao

also:
1. A body of mass m is projected vertically upwards with speed u. Air resistance is equal to k times the square of the speed where k is a constant. find the maximum height reached and the speed when next at the point of projection

2. Two equal forces of 10 N act on a particle. The angle between the 2 forces is 50 deg. Find the magnitude and direction of the single force which when applied will hold the particle in equilibrium.

So for Q 2 after some working the magnitude resultant force of the two 10 N forces is 18.126N. So we need 18.126 N of force to balance it. For equilibrium both the j and i components must be equal to 0. So resolving the j component we have $10 + 10cos50 +18.126cos(\alpha) = 0$

and resolving the i component we have $10 cos(40) + 18.126cos(\beta) = 0$

But solving for $\alpha$ for the j component has 2 answers and solving for $\beta$ for the i component also has 2 answers, which answer is right? And is this the right way to do this question?
Title: Re: TrueTears question thread
Post by: Mao on January 11, 2009, 06:36:03 pm: 1.
where is this question from? It's rather damn complicated if there's only pronumerals to deal with.

2.
not sure why you are doing it this way. Since the two forces are equal and acting at 50 degrees from each other, you should define i as the 25^o direction, hence the j component of the two forces cancel out. You will just have an 18.126N force acting at 180-25=155 degrees from either force.
Title: Re: TrueTears question thread
Post by: TrueTears on January 11, 2009, 06:49:20 pm: 1. yeah, its from the spesh essential book, only pronumerals no numbers., ive tried it for hours , the equations are annoying lol

2. o i see, stupid me ><
Title: Re: TrueTears question thread
Post by: /0 on January 11, 2009, 08:55:00 pm: I love equations likes this... part of the reason I like physics so much

Let up be Positive and down be Negative for this question.

$F_{net}=F_g+F_{air} =-mg- kv^2 \implies a_{net} = -g-\frac{kv^2}{m}$

$v\frac{dv}{dx} = -(g+\frac{kv^2}{m})$

$\frac{dv}{dx} = \frac{-(mg+kv^2)}{mv}$

$\frac{dx}{dv} = \frac{-mv}{mg+kv^2}$

$x=\frac{-m}{2k}\log_e|kv^2+mg|+C$

When $x=0$ , $v=u$ :

$\therefore C = \frac{m}{2k} \log_e|ku^2+mg|$

$\therefore x=\frac{-m}{2k}\log_e|kv^2+mg|+\frac{m}{2k}\log_e|ku^2+mg|=\frac{m}{2k}\log_e\left|\frac{ku^2+mg}{kv^2+mg}\right|$

For max height:

When $v=0$ ,

$x_H=\frac{m}{2k}\log_e\left|\frac{ku^2+mg}{mg}}\right|=\frac{m}{2k}\log_e\left|1+\frac{ku^2}{mg}\right|$

And since none of the pronumerals are 'likely' to be negative you might get away with round brackets...

For part b), we have $a_{net} = -g+\frac{kv^2}{m}$

$\frac{dv}{dx} = \frac{kv^2-mg}{mv}$

$\frac{dx}{dv}=\frac{mv}{kv^2-mg}$

$x = \frac{m}{2k}\log_e|kv^2-mg|+C$

When $x = x_H$ , $v=0$

$\implies C = \frac{m}{2k}\log_e\left|\frac{mg+ku^2}{m^2g^2}\right|$

$\therefore x = \frac{m}{2k}\log_e\frac{|(kv^2-mg)(ku^2+mg)|}{m^2g^2}$

When $x=0$ ,

$|(kv^2-mg)(ku^2+mg)| = m^2g^2$

$(kv^2-mg)(ku^2+mg) = \pm m^2g^2$

But $kv^2<mg \implies (kv^2-mg)(ku^2+mg) = -m^2g^2$

$\therefore v = u\sqrt{\frac{mg}{ku^2+mg}}$
Title: Re: TrueTears question thread
Post by: TrueTears on January 11, 2009, 09:33:34 pm: legendary ST
Title: Re: TrueTears question thread
Post by: TrueTears on February 07, 2009, 05:58:49 pm: yay spesh is finally back ! gonna repost some more Q in this thread nows XD
Title: Re: TrueTears question thread
Post by: lacoste on February 07, 2009, 06:59:50 pm: Quote from: /0 on January 11, 2009, 08:55:00 pm
I love equations likes this... part of the reason I like physics so much

But $kv^2<mg \implies (kv^2-mg)(ku^2+mg) = -m^2g^2$

$\therefore v = u\sqrt{\frac{mg}{ku^2+mg}}$

Damn, spesh looks hard, damn, man, and you love these q's
Title: Re: TrueTears question thread
Post by: TrueTears on February 07, 2009, 07:57:49 pm: Trivial q... for ellipses do we have to put the co-ordinates of the X and Y extremities?
Title: Re: TrueTears question thread
Post by: Mao on February 07, 2009, 08:07:50 pm: I would.
Title: Re: TrueTears question thread
Post by: TrueTears on February 07, 2009, 09:05:45 pm: so if in an exam, they didn't say anything about marking in X,Y extremities, would you lose marks if you don't?
Title: Re: TrueTears question thread
Post by: hard on February 07, 2009, 11:56:54 pm: at trutears, did you do any units 3/4 last year? if so how did you go?
Title: Re: TrueTears question thread
Post by: TrueTears on February 15, 2009, 06:19:05 pm: Quote from: hard on February 07, 2009, 11:56:54 pm
at trutears, did you do any units 3/4 last year? if so how did you go?
yeah i did music solo performance : 42
Chinese Second language advanced: 48
Title: Re: TrueTears question thread
Post by: TrueTears on February 15, 2009, 06:20:44 pm: Just a quick question :)

The series $S_n$ is given by $S_n = 2n^2 + 14n$ . The $n^{th}$ term $t_n$ is equal to

a) 4n+12
b) 4n-12
c) 2n+7
d) 2n-7
e) 2n+14

The answer is a) but i am not sure how to get it.

Many thanks !!
Title: Re: TrueTears question thread
Post by: hard on February 15, 2009, 06:28:35 pm: Quote from: TrueTears on February 15, 2009, 06:19:05 pm
Quote from: hard on February 07, 2009, 11:56:54 pm
at trutears, did you do any units 3/4 last year? if so how did you go?
yeah i did music solo performance : 42
Chinese Second language advanced: 48

wow
Title: Re: TrueTears question thread
Post by: TrueTears on February 15, 2009, 06:29:56 pm: and this one (sorry for bad paint job haha)

ABCD is a parallelogram, BE intersects CD at F. AB = 8 cm, BC = 10 cm. Given BC : DE = 5 : a, find a.

(http://img10.imageshack.us/img10/4057/testquestionze1.jpg)

thanks guys!
Title: Re: TrueTears question thread
Post by: TrueTears on February 15, 2009, 07:59:46 pm: Actually this questions is just a fun question i saw somewhere, anyone have an idea how to solve

cos(pi/11)cos(2pi/11)cos(3pi/11)cos(4pi/11)cos(5pi/11). Find the exact value.
Title: Re: TrueTears question thread
Post by: Mao on February 15, 2009, 09:22:07 pm: Quote from: TrueTears on February 15, 2009, 06:20:44 pm
Just a quick question :)

The series $S_n$ is given by $S_n = 2n^2 + 14n$ . The $n^{th}$ term $t_n$ is equal to

a) 4n+12
b) 4n-12
c) 2n+7
d) 2n-7
e) 2n+14

The answer is a) but i am not sure how to get it.

Many thanks !!

$t_n = S_n - S_{n-1} = 2n^2 + 14n - (2(n-1)^2 + 14(n-1)) = 2n^2 + 14n - (2n^2 + 10n -12) = 4n + 12$

but this is not on the specialist course :P
Title: Re: TrueTears question thread
Post by: Damo17 on February 15, 2009, 09:46:18 pm: Quote from: TrueTears on February 15, 2009, 07:59:46 pm
Actually this questions is just a fun question i saw somewhere, anyone have an idea how to solve

cos(pi/11)cos(2pi/11)cos(3pi/11)cos(4pi/11)cos(5pi/11). Find the exact value.

NOTE: I GOT THIS FROM: http://www.answerbag.com/q_view/589204 BUT DECIDED TO PUT IT INTO LATEX FROM.

Let $a = \pi/11$
Use the identity that $cos x = \frac{e^{i x} + e^{- i x}}{2}$
Let $p = e^{i a}$
So $cos\frac{\pi}{11} = cos(a) = \frac{p^1 + p^{-1}}{2}$

The value you are looking for is then:
$\frac{(p^1 + p^{-1})(p^2 + p^{-2})(p^3 + p^{-3})(p^4 + p^{-4})(p^5 + p^ {-5})}{32}$
Multiplying that out, a term at a time, from the right:
$\frac{(p^5 + p^{-5})}{32}$

$\frac{(p^9+p^1+p^{-1}+p^{-9})}{32}$

$\frac{(p^{12}+p^6+p^4+p^2+p^-2+p^{-4}+p^{-6}+p^{-12})}{32}$

$\frac{(p^{14}+p^{12}+p^8+2p^4+p^2+2p^0+p ^{-2}+2p^{-4}+p^{-8}+p^{-12}+p^{-14})}{32}$

$\frac{(p^{15}+p^{13}+p^{11}+2p^9+2p^7+3p^5 +3p^3+3p^1+3p^{-1}+3p^{-3}+3p^{-5}+2 p^{-7}+2p^{-9}+p^{-11}+p^{-13}+p^{-15})}{32}$

$p^{11}$ is $e^{i \pi}$ which by Euler is $-1$
So $p^{22}$ is $1$

The negative powers can be made postive by multiplying by $p^{22} = 1$
$\frac{(3p^{21}+3p^{19}+3p^{17}+3p^{15}+3p^{13} +2p^{11}+3p^9+3p^7+3p^5+3p^3+3p^ 1)}{32}$

The coefficient of $p^{11}$ is a touch small, I'll add and subtract $p^{11}$ :
$\frac{(3p^{21}+3p^{19}+3p^{17}+3p^{15}+3p^{13} +3p^{11}+3p^9+3p^7+3p^5+3p^3+3p^ 1-p^{11})}{32}$

Take out a common factor of 3p
$\frac{(3p(p^{20}+p^{18}+p^P{16}+p^{13}+p^{12}+p ^{10}+p^8+p^6+p^4+p^2+p^0) - p^{11})}{32}$

I claim that:
$p^{20}+p^{18}+...+p^0 = 0$
$p^{11}+1=0$
so $(p^{11}+1)(p^{10}+p^9+p^8+...+p^0)=0$
so $p^{21}+p^{20}+...+p^0=0$
so $(1+p)(p^{20}+p^{18}+p^{16}+...+p^0)=0$
and since $(1+p) \ne 0$
$(p^{20}+p^{18}+...+p^0)=0$

So that leads to:
$\frac{-p^{11}}{32}$
which is
$1/32$
Title: Re: TrueTears question thread
Post by: TrueTears on February 15, 2009, 10:10:03 pm: wow thanks mao and damo, lol i got that Q from a handout off my friend XD
Title: Re: TrueTears question thread
Post by: Mao on February 15, 2009, 10:12:01 pm: btw is that geometry question correct? I got a rather convoluted answer (convoluted being a=1.86784...) Where is the question from?
Title: Re: TrueTears question thread
Post by: TrueTears on February 15, 2009, 11:23:57 pm: not sure, i don't have an answer to it, a friend gave me the Q lol
Title: Re: TrueTears question thread
Post by: /0 on February 16, 2009, 06:30:39 pm: That geometry question is nasty. You can't even use similar triangles wtf
Title: Re: TrueTears question thread
Post by: Mao on February 16, 2009, 06:34:08 pm: Quote from: /0 on February 16, 2009, 06:30:39 pm
That geometry question is nasty. You can't even use similar triangles wtf

use the sine rule. answer is nasty
Title: Re: TrueTears question thread
Post by: TrueTears on February 16, 2009, 07:02:14 pm: Also...

O is any point on the diagonal BC of a parallelogram, prove that $\overrightarrow OA + \overrightarrow OB + \overrightarrow OC = \overrightarrow OD$
Title: Re: TrueTears question thread
Post by: /0 on February 16, 2009, 07:13:49 pm: If you look at a parallelogram, ABDC (labelled clockwise), then there is no way the directions OA, OB, and OC can add to get OD...?
Title: Re: TrueTears question thread
Post by: TrueTears on February 16, 2009, 07:17:44 pm: hmmm its a Q in the heinnemen book, it says exactly that
Title: Re: TrueTears question thread
Post by: Mao on February 16, 2009, 08:02:48 pm: how is BC a diagonal...?
Title: Re: TrueTears question thread
Post by: TrueTears on February 16, 2009, 08:13:45 pm: Quote from: Mao on February 16, 2009, 08:02:48 pm
how is BC a diagonal...?
i know ... there must be something wrong with the q.....
Title: Re: TrueTears question thread
Post by: TrueTears on February 17, 2009, 06:35:03 pm: Ok, nvm that question, my teacher just said that question was wrongly worded -_-
Title: Re: TrueTears question thread
Post by: TrueTears on February 17, 2009, 06:36:39 pm: Also this question:

if a = 2i - 2j + k and b = - i - j - 5k find a unit vector perpendicular to a and b
Title: Re: TrueTears question thread
Post by: Mao on February 17, 2009, 08:08:21 pm: let $\textbf{c} = c_1 \textbf{i} + c_2 \textbf{j} + c_3 \textbf{k}$

$\implies 2c_1 - 2c_2 + c_3 = 0$

$\implies -c_1 - c_2 - 5c_3 = 0$

$\implies c_1^2 + c_2^2 + c_3^2 = 1$

solving simultaneously to get the three coefficients.
Title: Re: TrueTears question thread
Post by: TrueTears on February 17, 2009, 08:24:08 pm: thanks mao
Title: Re: TrueTears question thread
Post by: Flaming_Arrow on February 17, 2009, 08:27:59 pm: sorry for hijacking the thread but how do you know they all equal 1?
Title: Re: TrueTears question thread
Post by: TrueTears on February 17, 2009, 08:28:24 pm: it says unit vector :P
Title: Re: TrueTears question thread
Post by: Flaming_Arrow on February 17, 2009, 08:29:21 pm: Quote from: TrueTears on February 17, 2009, 08:28:24 pm
it says unit vector :P

oops didn't pick that up
Title: Re: TrueTears question thread
Post by: TrueTears on February 23, 2009, 10:00:13 pm: prove these identities:

a) $sin^2xtanx + cos^2xcotx + 2sinxcosx = tanx + cotx$
b) $\frac{(cosxcotx - sinxtanx)(sinxcosx)}{cosx - sinx} = 1 + sinxcosx$

many thanks!
Title: Re: TrueTears question thread
Post by: vce08 on February 23, 2009, 11:26:57 pm: Quote from: TrueTears on February 23, 2009, 10:00:13 pm
prove these identities:

a) $sin^2xtanx + cos^2xcotx + 2sinxcosx = tanx + cotx$
b) $\frac{(cosxcotx - sinxtanx)(sinxcosx)}{cosx - sinx} = 1 + sinxcosx$

many thanks!

i got the first one

$sin^2xtanx + cos^2xcotx + 2sinxcosx$

$= (1-cos^2x)tanx + cos^2xcotx + 2sinxcosx$

$= tanx - sinxcosx + cos^2xcotx + 2sinxcosx$

$= tanx + sinxcosx + cos^2xcotx$

$= tanx + sinxcosx + \frac{cos^3x}{sinx}$

$= tanx + cosx(sinx + \frac{cos^2x}{sinx})$

$= tanx + cosx(sinx + \frac{1-sin^2x}{sinx})$

$= tanx + cosx(sinx - sinx + \frac{1}{sinx})$

$= tanx + cotx$

as required
Title: Re: TrueTears question thread
Post by: vce08 on February 24, 2009, 12:25:26 am: Quote from: TrueTears on February 23, 2009, 10:00:13 pm
prove these identities:

a) $sin^2xtanx + cos^2xcotx + 2sinxcosx = tanx + cotx$
b) $\frac{(cosxcotx - sinxtanx)(sinxcosx)}{cosx - sinx} = 1 + sinxcosx$

many thanks!

here is the second one

$\frac{(cosxcotx - sinxtanx)(sinxcosx)}{cosx - sinx}$

$=\cfrac{(\cfrac{cos^2x}{sinx} - \cfrac{sin^2x}{cosx})(sinxcosx)}{cosx - sinx}$

$=\cfrac{(\cfrac{1-sin^2x}{sinx} - \cfrac{1-cos^2x}{cosx})(sinxcosx)}{cosx - sinx}$

$=\cfrac{(\cfrac{1}{sinx} - sinx - \cfrac{1}{cosx} + cosx)(sinxcosx)}{cosx - sinx}$

$=\frac{cosx - sinx- sin^2xcosx + sinxcos^2x}{cosx - sinx}$

$=1 + \frac{-sin^2xcosx + sinxcos^2x}{cosx - sinx}$

$=1 + \frac{cosx(cosxsinx) - sinx(cosxsinx)}{cosx - sinx}$

$=1 + sinxcosx$

as required

i'm still new to this latex thing so i am quite slow at typing the answers up
Title: Re: TrueTears question thread
Post by: TrueTears on February 24, 2009, 10:24:05 am: thanks very much table!
Title: Re: TrueTears question thread
Post by: sb3700 on February 24, 2009, 04:52:33 pm: For the second one:

RHS = $\frac{(cosxcotx - sinxtanx)(sinxcosx)}{cosx - sinx}$
= $\frac{(\cfrac{cos^2x}{sinx} - \cfrac{sin^2x}{cosx})(sinxcosx)}{cosx-sinx}$
= $\frac{cos^3x - sin^3x}{cosx-sinx}$
= $\frac{(cosx-sinx)(cos^2x+sinxcosx + sin^2x)}{cosx-sinx}$ by difference of cubes factorisation
= $1 + sinxcosx$ = LHS
Title: Re: TrueTears question thread
Post by: TrueTears on February 24, 2009, 04:54:18 pm: Quote from: sb3700 on February 24, 2009, 04:52:33 pm
For the second one:

RHS = $\frac{(cosxcotx - sinxtanx)(sinxcosx)}{cosx - sinx}$
= $\frac{(\cfrac{cos^2x}{sinx} - \cfrac{sin^2x}{cosx})(sinxcosx)}{cosx-sinx}$
= $\frac{cos^3x - sin^3x}{cosx-sinx}$
= $\frac{(cosx-sinx)(cos^2x+sinxcosx + sin^2x)}{cosx-sinx}$ by difference of cubes factorisation
= $1 + sinxcosx$ = LHS

ARGH! damn i did exactly what you have done, but difference of cubes, z.z.z.z didn't realise that at all omg. I am so bad. -_-
Title: Re: TrueTears question thread
Post by: TrueTears on February 24, 2009, 06:53:51 pm: Let $sec \beta = b$ where $\beta \in [ \frac{\pi}{2},\pi]$ . Find, in terms of $\beta$ , 2 values of x in the range $[-\pi,\pi]$ which satisfies each of these equations:

a) $sec x = -b$

b) $cosec x = b$

Let $tan \gamma = c$ where $\gamma \in[ \pi,\frac{3\pi}{2}]$ . find in terms of $\gamma$ , 2 values of x in the range of $[0,2\pi]$ which satisfies

a) $cot x = c$

3. Given that $tan \alpha = p$ , where $\alpha$ is an acute angle, find each of the following in terms of $p$ .
a) $tan( \frac{3\pi}{2} + \alpha)$
Title: Re: TrueTears question thread
Post by: sb3700 on February 24, 2009, 09:49:45 pm: Quote from: TrueTears on February 24, 2009, 06:53:51 pm
Let $sec \beta = b$ where $\beta \in [ \frac{\pi}{2},\pi]$ . Find, in terms of $\beta$ , 2 values of x in the range $[-\pi,\pi]$ which satisfies each of these equations:

a) $sec x = -b$

b) $cosec x = b$

As $\beta \in [\frac{\pi}{2},\pi]$ , then $b = sec\beta \le 0$
Choose $\alpha = \pi - \beta$ (so $\alpha \in [0, \frac{\pi}{2}]$ and $cosec\alpha = cosec\beta$ and $sec\alpha = -sec\beta$ )

a) $sec x = -b \ge 0$
So $x = \alpha, -\alpha$ , ie: quadrant 1 or quadrant 4 (note you can add/subtract 2 $\pi$ to make sure end result is in domain).
$x = \pi - \beta, -\pi + \beta$ as $\beta = \pi - \alpha$
$x = \pm(\pi - \beta)$

b) $cosecx = b \le 0$
So $x = -\pi + \alpha, -\alpha$
So $x = -\pi + \pi - \beta, -\pi + \beta$
So $x = -\beta, -\pi + \beta$

I might do the others later
Title: Re: TrueTears question thread
Post by: TrueTears on February 25, 2009, 02:40:09 pm: thanks again sb3700, that really helped !
Title: Re: TrueTears question thread
Post by: sb3700 on February 25, 2009, 08:42:53 pm: Quote from: TrueTears on February 24, 2009, 06:53:51 pm
3. Given that $tan \alpha = p$ , where $\alpha$ is an acute angle, find each of the following in terms of $p$ .
a) $tan( \frac{3\pi}{2} + \alpha)$

This one is easier:
$tan( \frac{3\pi}{2} + \alpha) = -cot(\alpha) = -\frac{1}{tan(\alpha)} = -\frac{1}{p}$

Quote from: TrueTears on February 24, 2009, 06:53:51 pm
Let $tan \gamma = c$ where $\gamma \in[ \pi,\frac{3\pi}{2}]$ . find in terms of $\gamma$ , 2 values of x in the range of $[0,2\pi]$ which satisfies

a) $cot x = c$

So $tan(\gamma) = c = cot(x) = tan(\frac{\pi}{2} - x)$
Then $\frac{\pi}{2} - x = -2\pi + \gamma$ or $-\pi + \gamma$ (Q1 or Q3)
So $x = \frac{5\pi}{2} - \gamma$ or $x = \frac{3\pi}{2} - \gamma$
Title: Re: TrueTears question thread
Post by: TrueTears on February 27, 2009, 04:24:38 pm: thanks so much sb3700
Title: Re: TrueTears question thread
Post by: TrueTears on March 01, 2009, 06:26:01 pm: Prove $\frac{cos3x}{sinx} + \frac{sin3x}{cosx} = 2cot(2x)$
Title: Re: TrueTears question thread
Post by: TrueTears on March 01, 2009, 06:41:46 pm: and

Prove $cos4x = 8cos^4x - 8cos^2x +1$
Title: Re: TrueTears question thread
Post by: Mao on March 01, 2009, 06:44:57 pm: Quote from: TrueTears on March 01, 2009, 06:26:01 pm
Prove $\frac{cos3x}{sinx} + \frac{sin3x}{cosx} = 2cot(2x)$

multiplying by $\sin (x)\cos(x)$ on both sides

$\implies \cos x \cos 3x + \sin x \sin 3x = \sin 2x \cot 2x$

$\implies \cos (x-3x) = \sin 2x\cdot \frac{\cos 2x}{\sin 2x}$

$\implies \cos (-2x) = \cos 2x$

$LHS=RHS,\; QED$

Quote from: TrueTears on March 01, 2009, 06:41:46 pm
and

Prove $cos4x = 8cos^4x - 8cos^2x +1$

$LHS = \cos 4x = 2\cos^2 2x - 1 = 2 (2\cos^2 x-1)^2 -1 = 2 (4\cos^4 x - 4\cos^2 x + 1) -1 = 8\cos^4 x - 8\cos^2 x + 1=RHS$

QED
Title: Re: TrueTears question thread
Post by: TrueTears on March 01, 2009, 06:46:58 pm: Quote from: Mao on March 01, 2009, 06:44:57 pm
Quote from: TrueTears on March 01, 2009, 06:26:01 pm
Prove $\frac{cos3x}{sinx} + \frac{sin3x}{cosx} = 2cot(2x)$

multiplying by $\sin (x)\cos(x)$ on both sides

$\implies \cos x \cos 3x + \sin x \sin 3x = \sin 2x \cot 2x$

$\implies \cos (x-3x) = \sin 2x\cdot \frac{\cos 2x}{\sin 2x}$

$\implies \cos (-2x) = \cos 2x$

$LHS=RHS,\; QED$
nice thanks mao
Title: Re: TrueTears question thread
Post by: TrueTears on March 01, 2009, 06:50:47 pm: and last one:
$\frac{1-cosx}{sinx} = tan{\frac{x}{2}}$

thanks heaps for your help!
Title: Re: TrueTears question thread
Post by: Mao on March 01, 2009, 06:56:17 pm: Quote from: TrueTears on March 01, 2009, 06:50:47 pm
and last one:
$\frac{1-cosx}{sinx} = tan{\frac{x}{2}}$

thanks heaps for your help!

$LHS = \frac{1-\cos x}{\sin x} = \frac{1 - \left(1-2\sin^2 \frac{x}{2}\right)}{2\sin \frac{x}{2}\cos \frac{x}{2}} = \frac{2\sin^2 \frac{x}{2}}{2\sin \frac{x}{2}\cos \frac{x}{2}} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} = \tan\frac{x}{2} = RHS$

QED

btw, what's with the posting questions in trios :P you seem to be doing that a lot (and always in threes)
Title: Re: TrueTears question thread
Post by: TrueTears on March 01, 2009, 07:03:26 pm: thanks again mao

lol something me /0 come up with, triforce FTW!! ganges + banks = banges
Title: Re: TrueTears question thread
Post by: TrueTears on March 02, 2009, 06:45:10 pm: just a few more Q

1. Prove $\frac{sinx +1 - cosx}{sinx - 1 + cosx} = \frac{1+tan(\frac{x}{2})}{1-tan(\frac{x}{2})}$

2. $cot(x+y) = \frac{cotxcoty -1}{cotx + coty}$
Title: Re: TrueTears question thread
Post by: TrueTears on March 02, 2009, 09:56:05 pm: Can anyone help with Q 17 and Q 19? Many thanks!

(http://img299.imageshack.us/img299/386/ch3q.jpg)
Title: Re: TrueTears question thread
Post by: TrueTears on March 03, 2009, 01:13:53 pm: Given f:R ->R where f(x) = sin(2x-3). Show that a restriction of f, namly F, defined on [1,2] has an inverse $F^{-1}$ . Find the rule of
$F^{-1}$ and stating its domain and range
Title: Re: TrueTears question thread
Post by: kamil9876 on March 03, 2009, 02:49:03 pm: Question 19: In the second phase, u hav to treat the 20cm string in two seperate parts, one is the arc AP and the other is the remaining straight line segment PB'.

so we know $AP+PB'=20 (1)$

Familiarise urself with the arclength formula $l=r\theta$
You can use it to work out AP:

$AP=r(\frac{\pi}{2}-\theta)$

PB' can be worked out by applying elementary trig to the right angled triangle OPB' :

$\tan\theta=\frac{PB'}{10}$

Now sub these two expressions of PB' and AP into equation 1.
Title: Re: TrueTears question thread
Post by: kamil9876 on March 03, 2009, 03:09:46 pm: for question 17:

U must find an expression for the area of the blue area, and equate it the the area of the white section of the circle:

Let this area be $A$ and the radius of the circle be $r$

Blue area:

Draw the line $OX$ . Notice that the angle AOX is $\theta$ . Also, notice that the area is now split in half i.e: the area of triangle AOX is $\frac{A}{2}$

$\frac{A}{2}=\frac{1}{2} r AX$
$A=r AX$

$AX=r\tan\theta$

So $A=r^2 tan\theta$

The white area:

the area of the whole circle is $\pi r^2$ . To get the area of the white part you need to multiply the whole area by the fraction $\frac{2\pi-2\theta}{2\pi} = \frac{\pi-\theta}{\pi}$ . Hence:

$A = \frac{\pi-\theta}{\pi} \pi r^2$
$r^2 tan\theta = \frac{\pi-\theta}{\pi} \pi r^2$

$r^2$ cancels out as well as a $\pi$ .
Title: Re: TrueTears question thread
Post by: kamil9876 on March 03, 2009, 03:24:32 pm: And now for the question that isn't associated with a number:

A bit dissapointed with the lack of $\pi$ in inside the trig function domain.

Restricting the domain to [1,2] means that the thing inside the bracket will go from -1 to 1. SO the function exhibits the values: sin(-1), sin(-0.9), sin(-0.8).... sin(1) etc. (there are infinite possibilities since there are inifite real numbers between -1 and 1)

In other words: you could say that the range of f(x) for this domain is exactly the same as the range of sin(x) for domain [-1,1]. Such a range is monotonic (contains no turning points) since the turning points occur at $-pi/2$ and $pi/2$ and there are no other turning points between these two values. But the domain [-1,1] is in between these two values hence f(x) has no turning points, in fact it is always increasing. This means that for each x-value there is only one value of f(x) and conversely for each value of f(x) there is only one x-value. Hence the inverse function does exist.

Oh and for the other part of the question: the domain is trivially [1,2] while the range is [sin(-1),sin(1)] (as shown before) of f(x). SO for the inverse function, the domain and range will swap since the x and y values are being swapped (by definition of inverse).
Title: Re: TrueTears question thread
Post by: TrueTears on March 03, 2009, 04:24:11 pm: Thanks kamil!!

yeah also dw about Q 19, i've figured it out.

also these 2

1. Prove $\frac{sinx +1 - cosx}{sinx - 1 + cosx} = \frac{1+tan(\frac{x}{2})}{1-tan(\frac{x}{2})}$

2. $cot(x+y) = \frac{cotxcoty -1}{cotx + coty}$
Title: Re: TrueTears question thread
Post by: kamil9876 on March 03, 2009, 05:22:49 pm: Question1:

Multiply the numerater and denominator of RHS by $1+tan(x/2)$
$\frac{(1+tan(x/2))^2}{1-tan^2(x/2)}$

$\frac{1+2tan(x/2)+tan^2(x/2)}{1-tan^2(x/2)}$

$\frac{2tan(x/2)}{1-tan^2(x/2)}+\frac{1+tan^2(x/2)}{1-tan^2(x/2)}$

$\tan(x)+\frac{\sec^2(x/2)}{1-\frac{\sin^2(x/2)}{\cos^2(x/2)}}$

Multiply numerator and denominator by $cos^2(x/2)$

$\tan(x) + \frac{1}{cos^2(x/2)-sin^2(x/2)}$

$\tan(x) + \frac{1}{cos(x)}$

this may help.
Title: Re: TrueTears question thread
Post by: kamil9876 on March 03, 2009, 06:47:14 pm: and question 2 looks awfully familliar to the tan(x+y) expansion.

$\cot(x+y) = \frac{1}{tan(x+y)}$

$= \frac{1-\tan(x)\tan(y)}{tan(x)+tan(x)}$

$= \frac{1-\frac{1}{\cot(x)\cot(y)}}{\frac{1}{\cot(x)}+\frac{1}{\cot(y)}}$

Multiply numerator and denominator by $\cot(x)\cot(y)$
Title: Re: TrueTears question thread
Post by: TrueTears on March 05, 2009, 03:48:09 pm: nice thanks kamil i got it :)
Title: Re: TrueTears question thread
Post by: TrueTears on March 06, 2009, 06:16:01 pm: Just another question:

a) Find a suitable parametric equation for the ellipse $4(x+2)^2 + (y-5)^2 =4$

Make sure that the entire ellipse is given by your parametric equation

So i did this part. But can anyone just check if it is right because there are no answers to these questions :(

$(x+2)^2 + \frac{(y-5)^2}{4} = 1$

so $y= 2sint + 5$ and $x= cost - 2$ where $t \in [0,2\pi]$

b) i. The point $P (3cos\theta, 2sin\theta)$ , $\theta \in [0,2\pi]$ , is a point on the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ . S is the point $(\sqrt{5},0)$ .

The line segment SP is extended to the point Q where P is between S and Q and SQ = 3SP. Calculate the coordinates of Q ( in terms of $\theta$ )

ii. As $\theta$ varies, the point P moves around the ellipse. As P moves, so does Q. Find the cartesian equation of the locus of Q as $\theta$ varies. State the type of curve traced out by Q and give the coordinates of the centre of this curve.

If anyone could help with b) i and ii that would be great!
Title: Re: TrueTears question thread
Post by: kurrymuncher on March 06, 2009, 06:25:20 pm: yep, a) is right
Title: Re: TrueTears question thread
Post by: TrueTears on March 06, 2009, 06:49:27 pm: thanks ! i kinda have an idea for b) but just can't get it haha
Title: Re: TrueTears question thread
Post by: kamil9876 on March 06, 2009, 08:14:50 pm: If the line is extended by 3, then the rise is multiplied by 3, and hence so is the the run in order to keep the gradient the same. Think of similair triangles or parralel vectors. i.e: vector SQ=3SP and so the i component of SQ is 3 times as big as the i of SP and same for j.

Hence the rise for SP is $2\sin\theta$ while the rise for SQ is $y$ and so $y=6\sin\theta$

The run of SP is $3\cos\theta - \sqrt{5}$ but the run for SQ is $x - \sqrt{5}$ and so $3(3\cos\theta - \sqrt{5}) = x - \sqrt{5}$
Title: Re: TrueTears question thread
Post by: kamil9876 on March 06, 2009, 08:21:58 pm: for part 2, relate the two values for x and y that we got earlier (note that in my previous post, x and y are the x and y values of point Q). Now the way u relate them is by using the pythagorean identity the same way that u do for parametric equations of elipses, where in this case the parameter is $\theta$ and the parametric equations are the two equation for x and y.
Title: Re: TrueTears question thread
Post by: TrueTears on March 06, 2009, 08:33:50 pm: Quote from: kamil9876 on March 06, 2009, 08:14:50 pm
If the line is extended by 3, then the rise is multiplied by 3, and hence so is the the run in order to keep the gradient the same. Think of similair triangles or parralel vectors. i.e: vector SQ=3SP and so the i component of SQ is 3 times as big as the i of SP and same for j.

Hence the rise for SP is $2\sin\theta$ while the rise for SQ is $y$ and so $y=6\sin\theta$

The run of SP is $3\cos\theta - \sqrt{5}$ but the run for SQ is $x - \sqrt{5}$ and so $3(3\cos\theta - \sqrt{5}) = x - \sqrt{5}$

thanks kamil i get most of that but just near the end how do you get the run for SQ? where did the $x - \sqrt{5}$ come from?
Title: Re: TrueTears question thread
Post by: kamil9876 on March 06, 2009, 08:39:45 pm: because the run is the x value at Q minus the x value at S. It makes a lot of sense when u draw the diagram of the three points and the lines
Title: Re: TrueTears question thread
Post by: TrueTears on March 06, 2009, 08:43:08 pm: but how do you know the x value of Q is x?
Title: Re: TrueTears question thread
Post by: kamil9876 on March 06, 2009, 08:45:37 pm: i just called the point Q (x,y). Shouldve maybe added a subscript Q for it to make sense hah
Title: Re: TrueTears question thread
Post by: /0 on March 06, 2009, 08:54:38 pm: I used vectars:

Let $Q = (x,y)$

$\vec{SQ} = (x-\sqrt{5}) \mathbf{i}+y\mathbf{j}$

$\vec{SP} = (3\cos{\theta}-\sqrt{5})\mathbf{i}+2\sin{\theta}\mathbf{j}$

$\vec{SQ}=3\vec{SP}$

$\Rightarrow (x-\sqrt{5})\mathbf{i}+y\mathbf{j} = 3\left((3\cos{\theta}-\sqrt{5})\mathbf{i}+2\sin\theta\mathbf{j}\right)$

$x = 3(3\cos{\theta}-\sqrt{5})+\sqrt{5}$

$y = 6\sin\theta$

$\therefore Q = (9\cos{\theta}-2\sqrt{5},6\sin{\theta})$
Title: Re: TrueTears question thread
Post by: kamil9876 on March 06, 2009, 08:57:40 pm: hah yep that's what i did but without the notation. Linear equation, vectors and similair triangles are all similair in some cases (no pun intended)
Title: Re: TrueTears question thread
Post by: TrueTears on March 06, 2009, 09:24:40 pm: thanks i get i XD

yeah i did it another way and still got it lol
Title: Re: TrueTears question thread
Post by: TrueTears on March 06, 2009, 09:40:34 pm: another question.

1. a) Sketch the hyperbola $\frac{x^2}{9} - y^2 = 1$ clearly indicating its centre vertices and asymptotes.

This i can do.

b) By setting up a suitable quadratic equation, show that the line $y = \frac{5}{12}x + \frac{3}{4}$ is a tangent to the hyperbola $\frac{x^2}{9} - y^2 = 1$ .

This i can do.

c) Find the coordinates of the point of intersection of the tangent $y = frac\{5}{12}x + \frac{3}{4}$ and the hyperbola $\frac{x^2}{9} - y^2 = 1$ . Let this point be A.

This i can do.

d) The line $y = \frac{5}{12}x + \frac{3}{4}$ passes through the point B(3,2). By referring to the sketch you drew in part a, state the equation of the other tangent to the hyperbola $\frac{x^2}{9} - y^2 = 1$ that passes through B, and state the coordinates of the point of intersection, C, of this tangent with the hyperbola.

lol stuck on part d)
Title: Re: TrueTears question thread
Post by: /0 on March 06, 2009, 09:43:22 pm: I 'think' the other tangent is x = 3. Not sure though; although, it would fit the information.
Title: Re: TrueTears question thread
Post by: TrueTears on March 06, 2009, 09:43:46 pm: yeah , but how would you solve it ?
Title: Re: TrueTears question thread
Post by: TrueTears on March 06, 2009, 09:59:43 pm: Any ideas /0?
Title: Re: TrueTears question thread
Post by: TrueTears on March 06, 2009, 10:11:44 pm: actually /0, you know your vectors way of doing that question?

Quote from: /0 on March 06, 2009, 08:54:38 pm
I used vectars:

Let $Q = (x,y)$

$\vec{SQ} = (x-\sqrt{5}) \mathbf{i}+y\mathbf{j}$

$\vec{SP} = (3\cos{\theta}-\sqrt{5})\mathbf{i}+2\sin{\theta}\mathbf{j}$

$\vec{SQ}=3\vec{SP}$

$\Rightarrow (x-\sqrt{5})\mathbf{i}+y\mathbf{j} = 3\left((3\cos{\theta}-\sqrt{5})\mathbf{i}+2\sin\theta\mathbf{j}\right)$

$x = 3(3\cos{\theta}-\sqrt{5})+\sqrt{5}$

$y = 6\sin\theta$

$\therefore Q = (9\cos{\theta}-2\sqrt{5},6\sin{\theta})$

i know it works but... the question says SQ = 3SP meaning the magnitudes , where as you had them as vectors? If the magnitude of SQ is 3 times SP , does that also mean the vector SQ is also 3 times vector SP?
Title: Re: TrueTears question thread
Post by: /0 on March 06, 2009, 10:35:37 pm: What do you mean by 'solve' it?

CAN SOMEONE ELSE PLEASE CONFIRM THIS

I just kinda looked at the graph and saw x=3 as a possible tangent through (3,2). Is that how you're supposed to do it? Maybe? Seeing as it says "refer to your graph in part a)".

So C = (3,0)

Quote from: TrueTears on March 06, 2009, 10:11:44 pm

i know it works but... the question says SQ = 3SP meaning the magnitudes , where as you had them as vectors? If the magnitude of SQ is 3 times SP , does that also mean the vector SQ is also 3 times vector SP?

"The line segment SP is extended to the point Q where P is between S and Q and SQ = 3SP."

Since the line is extended, S, Q and P must be collinear, so you can use vectors.
Title: Re: TrueTears question thread
Post by: TrueTears on March 06, 2009, 10:36:59 pm: Quote from: /0 on March 06, 2009, 10:35:37 pm
What do you mean by 'solve' it?

CAN SOMEONE ELSE PLEASE CONFIRM THIS

I just kinda looked at the graph and saw x=3 as a possible tangent through (3,2). Is that how you're supposed to do it? Maybe? Seeing as it says "refer to your graph in part a)".

So C = (3,0)

Quote from: TrueTears on March 06, 2009, 10:11:44 pm

i know it works but... the question says SQ = 3SP meaning the magnitudes , where as you had them as vectors? If the magnitude of SQ is 3 times SP , does that also mean the vector SQ is also 3 times vector SP?

"The line segment SP is extended to the point Q where P is between S and Q and SQ = 3SP."

Since the line is extended, S, Q and P must be collinear, so you can use vectors.

nice i see, but even if they are collinear if their magnitudes have that relation ship doesn't mean the vectors also share that relationship? (SQ = 3SP)
Title: Re: TrueTears question thread
Post by: TrueTears on March 06, 2009, 10:39:31 pm: and also for that tangent question, yeah... i mean is there a more systematic way of doing it? Instead of just kinda guessing and looking at graph?
Title: Re: TrueTears question thread
Post by: kamil9876 on March 06, 2009, 11:05:52 pm: Quote from: TrueTears on March 06, 2009, 10:36:59 pm
Quote from: /0 on March 06, 2009, 10:35:37 pm
What do you mean by 'solve' it?

CAN SOMEONE ELSE PLEASE CONFIRM THIS

I just kinda looked at the graph and saw x=3 as a possible tangent through (3,2). Is that how you're supposed to do it? Maybe? Seeing as it says "refer to your graph in part a)".

So C = (3,0)

Quote from: TrueTears on March 06, 2009, 10:11:44 pm

i know it works but... the question says SQ = 3SP meaning the magnitudes , where as you had them as vectors? If the magnitude of SQ is 3 times SP , does that also mean the vector SQ is also 3 times vector SP?

"The line segment SP is extended to the point Q where P is between S and Q and SQ = 3SP."

Since the line is extended, S, Q and P must be collinear, so you can use vectors.

nice i see, but even if they are collinear if their magnitudes have that relation ship doesn't mean the vectors also share that relationship? (SQ = 3SP)

But like /0 said, the line is extended by a factor of 3, meaning that the gradient remains unchanged. ANother way of looking at it is imagine the line y=3x, now there exists a segment of this line that has magnitude one (an example is this line with a domain of $[0,\frac{1}{\sqrt{10}}$ . This line segment can also be described by the vector $\frac{1}{\sqrt{10}}$ i + $\frac{3}{\sqrt{10}}$ j. While this line extended by a factor of 3 is like the function y=3x for domain $[0,\frac{3}{\sqrt{10}}$ which is the same as the vector $\frac{3}{\sqrt{10}}$ i + $\frac{9}{\sqrt{10}}$ j.

Moral of the story: Vectors are like line segments, where the i component is the run, while the j is the rise. A vector being extended by a factor of 3 is like a line segment being extended, the rise over run doesn't change and so the factor that the rise was extended is the same as the factor that the run was extended. This factor is equal to the factor that the magnitude was extended, this can be proven using pythatgoras theorem or even more elegantly by imagining 3 coppies of the vector being placed head to tail. e.g: draw some vector a now the vector 3a=a + a + a so it's like 3 coppies being attatched head to tail. Now the new hypotenuse is the sum of the three individual hypotenuses, while the run is the sum of the three runs, and the rise is the sum of the three rises. This is explained in teh attatched image.
Title: Re: TrueTears question thread
Post by: TrueTears on March 06, 2009, 11:07:52 pm: ah yes yes yes, thank you so much.
Title: Re: TrueTears question thread
Post by: kamil9876 on March 07, 2009, 12:07:33 am: /0 is right, but analytically speaking we havn't ruled out the possibility that there may be more tangents fiting this information. It's probably the only one as the question said to 'use ur diagram' and other heurestical reasoning about the geometry of the situation sort of leads to this conclusion. A way that this may be done analytically, if ur interested, is to first of all find dy/dx (which i assume u hav already done) and then consider the tangent's gradient. The tangent goes through some point on the hyperbola (x,y) and it satisfies the conditions specified by dy/dx. hence u must now find values that satisfy the equation:

$\frac{dy}{dx} = \frac{y-2}{x-3}$

You must also make the substituion using the equation of the hyperbola to get rid of either x or y so that only one variable is involved. This is a fuckload of algebra that is beyond what the question was probably designed for, I think some geometrical reasoning can show that indeed x=3 is the only one. Also, the word "the" in "state the equation of the other tangent to the hyperbola" hints that it is the only one.
Title: Re: TrueTears question thread
Post by: TrueTears on March 07, 2009, 12:20:53 am: nice thanks
Title: Re: TrueTears question thread
Post by: TrueTears on March 07, 2009, 12:38:32 am: 1. In the triangle OAB, let M, N and P be the midpoint of the sides OB, AB and OA respectively. Let the perpendicular bisectors of sides AB and OA meet at G as shown. Let the position vector of P and M be p and m respectively and let $\overrightarrow {AN}$ be denoted by r as shown.

(http://img21.imageshack.us/img21/5499/sach.jpg)

a) show that m = p + r

this i have done

b) Let $\overrightarrow {GP}$ be detnoed by v and $\overrightarrow {GN}$ be denoted by w, as shown. One possible expression for $\overrightarrow {GM}$ is $\overrightarrow {GM} = \overrightarrow {GP} +\overrightarrow {PO} + \overrightarrow {OM}$ = v - p + m = v+r, by a)
Use this expression to show that $\overrightarrow {GM}$ . p = r . p

how do you do b)?

2. A radar tracking facility tracks a plane flying in a straight line. With the radar facility as the origin, the plane's initial and final position vectors are a = 2i + 8j + k and b = 8i -4j + 13k respectively.
a) Find a unit vector , u , in the direction of the motion of the plane.

this i found to be $u = \frac{1}{18}(6i - 12j +12k)$

b) The position vector of the plane, p, when it is closest to the radar can be expressed as p = a + $\gamma$ u, where $\gamma$ is a real number. find the value of $\gamma$ and hence the plane's position when it is closest to the radar facility.

Any help on b)? I've tried letting p = ai +bj +ck but that just gets too many equations and can't be solved.

thanks guys!
Title: Re: TrueTears question thread
Post by: kamil9876 on March 07, 2009, 12:47:12 am: for b, use the distributive properties of the dot product:

GM.p=p.(v+r)
=p.v + p.r
but p.v=0 as indicated on the diagram(perpendicular to each other)
Title: Re: TrueTears question thread
Post by: kamil9876 on March 07, 2009, 01:18:25 am: for the next question, refer to attatched diagram. OA=a, OP=p and OB=b.

OP.AB=0
0=(a+ $\gamma$ u)(b-a)
0=a.b-a.a+ $\gamma$ u.b- $\gamma$ u.a

because u know a, b, u you can evaluate all those dot products and end up getting some number, with some $\gamma$ s in there that you can then work out as they will be the only unkown.
Title: Re: TrueTears question thread
Post by: TrueTears on March 07, 2009, 11:47:45 am: thanks kamil for all your help, I got both of them now XD
Title: Re: TrueTears question thread
Post by: TrueTears on March 07, 2009, 12:21:24 pm: Another question:

Lines y = 2x and y = 10-2x are the asymptotes of a hyperbola of the form. $\frac{(x-p)^2}{a^2} - \frac{(y-q)^2}{b^2} = 1$

a) use the equations of the asymptotes to establish a relation between a and b

Just stuck on this part lol
Title: Re: TrueTears question thread
Post by: Mao on March 07, 2009, 06:12:35 pm: the asymptotes are:

$y=\pm \frac{b}{a}(x-p) + q$

from the given gradients of the asymptotes, $2=\frac{b}{a}\implies a=2b$ (q=5, p=2.5)
Title: Re: TrueTears question thread
Post by: TrueTears on March 07, 2009, 06:13:31 pm: oh i see lol, stupid me
Title: Re: TrueTears question thread
Post by: TrueTears on March 07, 2009, 06:25:52 pm: another one

Points A, B and C have position vectors $5j + 5k, 4i + j+ k$ and $\frac{5}{2}(i + j + k)$ respectively.

Find AC : BC

i keep getting 5:-3

but can you have a negative ratio? Or is that wrong lols
Title: Re: TrueTears question thread
Post by: Mao on March 07, 2009, 06:49:57 pm: AC and BC are scalar quantities, denoting the length of $\vec{AC}$ and $\vec{BC}$

so yah, shouldn't be negative.
Title: Re: TrueTears question thread
Post by: TrueTears on March 07, 2009, 06:52:36 pm: z.z.z. lol its meant to be the magnitude ratio not their vectors hahahaha
Title: Re: TrueTears question thread
Post by: Over9000 on March 07, 2009, 10:56:50 pm: For that big triangle question, how do you do part c
show that GM.m =0
Title: Re: TrueTears question thread
Post by: TrueTears on March 07, 2009, 10:58:35 pm: so $\overrightarrow {GM}.m = \overrightarrow {GM}.(p+r) = \overrightarrow {GM}.p + \overrightarrow {GM}.r = r.p + (-r.p) = 0$
Title: Re: TrueTears question thread
Post by: Over9000 on March 07, 2009, 11:00:25 pm: thanks
Title: Re: TrueTears question thread
Post by: TrueTears on March 07, 2009, 11:03:36 pm: Quote from: TrueTears on March 07, 2009, 06:25:52 pm
another one

Points A, B and C have position vectors $5j + 5k, 4i + j+ k$ and $\frac{5}{2}(i + j + k)$ respectively.

Find AC : BC

i keep getting 5:-3

but can you have a negative ratio? Or is that wrong lols
this is the next part to that question

Find $|\overrightarrow {OB}|$ . $cos\theta^{o}$ , where $\angle AOB = \theta^{o}$

how can you dot a magnitude with a $cos\theta^{o}$ ?

And how would you do this question

thanks!
Title: Re: TrueTears question thread
Post by: kamil9876 on March 07, 2009, 11:23:42 pm: Generally, a dot means multiplication, it is used in certain books and in some countries in order to avoid ambiguity as 6x3(6 times 3) can be confused with 6 multiplied by some number x then multiplied by 3. You can see that $|\overrightarrow {OB}|$ . $cos\theta^{o}$ looks like the dot product without the other magnitude, so it makes sense that this question involves the dot product and hence further shows that the dot means multiplication.
Title: Re: TrueTears question thread
Post by: TrueTears on March 07, 2009, 11:59:08 pm: Quote from: Mao on March 07, 2009, 06:12:35 pm
the asymptotes are:

$y=\pm \frac{b}{a}(x-p) + q$

from the given gradients of the asymptotes, $2=\frac{b}{a}\implies a=2b$ (q=5, p=2.5)
i think you mean 2a = b :P
Title: Re: TrueTears question thread
Post by: TrueTears on March 08, 2009, 12:40:50 am: 1.

$x = 1.5 sec(t - \frac{\pi}{3}) + 2.5$
$y = 3tan(t - \frac{\pi}{3}) + 5$

are the 2 parametric equations for the hyperbola with equation $\frac{(2x-5)^2}{9} - \frac{(y-5)^2}{9} = 1$ . Determine the range of values of t which will represent the hyperbola without duplication.

Basically i've worked out the range of the hyperbola is R and domain is R\(1,4).

so range of the y = parametric equation must = range of cartesian = R

and range of the x = parametric equation must = domain of cartesian = R\(1,4)

What do i do from here? Like if i have $t \in [0,\pi]$ works for the $y = 3tan(t - \frac{\pi}{3}) + 5$ parametric equation but then for the $x = 1.5 sec(t - \frac{\pi}{3}) + 2.5$ parametric equation is not enough to draw a whole hyperbola. But if i have $t \in [0,\frac{3\pi}{4}]$ then that allows the $x = 1.5 sec(t - \frac{\pi}{3}) + 2.5$ equation to work but it will have made part of the hyperbola duplicated. What do i do?
Title: Re: TrueTears question thread
Post by: TrueTears on March 08, 2009, 01:50:40 am: Quote from: TrueTears on March 07, 2009, 06:25:52 pm
another one

Points A, B and C have position vectors $5j + 5k, 4i + j+ k$ and $\frac{5}{2}(i + j + k)$ respectively.

2. this is the next part to that question

the position vector of P is $\overrightarrow {OP} = j + k$ . P is a point on $\overrightarrow {OA}$ nearest to B.

Find the position vector of D, the centre of the circle through A, B and P
Title: Re: TrueTears question thread
Post by: TrueTears on March 08, 2009, 01:58:18 am: 3. Engineers are trying to make a chute, they consider the equation $y = (6cos(\frac{\pi}{10}x) + 6) \times 2(1 - cos(\frac{\pi}{10}x))$ . Express the rule for this curve in the form
$y = r (1 - cos(\frac{\pi}{5}x))$ , where r is a positive interger.
Title: Re: TrueTears question thread
Post by: TrueTears on March 08, 2009, 02:08:33 am: 4. In the triangle OAB, let M, N and P be the midpoint of the sides OB, AB and OA respectively. Let the perpendicular bisectors of sides AB and OA meet at G as shown. Let the position vector of P and M be p and m respectively and let $\overrightarrow {AN}$ be denoted by r as shown.

(http://img21.imageshack.us/img21/5499/sach.jpg)

a) show that m = p + r

this i have done

b) i. Let $\overrightarrow {GP}$ be detnoed by v and $\overrightarrow {GN}$ be denoted by w, as shown. One possible expression for $\overrightarrow {GM}$ is $\overrightarrow {GM} = \overrightarrow {GP} +\overrightarrow {PO} + \overrightarrow {OM}$ = v - p + m = v+r, by a)
Use this expression to show that $\overrightarrow {GM}$ . p = r . p

ii. In a similar way show that $\overrightarrow {GM} = w - p$ , and hence that $\overrightarrow {GM}.r = -r.p$

this i have done

c)Use your results from a) and b) to show that $\overrightarrow {GM} . m = 0$

this i have done

d) What conclusion can you draw from a) , b) and c)?

How do you do d) ?
Title: Re: TrueTears question thread
Post by: /0 on March 08, 2009, 06:21:43 am: Quote from: TrueTears on March 08, 2009, 01:50:40 am
Quote from: TrueTears on March 07, 2009, 06:25:52 pm
another one

Points A, B and C have position vectors $5j + 5k, 4i + j+ k$ and $\frac{5}{2}(i + j + k)$ respectively.

2. this is the next part to that question

the position vector of P is $\overrightarrow {OP} = j + k$ . P is a point on $\overrightarrow {OA}$ nearest to B.

Find the position vector of D, the centre of the circle through A, B and P

If you draw a diagram you'll notice that OP must be the vector resolute of OB in the direction of OA, since that makes OA perpendicular to PB, which gives the shortest distance.

Then, since the circle goes through A, B and P, and $\angle APB = 90^{\circ}$ , AB must be a diameter, with D the midpoint of AB.
Title: Re: TrueTears question thread
Post by: TrueTears on March 08, 2009, 04:00:16 pm: thanks /0
Title: Re: TrueTears question thread
Post by: TrueTears on March 08, 2009, 04:02:27 pm: oh nvm i got Q 3
Title: Re: TrueTears question thread
Post by: TrueTears on March 08, 2009, 05:55:24 pm: and another one

points A B C and D are defined by position vectors a b c and d respectively. If $\overrightarrow {AB} + \overrightarrow {CD} = 0$
a) express d in terms of a, b and c

so $\overrightarrow {AB} = b - a$
and $\overrightarrow {CD} = d - c$

b - a + d - c = 0

d = c + a -b

b) Show that AC and BD bisect each other
How do you do this part?

Thanks
Title: Re: TrueTears question thread
Post by: Mao on March 08, 2009, 06:04:53 pm: AC = c - a
BD = d - b = c + a - 2b

let X be the midpoint of AC, then, AX = XC = 0.5 (c - a)
BX = BA + AX = (a - b) + 0.5 (c - a) = 0.5 c + 0.5 a - b = 0.5 BD

implies X is also the midpoint of BD

hence, AC and BD bisect each other at X
Title: Re: TrueTears question thread
Post by: TrueTears on March 08, 2009, 06:32:11 pm: Thanks mao

also got any ideas on the other 2? Q1 and Q4 d)?
Title: Re: TrueTears question thread
Post by: kamil9876 on March 08, 2009, 09:58:33 pm: Q1.

Because you are considering the maximum domain and range of the hyperbola, then the only way to yield this is to ensure that the parametric equations have maximum range( I am pretty sure). Hence the $\tan(t-\frac{pi}{3})$ and $\sec(t - \frac{pi}{3})$ bits must hav their maximum ranges. The maximum ranges for $\tan\theta$ and $\sec\theta$ occur when $\theta$ is [0,pi)/{pi/2} for both functions. Now in this case $\theta=t-\frac{pi}{3}$ so t has a domain of $[\pi/3,\pi + \pi/3) without {\pi/2 + \pi/3}$ i.e: a domain translated $\pi/3$ units to the positive side. No duplication arises since $\pi$ is just one period for both of those trig functions and the values don't repeat.
Title: Re: TrueTears question thread
Post by: TrueTears on March 09, 2009, 01:40:48 am: Quote from: kamil9876 on March 08, 2009, 09:58:33 pm
Q1.

Because you are considering the maximum domain and range of the hyperbola, then the only way to yield this is to ensure that the parametric equations have maximum range( I am pretty sure). Hence the $\tan(t-\frac{pi}{3})$ and $\sec(t - \frac{pi}{3})$ bits must hav their maximum ranges. The maximum ranges for $\tan\theta$ and $\sec\theta$ occur when $\theta$ is [0,pi)/{pi/2} for both functions. Now in this case $\theta=t-\frac{pi}{3}$ so t has a domain of $[\pi/3,\pi + \pi/3) without {\pi/2 + \pi/3}$ i.e: a domain translated $\pi/3$ units to the positive side. No duplication arises since $\pi$ is just one period for both of those trig functions and the values don't repeat.
ah yes i fully understand now,

thanks for the yet again awesome explanation kamil !
Title: Re: TrueTears question thread
Post by: kamil9876 on March 09, 2009, 06:55:25 pm: Also, make sure (if u havnt already noticed), that the domain of t has one open bracket, and one closed bracket, as tan(pi) and tan(0) are equal to each other and same for sec. Hence if both endpoints of the domain are included then one point is duplicated, if both are not included then one point is missing. So, have one endpoint included, one not.
Title: Re: TrueTears question thread
Post by: TrueTears on March 09, 2009, 06:56:23 pm: yeap, noticed that ^^
Title: Re: TrueTears question thread
Post by: TrueTears on March 09, 2009, 10:39:58 pm: also this Q7 lol just stumped on it -_-"

Many thanks guys!!

(http://img17.imageshack.us/img17/196/proofvxl.jpg)
Title: Re: TrueTears question thread
Post by: /0 on March 09, 2009, 11:25:57 pm: $|\vec{AB}|=|u|$

$|\vec{AC}|=|u+v|$

If they are equal, $|u|^2=|u+v|^2$

$|u|^2= (u+v)\cdot (u+v)$

$u\cdot u= u \cdot u + 2 u \cdot v + v \cdot v$

Also, $|\vec{AB}| = |\vec{CD}|\Rightarrow |u| = |v|, \Rightarrow u \cdot u = v\cdot v$

$\Righatarrow u \cdot u = 2 u \cdot u + 2u\cdot v$

$2u\cdot v +u\cdot u=0$

$u\cdot (2v + u)=0$

And since, $\vec{AD} = 2v+u$ , we have $AB \perp AD$
Title: Re: TrueTears question thread
Post by: TrueTears on March 09, 2009, 11:45:20 pm: ahh nice thanks man
Title: Re: TrueTears question thread
Post by: TrueTears on March 10, 2009, 12:01:28 am: and also for Q8, i've worked out $\overrightarrow {BC} = 2i-3j-4k$

so the shortest distance must be a straight line perpendicular to $\overrightarrow {OC}$ , but the question says "...between the POINT with coordinates..." How can you get something perpendicular to a point? Or maybe i'm just misunderstanding the question lol

EDIT: nvm i figured it out :)
Title: Re: TrueTears question thread
Post by: TrueTears on March 10, 2009, 04:55:36 pm: Ok just a simple question which I'm not sure about.

simplify:
$tan(sin^{-1}(\frac{-2}{3}))$

so let $sin^{-1}(\frac{-2}{3}) = x$

$sinx = \frac{-2}{3}$

using the identity $\frac{1}{sin^2x} = 1 + \frac{1}{tan^2x}$

solving for tanx yields $tanx = \frac{2\sqrt{5}}{5}$ or $-\frac{2\sqrt{5}}{5}$

But how do you know which value to take? because sin is negative in 3rd and 4th quadrants, but tan is positive in 3rd and negative in 4th. What value do you take for tanx? the +ve or -ve?
Thanks guys!
Title: Re: TrueTears question thread
Post by: GerrySly on March 10, 2009, 06:48:10 pm: Quote from: TrueTears on March 10, 2009, 04:55:36 pm
Ok just a simple question which I'm not sure about.

simplify:
$tan(sin^{-1}(\frac{-2}{3}))$

so let $sin^{-1}(\frac{-2}{3}) = x$

$sinx = \frac{-2}{3}$

using the identity $\frac{1}{sin^2x} = 1 + \frac{1}{tan^2x}$

solving for tanx yields $tanx = \frac{2\sqrt{5}}{5}$ or $-\frac{2\sqrt{5}}{5}$

But how do you know which value to take? because sin is negative in 3rd and 4th quadrants, but tan is positive in 3rd and negative in 4th. What value do you take for tanx? the +ve or -ve?
Thanks guys!

I would take a different approach and think about right angle triangles. sin is $\frac{O}{H}$ , so that means that O = 2, and H = 3, so therefore we can find

$\begin{align*}A &= \sqrt {3^2 - 2^2}\\&= \sqrt{5}\end{align*}$

Now we know that $\tan (\theta) = \frac{O}{H}$ , so therefore, $\tan (sin^{-1} (\frac{-2}{3})) = \frac{2}{\sqrt{5}}$

Just visualise the triangles, then you can derive any trigonometric function from that

(http://img11.imageshack.us/img11/1716/tringles.jpg)
Title: Re: TrueTears question thread
Post by: TrueTears on March 10, 2009, 06:52:20 pm: yeap, cool thanks gerrysly
Title: Re: TrueTears question thread
Post by: TrueTears on March 10, 2009, 07:50:11 pm: Also /0 why is AB = CD?

Quote from: /0 on March 09, 2009, 11:25:57 pm

Also, $|\vec{AB}| = |\vec{CD}|\Rightarrow |u| = |v|, \Rightarrow u \cdot u = v\cdot v$
Title: Re: TrueTears question thread
Post by: Mao on March 10, 2009, 08:06:49 pm: Quote from: TrueTears on March 10, 2009, 04:55:36 pm
Ok just a simple question which I'm not sure about.

simplify:
$tan(sin^{-1}(\frac{-2}{3}))$

so let $sin^{-1}(\frac{-2}{3}) = x$

$sinx = \frac{-2}{3}$

using the identity $\frac{1}{sin^2x} = 1 + \frac{1}{tan^2x}$

solving for tanx yields $tanx = \frac{2\sqrt{5}}{5}$ or $-\frac{2\sqrt{5}}{5}$

But how do you know which value to take? because sin is negative in 3rd and 4th quadrants, but tan is positive in 3rd and negative in 4th. What value do you take for tanx? the +ve or -ve?
Thanks guys!

Remembering that $x\in \left[ -\frac{\pi}{2},\frac{\pi}{2}\right]$ for arcsin, since you are taking an arcsin of a negative number, your answer will be in quadrant 4.

Hence, tan of that angle will be negative.

This also applies to GerrySly's response.
Title: Re: TrueTears question thread
Post by: TrueTears on March 10, 2009, 08:07:51 pm: yeah thanks mao
Title: Re: TrueTears question thread
Post by: Mao on March 10, 2009, 08:12:13 pm: Quote from: TrueTears on March 10, 2009, 07:50:11 pm
Also /0 why is AB = CD?

Quote from: /0 on March 09, 2009, 11:25:57 pm

Also, $|\vec{AB}| = |\vec{CD}|\Rightarrow |u| = |v|, \Rightarrow u \cdot u = v\cdot v$

not sure why he did that but it's incorrect.

$|AC| = |BC|\implies (u + v) \cdot (u + v) = v \cdot v \implies 2 u \cdot v = - u \cdot u$

$DA \cdot AB = (u + 2v) \cdot u = u \cdot u + 2 u \cdot v = u \cdot u - u \cdot u = 0$

$\implies DA \perp AB$
Title: Re: TrueTears question thread
Post by: TrueTears on March 10, 2009, 08:15:56 pm: Quote from: Mao on March 10, 2009, 08:12:13 pm
Quote from: TrueTears on March 10, 2009, 07:50:11 pm
Also /0 why is AB = CD?

Quote from: /0 on March 09, 2009, 11:25:57 pm

Also, $|\vec{AB}| = |\vec{CD}|\Rightarrow |u| = |v|, \Rightarrow u \cdot u = v\cdot v$

not sure why he did that but it's incorrect.

$|AC| = |BC|\implies (u + v) \cdot (u + v) = v \cdot v \implies 2 u \cdot v = - u \cdot u$

$DA \cdot AB = (u + 2v) \cdot u = u \cdot u + 2 u \cdot v = u \cdot u - u \cdot u = 0$

$\implies DA \perp AB$
yeah that's what i did XD
Title: Re: TrueTears question thread
Post by: TrueTears on March 10, 2009, 08:20:14 pm: $sin(t + g) = k \times cos(t - g)$

show that $tan(t) = \frac{k - tan(g)}{1 - k \times tan(g)}$
Title: Re: TrueTears question thread
Post by: Mao on March 10, 2009, 08:39:46 pm: $\sin (t+g) = k \cos (t-g)$

$\sin t \cos g + \sin g \cos t = k (\cos t \cos g + \sin t \sin g)$

$\sin t (\cos g - k \sin g) = \cos t (k\cos g - \sin g)$

$\frac{\sin t}{\cos t} = \frac{k \cos g - \sin g}{\cos g - k \sin g}$

$\tan t = \frac{k \cos g - \sin g}{\cos g - k\sin g} \times \frac{\sec g}{\sec g}$

$\tan t = \frac{k - \tan g}{1 - k\tan g}$

as required
Title: Re: TrueTears question thread
Post by: TrueTears on March 10, 2009, 08:42:57 pm: nice haha
Title: Re: TrueTears question thread
Post by: TrueTears on March 16, 2009, 06:57:21 pm: Just a quick question regarding complex conjugates

say you have $z = -7 + 3i$

now the conjugate here would be $-7 - 3i$ right?

but if you have $+7 + 3i$ , when this is multiplied with $z$ it also gives a real number

ie, $(-7+3i)(7+3i) = -49 - 21i + 21i + 9i^2 = -49-9 = -58$

however... the conjugate i always thought would just be -7-3i,

ie $(-7+3i)(-7-3i) = 49 +21i - 21i-9i^2 = 49+9 = 58$

Now I'm wondering, does this mean $z$ has 2 conjugates? I mean one yields a positive answer and the other is a negative, but they are still different numbers...

Thanks guys

EDIT: I always thought each complex number had only ONE conjugate?
Title: Re: TrueTears question thread
Post by: Damo17 on March 16, 2009, 07:10:12 pm: Quote from: TrueTears on March 16, 2009, 06:57:21 pm
Just a quick question regarding complex conjugates

say you have $z = -7 + 3i$

now the conjugate here would be $-7 - 3i$ right?

but if you have $+7 + 3i$ , when this is multiplied with $z$ it also gives a real number

ie, $(-7+3i)(7+3i) = -49 - 21i + 21i + 9i^2 = -49-9 = -58$

however... the conjugate i always thought would just be -7-3i,

ie $(-7+3i)(-7-3i) = 49 +21i - 21i-9i^2 = 49+9 = 58$

Now I'm wondering, does this mean $z$ has 2 conjugates? I mean one yields a positive answer and the other is a negative, but they are still different numbers...

Thanks guys

EDIT: I always thought each complex number had only ONE conjugate?

Well no, z can only have 1 conjugate as the conjugate signifies that there is "imaginary refection" meaning the same size but opposite angle.
(http://i232.photobucket.com/albums/ee101/stifla_2007/complex_conjugates.png)

Well this is what I think.
Title: Re: TrueTears question thread
Post by: TrueTears on March 16, 2009, 07:13:11 pm: ohh nice thanks Damo17, so basically a conjugate is just the imaginary part reflected in the x axis?

I always thought it was just you multiply the conjugate by the original complex number and as long as it yields a real number then it is a conjugate.

But now that you've cleared it up thanks :)
Title: Re: TrueTears question thread
Post by: Damo17 on March 16, 2009, 07:19:10 pm: Quote from: TrueTears on March 16, 2009, 07:13:11 pm
ohh nice thanks Damo17, so basically a conjugate is just the imaginary part reflected in the x axis?

I always thought it was just you multiply the conjugate by the original complex number and as long as it yields a real number then it is a conjugate.

But now that you've cleared it up thanks :)

No problem. That's right, the conjugate is just the imaginary part reflected in the x-axis. Remember if $z= a cis \theta$ then $\tilde{z}=a cis (-\theta)$
When you multiply a complex number by its complex conjugate, you always get a real positive number.
Title: Re: TrueTears question thread
Post by: Mao on March 16, 2009, 10:12:04 pm: the conjugate of -7+3i is -7-3i

if you happen to multiply it by -1:

$(-7+3i)\cdot (7+3i) = (-7+3i)\cdot (-1\cdot (-7-3i)) = -|-7+3i|^2$

it's not another conjugate, it's just the negative of the conjugate.
Title: Re: TrueTears question thread
Post by: TrueTears on March 16, 2009, 10:17:26 pm: Quote from: Mao on March 16, 2009, 10:12:04 pm
the conjugate of -7+3i is -7-3i

if you happen to multiply it by -1:

$(-7+3i)\cdot (7+3i) = (-7+3i)\cdot (-1\cdot (-7-3i)) = -|-7+3i|^2$

it's not another conjugate, it's just the negative of the conjugate.
oh i get it now, yeah factorising out that -1 haha
Title: Re: TrueTears question thread
Post by: Mao on March 16, 2009, 10:26:17 pm: A complex conjugate is defined such that:
$z \cdot \tilde{z} = |z|^2$

There is an infinite set of complex numbers that can multiply a complex number to get a real number (or positive real number), any scalar multiple works. (z=-7+3i, another vector could be 14+6i, or -21-9i, etc)
Title: Re: TrueTears question thread
Post by: Damo17 on March 16, 2009, 10:28:02 pm: And also, the reason why $z\tilde{z}$ cannot be negative is because:
$z \tilde{z} = (x + iy)(x - iy) = x^2 + iyx - ixy -(i^2)y^2 = x^2 + y^2$
Now as $x^2+y^2 = r^2$ , $z\tilde{z}$ has to always be positive.
Title: Re: TrueTears question thread
Post by: TrueTears on March 16, 2009, 10:34:42 pm: Thanks mao and damo appreciate all your help
Title: Re: TrueTears question thread
Post by: TrueTears on March 28, 2009, 06:26:12 pm: Find real values of 'a' for which 'ai' is a solution of the polynomial equation: $z^4 - 2z^3 +7z^2 -4z + 10 = 0$
Title: Re: TrueTears question thread
Post by: Over9000 on March 28, 2009, 06:41:16 pm: With that question, we are basically trying to find a value of z that makes the equation = 0
Let z =xi

$(xi)^{4} - 2(xi)^{3} + 7(xi)^{2} - 4xi + 10=0$
$x^{4} + 2x^{3}i - 7x^{2} - 4xi + 10 =0$
$(2x^{3}-4x)i=0$ and $(x^{4} - 7x^{2} +10)=0$
solve this $x = \pm{\sqrt{2}}$
Title: Re: TrueTears question thread
Post by: TrueTears on March 28, 2009, 06:42:44 pm: Quote from: Over9000 on March 28, 2009, 06:41:16 pm
With that question, we are basically trying to find a value of z that makes the equation = 0
Let z =xi

$(xi)^{4} - 2(xi)^{3} + 7(xi)^{2} - 4xi + 10=0$
$x^{4} + 2x^{3}i - 7x^{2} - 4xi + 10 =0$
$(2x^{3}-4x)i=0$ and $(x^{4} - 7x^{2} +10)=0$
solve this $x = \pm{\sqrt{2}}$
Oh yeah that was my thought process, but I expanded that $(xi)^{4} - 2(xi)^{3} + 7(xi)^{2} - 4xi + 10=0$ out wrong, damn!

Thanks >9000
Title: Re: TrueTears question thread
Post by: Over9000 on March 28, 2009, 06:43:40 pm: No probs, people make mistakes, thats what were here for ;)
Title: Re: TrueTears question thread
Post by: TrueTears on March 30, 2009, 04:44:36 pm: Let w = 2z. Describe the locus of w if z describes a circle with centre (1,2) and radius 3.

Basically, I did this question by inspection. w is just z doubled. so the new centre is (2,4) and radius is 6.

But how do you do this problem algebraically?
Title: Re: TrueTears question thread
Post by: kamil9876 on March 30, 2009, 06:28:43 pm: Write out an equation for z. and sub in z=w/2.

|z-1-2i|=3
|w/2 -1-2i|=3
and now multiply everything by 2 to get w by itself.
Title: Re: TrueTears question thread
Post by: TrueTears on March 30, 2009, 06:32:33 pm: Quote from: kamil9876 on March 30, 2009, 06:28:43 pm
Write out an equation for z. and sub in z=w/2.

|z-1-2i|=3
|w/2 -1-2i|=3
and now multiply everything by 2 to get w by itself.
nice thanks kamil.
Title: Re: TrueTears question thread
Post by: /0 on March 30, 2009, 07:41:00 pm: haha good solution kamil, much better than my solution
Title: Re: TrueTears question thread
Post by: kamil9876 on March 30, 2009, 08:04:03 pm: lol may i see it? curious
Title: Re: TrueTears question thread
Post by: TrueTears on March 30, 2009, 08:04:17 pm: lol just stomped on this Q over C

Factorise $x^4+x^3+x^2+x+1$
Title: Re: TrueTears question thread
Post by: Damo17 on March 30, 2009, 08:08:07 pm: Quote from: TrueTears on March 30, 2009, 08:04:17 pm
lol just stomped on this Q over C

Factorise $x^4+x^3+x^2+x+1$

http://vcenotes.com/forum/index.php/topic,10575.0.html
Title: Re: TrueTears question thread
Post by: TrueTears on March 30, 2009, 08:12:51 pm: thanks damo !
Title: Re: TrueTears question thread
Post by: TrueTears on April 07, 2009, 04:31:57 pm: Just this one question, bit stomped on it.

Thanks guys

(http://img22.imageshack.us/img22/333/questiont.jpg)
Title: Re: TrueTears question thread
Post by: kamil9876 on April 07, 2009, 08:36:30 pm: a good place to start is to start with dilations in x axis. The thing has a domain that is 6 units long. Whereas the original domain is 2pi/3 units long. SO the dilation mustve been 6/(2pi/3) and so the value of b is the reciprocal of that.

$b=\frac{\pi}{9}$

Due to the symetricity of the graph, the endpoints of the domain are the same. And they are simply sec(pi/3)=2. You want the endpoints to be at zero so you shift the graph 2 units down. Then you translate 3 units to the right. Then reflect in y axis and you've got the basic shape, all you have to do now is amplify (dilate in y axis to get a maximum of 4.
Title: Re: TrueTears question thread
Post by: TrueTears on May 01, 2009, 10:45:26 pm: thanks kamil for that

Now a very interesting question:

Implicit differentiate with respect to x $x^2-2xy+y^2 = 0$

$2x - 2(y + x\frac{dy}{dx}) + 2y\frac{dy}{dx} = 0$

solving for $\frac{dy}{dx}$ leads to $\frac{dy}{dx} = \frac{y-x}{y-x}$ $,$ $y \neq x$

HOWEVER the original equation can be simplified to $x^2-2xy+y^2 = 0 \implies (x-y)^2 = 0 \implies y = x$

yet the derivative says $y \neq x$ ?

Thanks guys!
Title: Re: TrueTears question thread
Post by: kamil9876 on May 01, 2009, 10:53:42 pm: in ur implicit differentiation u divided both sides by y-x, which is the same as dividing by 0.

Consider the argument:

$y=x$
$y-x=0$
divide both sides by y-x:

$\frac{y-x}{y-x}=0$ and hence y does not equal x.

The flaw is in dividing both sides by an unkown quantity without ensuring it was non-zero.

btw, if y=x u get 0/0 which isnt as bad as 1/0. If you ever encounter 0/0 u most likely did that by having a true statement such as ab=c (where a and c are zero) and divided both sides by a to get b by itself. Hence try to trackback to any division.
Title: Re: TrueTears question thread
Post by: TrueTears on May 01, 2009, 10:59:35 pm: Thanks for that, but how do I implicit differentiate it without dividing by 0?
Title: Re: TrueTears question thread
Post by: kamil9876 on May 01, 2009, 11:14:45 pm: just do a quick test to see if the thing u are dividing is not zero(that is generally true, not just implicit differentiation):

say:
$ e^x(x^2+2x-2)=e^x$

I can divide both sides by e^x because i know it is never 0.

However if we have:
$ (cos(x))(x^2+2x-2)=(cos(x))$
You must take into account the possibility that cos(x)=0 Hence do a seperate solution for the case cos(x)=0 and a seperate solution if it does not equal 0(where division is good).
Title: Re: TrueTears question thread
Post by: TrueTears on May 01, 2009, 11:22:52 pm: Thanks kamil and Mao, I understand it now.
Title: Re: TrueTears question thread
Post by: kamil9876 on May 01, 2009, 11:26:24 pm: Also, general rule: You cannot change the domain of a function by algebraically manipulating it:

consider y=x with domain R.
$ y=\frac{x^2}{x}$ has domain R/{0}

by writing $x \neq 0$ we havnt changed the domain of the original function, just said that our new form is only true for $x \neq 0$ and hence our new form is not the complete picture:

This is analogous to ur case. $y \neq x$ does not mean that u magically changed the domain of the function but that ur new form only allow values for which the condition $y \neq x$ holds.
Title: Re: TrueTears question thread
Post by: TrueTears on May 01, 2009, 11:28:02 pm: Quote from: kamil9876 on May 01, 2009, 11:26:24 pm
Also, general rule: You cannot change the domain of a function by algebraically manipulating it:

consider y=x with domain R.
$ y=\frac{x^2}{x}$ has domain R/{0}

by writing $x \neq 0$ we havnt changed the domain of the original function, just said that our new form is only true for $x \neq 0$ and hence our new form is not the complete picture:

This is analogous to ur case. $y \neq x$ does not mean that u magically changed the domain of the function but that ur new form only allow values for which the condition $y \neq x$ holds.
True, true. very true.

Thanks for explanation again! :coolsmiley:
Title: Re: TrueTears question thread
Post by: TrueTears on May 03, 2009, 11:35:22 pm: And also this question:

The equation of a curve C is $x^3+xy+2y^3 = k$ , where $k$ is a constant

a) find $\frac{dy}{dx}$ $\implies$ $\frac{dy}{dx} = \frac{-3x^2-y}{x+6y^2}$

C does have a tangent parallel to the y axis

b) show that the y coordinate at the point of contact satisfies $216y^6+4y^3+k = 0$

c) Hence show that $k \le \frac{1}{54}$

stomped on part c lol
Title: Re: TrueTears question thread
Post by: Mao on May 03, 2009, 11:44:21 pm: the equation in b is a quadratic, it has at least one solution (as per part a), hence the discriminant must be greater or equal to 0

$\implies \Delta \ge 0$
Title: Re: TrueTears question thread
Post by: TrueTears on May 03, 2009, 11:49:54 pm: Thanks Mao and kamil (MSN)
Title: Re: TrueTears question thread
Post by: TrueTears on May 03, 2009, 11:58:24 pm: $f : [\pi , \frac{3\pi}{2}] \to R , f(x) = sin(x)$

Find the derivative of $f^{-1}(x)$ with respect to $x$
Title: Re: TrueTears question thread
Post by: TrueTears on May 04, 2009, 12:23:11 am: Consider $f(x) = tan^{-1}(x) + tan^{-1}(\frac{1}{x})$ , $x \neq 0$

If x > 0 , find $f(x)$
Title: Re: TrueTears question thread
Post by: humph on May 04, 2009, 02:14:14 am: Quote from: TrueTears on May 04, 2009, 12:23:11 am
Consider $f(x) = tan^{-1}(x) + tan^{-1}(\frac{1}{x})$ , $x \neq 0$

If x > 0 , find $f(x)$
I remember this from one of my UMEP tests :)
So for $x \neq 0$ , we have that
$f'(x) = \frac{1}{x^2 + 1} + \frac{-1}{x^2} \frac{1}{(1/x)^2 + 1} = \frac{1}{x^2 + 1} - \frac{1}{1 + x^2} = 0$ .
Thus $f$ is constant. Taking $x=1$ gives
$f(1) = \arctan(1) + \arctan(1/1) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$
and so for all $x \neq 0$ ,
$f(x) = \frac{\pi}{2}$ .

Other interesting inverse trigonometric identities:
$ \begin{align*} \arcsin x + \arccos x & = \pi/2 , \qquad \mathrm{arccsc}\: x & = \arcsin(1/x) \\ \mathrm{arccsc}\: x + \mathrm{arcsec}\: x & = \pi/2 , \qquad \mathrm{arcsec}\: x & = \arccos(1/x) \\ \arctan x + \mathrm{arccot}\: x & = \pi/2 , \qquad \mathrm{arccot}\: x & = \arctan(1/x) \\ \end{align*} $
Title: Re: TrueTears question thread
Post by: TrueTears on May 04, 2009, 08:31:31 am: Thanks humph!
Title: Re: TrueTears question thread
Post by: kamil9876 on May 05, 2009, 01:55:05 pm: To do that without calculus:

since x>0 and arctan has domain $(-\frac{\pi}{2},\frac{\pi}{2})$ , $0<x<\frac{\pi}{2}$

Let $tan\theta=x$
$ tan(\frac{\pi}{2}-\theta)=\frac{1}{x}$ , this can be verified by constructing a right angle triangle with adjacent and opposite sides of $1$ and $x$ respecticely, and an angle of $\theta$

taking inverse tan of both equations and adding them gives QED.
Title: Re: TrueTears question thread
Post by: TrueTears on May 05, 2009, 04:08:13 pm: ^^^ thanks kamil!
Title: Re: TrueTears question thread
Post by: TrueTears on May 15, 2009, 04:23:23 pm: $\int cot^2(x) dx$

I don't think there is a way to integrate it without using $cot(x) = tan(\frac{\pi}{2} - x)$

What do you think?
Title: Re: TrueTears question thread
Post by: Mao on May 15, 2009, 10:00:22 pm: $\int \cot^2 (x)\; dx = \int \csc^2 (x) - 1\; dx = -\cot (x) - x + C$

remember that $\frac{d}{dx} \cot (x) = -\csc^2 (x)$

the relationship between cot/csc in many respects resembles tan/sec
Title: Re: TrueTears question thread
Post by: TrueTears on May 16, 2009, 01:01:20 pm: Thank you so much Mao
Title: Re: TrueTears question thread
Post by: TrueTears on May 17, 2009, 10:36:03 pm: $\int \sqrt{1-sin^2(x)} dx$
Title: Re: TrueTears question thread
Post by: pHysiX on May 17, 2009, 10:39:00 pm: 1-sin^2(x) = cos^2(x)

so it's just the integral of cos(x), which is sin(x) +c =]
Title: Re: TrueTears question thread
Post by: Over9000 on May 17, 2009, 10:42:12 pm: But its mod cos(x), I dont think you can do that
Title: Re: TrueTears question thread
Post by: TrueTears on May 17, 2009, 10:42:45 pm: Quote from: pHysiX on May 17, 2009, 10:39:00 pm
1-sin^2(x) = cos^2(x)

so it's just the integral of cos(x), which is sin(x) +c =]
yeap, but its |cosx| which is different from cosx
Title: Re: TrueTears question thread
Post by: pHysiX on May 17, 2009, 10:47:53 pm: hmmm. then i do stand corrected.

because what my reasoning is:

cos^2(x) is greater than or equal to zero for all x

hence the surd can work just like normal. but yup, you do get a +ve and -ve part.

back to the drawing boards. sorry guys =]

-----
if you do it by calc, it's sin(x).sgn(cos(x)) minus some chunk

the first part, the sgn thingo makes sense, but i'm trying to figure out the latter part =]
Title: Re: TrueTears question thread
Post by: kamil9876 on May 17, 2009, 11:22:57 pm: basically you can write it as.

$\int \pm cosx dx$

And so u get:

$\pm sinx +c$
it is +sinx+c if x is between 0,pi/2. -sinx if x is between pi/2 and pi, etc. (you have to look at the original cos)
-----------------------------------------------------------------------------
If it was a definite integral you would have to be more careful and split it up:

$\int_{0}^{pi} \sqrt{1-sin^2x}dx$
$=\int_{0}^{\frac{\pi}{2}} cosx + \int_{\frac{\pi}{2}}^{\pi} -cosx dx$

Basically a modulus sign indicates that it is a hybrid function, and so the actual function that you use (cosx, or -cosx) depends on the domain you are considering.
Title: Re: TrueTears question thread
Post by: Mao on May 17, 2009, 11:59:47 pm: A bit tricky, but here it is

$\frac{d}{dx} \sin(x)\cdot \mbox{sign}(\cos(x)) = \frac{d}{dx} \sqrt{\cos^2(x)} \tan(x) = -\sin(x)\cos(x)\tan(x) \frac{1}{\sqrt{\cos^2 (x)}} + \frac{\sqrt{\cos^2(x)}}{\cos^2(x)} = -\frac{\sin^2(x)}{|cos(x)|} + \frac{1}{|\cos(x)|} = |\cos(x)|$

[think of it as, if you take the area under the |cos(x)| graph, it's just like when you take the area under cos(x), except where it usually is negative, you slap another negative on top.

In general, $\frac{d}{dx} \mbox{sign}(f(x)) F(x) = 0 + \mbox{sign}(f(x))f(x) = |f(x)|$

However, if you are doing a definite integral with that, you need to count how many half-periods are between the two terminals, because f(x) in this case is an oscillating function. Hence you need the 'floor' function. The actual derivation is messy and requires a diagram.. it's getting a little too late for that :)

so I hereby slap the NOT IN COURSE LOL sticker on this question. :) good night.
Title: Re: TrueTears question thread
Post by: Mao on May 18, 2009, 12:04:13 am: More rigorous derivation of $\frac{d}{dx} \mbox{sign}(f(x)) F(x) = |f(x)|$

Given that $\frac{d}{dx} |f(x)| = \frac{d}{dx} \sqrt{f^2(x)} = \frac{f'(x)f(x)}{\sqrt{f^2(x)}} = f'(x)\frac{f(x)}{|f(x)|} = f'(x)\frac{|f(x)|}{f(x)}$

$\frac{d}{dx} \mbox{sign}(f(x)) F(x) = \frac{d}{dx} \frac{|f(x)|}{f(x)} F(x) = \frac{f'(x)\cdot \frac{|f(x)|}{f(x)}\cdot f(x) - f'(x) |f(x)|}{f^2(x)} F(x) + f(x) \frac{|f(x)|}{f(x)} = \mbox{sign}(f(x))f(x) = \mbox{sign}^2(f(x)) |f(x)| = |f(x)|$
Title: Re: TrueTears question thread
Post by: TrueTears on May 18, 2009, 12:19:20 am: Thank you Mao, kamil and physix!
Title: Re: TrueTears question thread
Post by: TrueTears on May 18, 2009, 05:31:09 pm: Another one:
$\int \frac{1}{x^3-1}dx$

thanks =]
Title: Re: TrueTears question thread
Post by: Over9000 on May 18, 2009, 06:26:31 pm: Lol, what spastic teacher gave you that
I shall help you by at least giving you the partial fractions, try and integrate them if not...... I will help

$\int\frac{1}{x^{3}-1}dx$
$x^{3} - 1 = (x-1)(x^{2} + x + 1)$ by factorising (x=1 makes it 0 then long divide...)
so $\int\frac{1}{x^{3}-1}dx$ = $\int\frac{1}{(x-1)(x^{2} + x + 1)}dx$
Now time for partial fractions
$\frac{1}{(x-1)(x^{2} + x + 1)} = \frac{A}{x-1} + \frac{Bx + c}{x^{2} + x + 1}$
$= \frac{A(x^{2} + x + 1) + (Bx + c)(x - 1)}{(x-1)(x^{2} + x + 1)}$
$\therefore 1 = A(x^{2} + x + 1) + (Bx + c)(x - 1)$
sub x = 1, 1= 3A
$\therefore A = \frac{1}{3}$
$1 = \frac{1}{3}(x^{2} + x + 1) + (Bx + c)(x - 1)$
$0 = \frac{1}{3} x^{2} + \frac{-2}{3}x + \frac{1}{3} + Bx^{2} - Bx + cx - c$ (expanding all out)
$0 = (B + \frac{1}{3})x^{2} + (\frac{1}{3} - B + c)x - (\frac{2}{3} + c)$
when x=0, $c= \frac{-2}{3}$
$\therefore$ $0 = (B + \frac{1}{3})x^{2} + (\frac{1}{3} - B + \frac{-2}{3})x - (\frac{2}{3} + \frac{-2}{3})$
$\therefore$ B = $\frac{-1}{3}$
so $\int\frac{1}{(x-1)(x^{2} + x + 1)}dx = \int\frac{1}{3(x-1)} + \frac{-x - 2}{3(x^{2} + x + 1)}dx$

EDIT:asked to do more

So to get the three integrals
$\int\frac{1}{3(x-1)} + \frac{-x - 2}{3(x^{2} + x + 1)}dx = \int\frac{-x-\frac{1}{2}}{3(x^{2} + x + 1)}dx$ (notice how this is just a simple log now?) + $\int\frac{-\frac{3}{2}}{3(x^{2} + x + 1)}dx$ $+ \int\frac{1}{3(x-1)}dx$
$\int\frac{-\frac{1}{2}(2x + 1)}{3(x^{2} + x + 1)}dx$ = $\int\frac{-(2x + 1)}{6(x^{2} + x + 1)}dx$ $\frac{dy}{dx} x^{2} + x + 1 = 2x + 1$
$\therefore \int\frac{-x-\frac{1}{2}}{3(x^{2} + x + 1)}dx = \frac{-1}{6}log_e(|x^{2} + x + 1|)dx$

Try the rest
Title: Re: TrueTears question thread
Post by: TrueTears on May 18, 2009, 07:01:03 pm: Quote from: Over9000 on May 18, 2009, 06:26:31 pm
Lol, what spastic teacher gave you that
I shall help you by at least giving you the partial fractions, try and integrate them if not...... I will help

$\int\frac{1}{x^{3}-1}dx$
$x^{3} - 1 = (x-1)(x^{2} + x + 1)$ by factorising (x=1 makes it 0 then long divide...)
so $\int\frac{1}{x^{3}-1}dx$ = $\int\frac{1}{(x-1)(x^{2} + x + 1)}dx$
Now time for partial fractions
$\frac{1}{(x-1)(x^{2} + x + 1)} = \frac{A}{x-1} + \frac{Bx + c}{x^{2} + x + 1}$
$= \frac{A(x^{2} + x + 1) + (Bx + c)(x - 1)}{(x-1)(x^{2} + x + 1)}$
$\therefore 1 = A(x^{2} + x + 1) + (Bx + c)(x - 1)$
sub x = 1, 1= 3A
$\therefore A = \frac{1}{3}$
$1 = \frac{1}{3}(x^{2} + x + 1) + (Bx + c)(x - 1)$
$0 = \frac{1}{3} x^{2} + \frac{-2}{3}x + \frac{1}{3} + Bx^{2} - Bx + cx - c$ (expanding all out)
$0 = (B + \frac{1}{3})x^{2} + (\frac{1}{3} - B + c)x - (\frac{2}{3} + c)$
when x=0, $c= \frac{-2}{3}$
$\therefore$ $0 = (B + \frac{1}{3})x^{2} + (\frac{1}{3} - B + \frac{-2}{3})x - (\frac{2}{3} + \frac{-2}{3})$
$\therefore$ B = $\frac{-1}{3}$
so $\int\frac{1}{(x-1)(x^{2} + x + 1)}dx = \int\frac{1}{3(x-1)} + \frac{-x - 2}{3(x^{2} + x + 1)}dx$

EDIT:asked to do more

So to get the three integrals
$\int\frac{1}{3(x-1)} + \frac{-x - 2}{3(x^{2} + x + 1)}dx = \int\frac{-x-\frac{1}{2}}{3(x^{2} + x + 1)}dx$ (notice how this is just a simple log now?) + $\int\frac{-\frac{3}{2}}{3(x^{2} + x + 1)}dx$ $+ \int\frac{1}{3(x-1)}dx$
$\int\frac{-\frac{1}{2}(2x + 1)}{3(x^{2} + x + 1)}dx$ = $\int\frac{-(2x + 1)}{6(x^{2} + x + 1)}dx$ $\frac{dy}{dx} x^{2} + x + 1 = 2x + 1$
$\therefore \int\frac{-x-\frac{1}{2}}{3(x^{2} + x + 1)}dx = \frac{-1}{6}log_e(|x^{2} + x + 1|)dx$

Thanks heaps over9000, genius.
Title: Re: TrueTears question thread
Post by: dcc on May 19, 2009, 10:00:22 pm: How about $\int_0^{\infty} \frac{\sin x}{e^x}\ dx$ (definite integrals are dead boring).
Title: Re: TrueTears question thread
Post by: humph on May 20, 2009, 12:34:15 am: Quote from: dcc on May 19, 2009, 10:00:22 pm
How about $\int_0^{\infty} \frac{\sin x}{e^x}\ dx$ (definite integrals are dead boring).

$ \int_0^{\infty} \frac{\sin x}{e^x}\: dx = \int_0^{\infty} e^{-x}\Im (e^{ix})\: dx = \Im \left(\int_0^{\infty}e^{(i-1)x}\: dx\right) = \Im \left(\left[\frac{1}{i-1}e^{(i-1)x}\right]^{\infty}_{0}\right) $
$ \Longrightarrow \int_0^{\infty} \frac{\sin x}{e^x}\: dx = \left[\Im \left(\frac{-1-i}{2} e^{-x}e^{ix}\right)\right]^{\infty}_{0} = -\frac{1}{2}\left[e^{-x}\Im \left((1+i)(\cos x + i \sin x)\right)\right]^{\infty}_{0} $
$ \Longrightarrow \int_0^{\infty} \frac{\sin x}{e^x}\: dx = -\frac{1}{2}\left[e^{-x}\left(\cos x + \sin x\right)\right]^{\infty}_{0} = -\frac{1}{2}\left(0 - 1\right) = \frac{1}{2}. $
Title: Re: TrueTears question thread
Post by: BlueYoHo on May 21, 2009, 09:42:18 pm: What the hell are you guys doing? I feel like I shouldn't even be doing spec...lol

EDIT: sorry for random, off-topic post :P
Title: Re: TrueTears question thread
Post by: TonyHem on May 21, 2009, 09:42:46 pm: it's not in spesh :l
Title: Re: TrueTears question thread
Post by: BlueYoHo on May 21, 2009, 09:51:14 pm: Nah but even the partial fractions stuff. It was like 20 lines. Intimidating.

We haven't started it yet though, so hopefully it's easier then it looks.
Title: Re: TrueTears question thread
Post by: TrueTears on May 21, 2009, 09:56:41 pm: Quote from: BlueYoHo on May 21, 2009, 09:51:14 pm
Nah but even the partial fractions stuff. It was like 20 lines. Intimidating.

We haven't started it yet though, so hopefully it's easier then it looks.
over9000 likes to making long workings :P
Title: Re: TrueTears question thread
Post by: Over9000 on May 21, 2009, 09:58:29 pm: I pride myself in making an easy 2 line question into 20 lines of working so it looks better.
Jks, it was just a long question.
Title: Re: TrueTears question thread
Post by: BlueYoHo on May 21, 2009, 09:59:44 pm: lol
Title: Re: TrueTears question thread
Post by: dejan91 on May 26, 2009, 12:34:30 am: Quote from: Over9000 on May 21, 2009, 09:58:29 pm
I pride myself in making an easy 2 line question into 20 lines of working so it looks better.

Haha true that :P
Title: Re: TrueTears question thread
Post by: TrueTears on June 12, 2009, 08:46:44 pm: The region bounded by the graph of $y = \frac{1}{\sqrt{x^2+9}}$ the x axis , the y axis and the line x = 4 is rotated about the y axis. Find the volume generated.

Thanks.
Title: Re: TrueTears question thread
Post by: hyperblade01 on June 12, 2009, 10:24:43 pm: My first attempt at using latex so I do apologise. Also may not be the best explanation as I haven't done specialist for a week and a half but I hope it helps:

$y = \frac{1}{\sqrt{x^2+9}}$
$\therefore x^2 = \frac{1}{y^2} - 9$
When $x=0\ y =\frac{1}{3}$
When $x=4\ y =\frac{1}{5}$

Taking into account the restriction of x=4 you have to break up the graph so find the volume between y=1/5 and y=1/3

$V = \pi\int^b_a x^2 dy$
$= \pi\int^\frac{1}{3}_\frac{1}{5} \frac{1}{y^2} - 9dy$
$= \frac{4\pi}{5}$

Between y = 0 and y = 1/5 and x = 4 there is a rectangle so to find the volume its just Area of circle X height

$V= \pi r^2 h$
$= \pi *(4)^2 * 1/5$
$= \frac {16\pi}{5}$

Add the two volumes and you get $4\pi$ cubic units
Title: Re: TrueTears question thread
Post by: TrueTears on June 12, 2009, 10:26:29 pm: Thanks, was just thinking about breaking it up right now :)
Title: Re: TrueTears question thread
Post by: TrueTears on June 12, 2009, 11:34:01 pm: If $A = \int_0^k cos(x^2)dx$ and $B = \int_0^k tan(x)dx$ where $0<k<\frac{\pi}{2}$ , then $\int_{-k}^k[cos(x^2) + tan(x)]dx$ is equal to? (In terms of A and B)
Title: Re: TrueTears question thread
Post by: kamil9876 on June 12, 2009, 11:50:35 pm: I remember doing this:

http://vcenotes.com/forum/index.php/topic,11962.msg149500.html#msg149500
Title: Re: TrueTears question thread
Post by: TrueTears on June 12, 2009, 11:54:36 pm: Thanks kamil but that has already been answered.
Title: Re: TrueTears question thread
Post by: kamil9876 on June 13, 2009, 12:01:41 am: oh right, and now the next question:

Quote from: TrueTears on June 12, 2009, 11:34:01 pm
If $A = \int_0^k cos(x^2)dx$ and $B = \int_0^k tan(x)dx$ where $0<k<\frac{\pi}{2}$ , then $\int_{-k}^k[cos(x^2) + tan(x)]dx$ is equal to? (In terms of A and B)
You should think of this in a geometrical-sort-of-way:

$\int_0^k cos(x^2) dx = \int_{-k}^0 cos(x^2)$ since the graph is symetrical about the y-axis

$\int_0^k tan(x)dx=-\int_{-k}^0 tan(x)dx$ because again symmetrical about the y-axis however flipped over. tan(x)=-tan(-x).

Therfore out integral:

$\int_{-k}^k[cos(x^2) + tan(x)]dx$
$=\int_{-k}^0 cos(x^2)dx + \int_0^{k} cos(x^2) + \int_{-k}^0 tan(x )dx + \int_0^k tan(x )dx$
$=A+A-B+B$
$=2A$
Title: Re: TrueTears question thread
Post by: TrueTears on June 13, 2009, 01:18:44 am: Awesome thanks.

The section of a straight line with equation $y = \frac{1}{2}x - 2$ between $x = 4$ and $x = 6$ is to be rotated about the y axis to form a container. What is the exact area of the water surface when the container is filled with $\frac{273\pi}{16}$ $unit^3$ of water?
Title: Re: TrueTears question thread
Post by: kamil9876 on June 13, 2009, 01:48:11 am: edit: did the problem for x-axis :P
Title: Re: TrueTears question thread
Post by: TrueTears on June 13, 2009, 02:01:10 am: Think you did it for rotated around x axis :P

but I got it now, thank you kamil!
Title: Re: TrueTears question thread
Post by: TrueTears on June 13, 2009, 02:59:55 pm: If $\frac{dy}{dx} = \sqrt{sinx}$ and y = 1 when $x = 0$ , then the value of y when $x = \frac{\pi}{3}$ can be found by evaluating?

Show all working.

Thanks.
Title: Re: TrueTears question thread
Post by: Winston on June 13, 2009, 03:05:50 pm: Hey Truetears. Which program do you use to write your equations with?
Cheers
Title: Re: TrueTears question thread
Post by: TrueTears on June 13, 2009, 03:13:20 pm: http://vcenotes.com/forum/index.php/topic,10280.0.html
Title: Re: TrueTears question thread
Post by: Winston on June 13, 2009, 03:17:36 pm: thanks
Title: Re: TrueTears question thread
Post by: Damo17 on June 13, 2009, 03:21:51 pm: Quote from: TrueTears on June 13, 2009, 02:59:55 pm
If $\frac{dy}{dx} = \sqrt{sinx}$ and y = 1 when $x = 0$ , then the value of y when $x = \frac{\pi}{3}$ can be found by evaluating?

Show all working.

Thanks.

NOTE: taking a stab at this.

$\frac{dy}{dx}= \sqrt{sinx}$

$\int \sqrt{sinx}= \int (sinx)^{\frac{1}{2}}= -\frac{cos^{\frac{3}{2}}x}{\frac{3}{2}} +c$
$= -\frac{2cos^{\frac{3}{2}}x}{3} +c$

$y=1$ when $x=0$

$1=-\frac{2cos^{\frac{3}{2}}0}{3} +c$
$1=-\frac{2}{3} +c$
$c=\frac{5}{3}$

when $x=\frac{\pi}{3}$

$y= -\frac{2cos^{\frac{3}{2}}(\frac{\pi}{3})}{3} + \frac{5}{3}$
$y=-\frac{2(\frac{\sqrt{2}}{4})}{3}+\frac{5}{3}$
$y=-\frac{\sqrt{2}}{6}+\frac{10}{6}$
$y=\frac{10-\sqrt{2}}{6}$
Title: Re: TrueTears question thread
Post by: TrueTears on June 13, 2009, 03:35:27 pm: Thanks for having a go but I don't think $\sqrt{sin(x)}$ is integrable XD

I'm looking for more of an expression which can be used to evaluate it. (If you know what I mean lols)
Title: Re: TrueTears question thread
Post by: Damo17 on June 13, 2009, 03:46:21 pm: Quote from: TrueTears on June 13, 2009, 03:35:27 pm
Thanks for having a go but I don't think $\sqrt{sin(x)}$ is integrable XD

I'm looking for more of an expression which can be used to evaluate it. (If you know what I mean lols)

Just thought I would give it a go.

Just curious as to where you got this question?
Title: Re: TrueTears question thread
Post by: TrueTears on June 13, 2009, 04:03:41 pm: NEAP topic test.
Title: Re: TrueTears question thread
Post by: Damo17 on June 13, 2009, 04:04:22 pm: Quote from: TrueTears on June 13, 2009, 04:03:41 pm
NEAP topic test.

Do you have the answer?
Title: Re: TrueTears question thread
Post by: TrueTears on June 13, 2009, 04:06:53 pm: No answers sorry :(
Title: Re: TrueTears question thread
Post by: Damo17 on June 13, 2009, 04:12:39 pm: Have a look here:

http://www.physicsforums.com/archive/index.php/t-8259.html%253Cbr%2520/t-83012.html
http://en.wikipedia.org/wiki/Elliptic_integral
Title: Re: TrueTears question thread
Post by: ed_saifa on June 13, 2009, 04:14:08 pm: Maybe this question requires Euler?
Title: Re: TrueTears question thread
Post by: Damo17 on June 13, 2009, 04:39:37 pm: this questions looks like it can be done only by calculator (in the spesh course).

From calc:
$fnInt(\sqrt{sinx},x,0,\frac{\pi}{3}) = 0.94$
Title: Re: TrueTears question thread
Post by: TrueTears on June 13, 2009, 04:46:26 pm: Alright cool thanks guys.
Title: Re: TrueTears question thread
Post by: TrueTears on June 14, 2009, 01:16:49 am: $\int \sqrt{9-x^2}dx$

Thanks.
Title: Re: TrueTears question thread
Post by: ryley on June 14, 2009, 01:34:18 am: I think if you make the substitution x = 3cos(theta) it should work.
Title: Re: TrueTears question thread
Post by: /0 on June 14, 2009, 02:03:30 am: Yeah or $x=3\sin{\theta}$ .
In general when you get stuff like $\sqrt{a^2 - x^2}$ you can substitute $x = a\sin{\theta}$ or $a\cos{\theta}$

$x = 3\sin{\theta} \Rightarrow dx = 3\cos{\theta}d\theta$

$\Rightarrow \int \sqrt{9-x^2}dx = \int \sqrt{9-(3\sin{\theta})^2}dx = \int 3\sqrt{1-\sin^2{\theta}}dx = \int 3\cos{\theta}dx$

$= \int (3\cos\theta)(3\cos\theta d\theta) =9 \int \cos^2\theta d\theta = 9\int \frac{\cos{2\theta}+1}{2}d\theta = \frac{9}{2}\int(\cos{2\theta}+1)d\theta = \frac{9}{2}(\frac{1}{2}\sin{2\theta}+\theta) = \frac{9}{4}\sin2\theta + \frac{9}{2}\theta +C$

$\theta = \sin^{-1}\left(\frac{x}{3}\right)$
$\sin{2\theta} = 2\cos\theta\sin\theta = 2\sqrt{1-\left(\frac{x}{3}\right)^2}(\frac{x}{3})=\frac{2}{9}x\sqrt{9-x^2}$

$\therefore \int \sqrt{9-x^2}dx = \frac{9}{4}\left(\frac{2}{9}x\sqrt{9-x^2}\right)+\frac{9}{2}\sin^{-1}\left(\frac{x}{3}\right)=\frac{1}{2}x\sqrt{9-x^2}+\frac{9}{2}\sin^{-1}\left(\frac{x}{3}\right)+C$

I don't think this is in the course though
Title: Re: TrueTears question thread
Post by: TrueTears on June 14, 2009, 02:07:24 am: Alright cool thanks ryley and /0.
Title: Re: TrueTears question thread
Post by: TrueTears on June 14, 2009, 02:19:00 am: Quote from: /0 on June 14, 2009, 02:03:30 am
Yeah or $x=3\sin{\theta}$ .
In general when you get stuff like $\sqrt{a^2 - x^2}$ you can substitute $x = a\sin{\theta}$ or $a\cos{\theta}$

$x = 3\sin{\theta} \Rightarrow dx = 3\cos{\theta}d\theta$

$\Rightarrow \int \sqrt{9-x^2}dx = \int \sqrt{9-(3\sin{\theta})^2}dx = \int 3\sqrt{1-\sin^2{\theta}}dx = \int 3\cos{\theta}dx$

$= \int (3\cos\theta)(3\cos\theta d\theta) =9 \int \cos^2\theta d\theta = 9\int \frac{\cos{2\theta}+1}{2}d\theta = \frac{9}{2}\int(\cos{2\theta}+1)d\theta = \frac{9}{2}(\frac{1}{2}\sin{2\theta}+\theta) = \frac{9}{4}\sin2\theta + \frac{9}{2}\theta +C$

$\theta = \sin^{-1}\left(\frac{x}{3}\right)$
$\sin{2\theta} = 2\cos\theta\sin\theta = 2\sqrt{1-\left(\frac{x}{3}\right)^2}(\frac{x}{3})=\frac{2}{9}x\sqrt{9-x^2}$

$\therefore \int \sqrt{9-x^2}dx = \frac{9}{4}\left(\frac{2}{9}x\sqrt{9-x^2}\right)+\frac{9}{2}\sin^{-1}\left(\frac{x}{3}\right)=\frac{1}{2}x\sqrt{9-x^2}+\frac{9}{2}\sin^{-1}\left(\frac{x}{3}\right)+C$

I don't think this is in the course though
Wait isn't $\sqrt{1-sin^2\theta} = \sqrt{cos^2\theta} = |cos\theta|$ ?

What about the $\sqrt{9} = \pm 3$ ? Discard the negative?
Title: Re: TrueTears question thread
Post by: /0 on June 14, 2009, 02:51:03 am: The function we are analysing is defined between $x=-3$ and $x=3$ . Therefore, $\theta$ is defined between $3\sin{\theta} = -3\Rightarrow \theta = -\frac{\pi}{2}$ and $3\sin{\theta}=3 \Rightarrow \theta = \frac{\pi}{2}$ .
For this domain, $\cos{\theta}$ is positive, so there is no need for absolute value signs or the negative part of the square root.
Title: Re: TrueTears question thread
Post by: TrueTears on June 14, 2009, 11:10:35 pm: thx

(http://img25.imageshack.us/img25/7615/speshhelp.jpg)

The shaded region shown in the diagram above is rotated around the x axis to form a solid of revolution. $g'(x) > 9$ and $g"(x) > 0$ for all $x \in [a,b]$ and the volume of the solid of revolution is V cubic units.

Which of the following is FALSE:

a) $V < \pi (g(b))^2(b-a)$
b) $V > \pi (g(a))^2(b-a)$
c) $V = \pi \int_a^b (g(x))^2 dx$
d) $V = \pi((G(b))^2 - (G(a))^2)$ , where $G'(x) = g(x)$
e) $V < \pi ((g(b))^2b - (g(a))^2a)$
Title: Re: TrueTears question thread
Post by: /0 on June 14, 2009, 11:55:35 pm: I say d) because while $\int_a^b g(x) dx = [G(x)]_a^b$ , this does not imply that $\int_a^b [g(x)]^2dx = [\ [G(x)]^2\ ]_a^b$
Title: Re: TrueTears question thread
Post by: TrueTears on June 14, 2009, 11:55:57 pm: Thanks I just got that as well.
Title: Re: TrueTears question thread
Post by: TrueTears on June 28, 2009, 01:26:21 pm: Quote from: kevinn on June 28, 2009, 01:08:46 pm
Hi guys, I have a mind blank
Can someone please do this for me :)
It's question 4g from Essential, exercise 7E

$\int sin(x)^2cos(x)^4$

Thanks :)
As a general rule next time please create another thread to post your questions to avoid ambiguity :)
Title: Re: TrueTears question thread
Post by: Mao on June 28, 2009, 11:42:52 pm: Quote from: Flaming_Arrow on June 28, 2009, 01:17:33 pm
Quote from: kevinn on June 28, 2009, 01:08:46 pm
Hi guys, I have a mind blank
Can someone please do this for me :)
It's question 4g from Essential, exercise 7E

$\int sin(x)^2cos(x)^4$

Thanks :)

$\int (1-\cos^2 x) cos ^4 (x)$

let u = cos x

i think u can do it from there

No, that won't do it, you will have a sin(x) term from substitution which will be difficult (but not impossible) to get rid of.

$\begin{align*} \int \sin^2(x)\cos^4(x)\; dx &= \int (\sin (x)\cos(x))^2 \cdot \cos^2(x)\; dx \\ &= \int \frac{\sin^2(2x)}{4}\cdot \frac{1+\cos(2x)}{2}\; dx \\ &= \frac{1}{8}\int \sin^2 (2x) + \sin^2(2x)\cos(2x)\; dx\\ &= \frac{1}{8}\left(\int \frac{1-\cos(4x)}{2}\; dx + \int \sin^2(2x)\cos(2x)\; dx\right)\\ &= \frac{1}{16}\left(x - \frac{\sin(4x)}{4}\right) + \frac{\sin^3(2x)}{48} + C \\ I & = -\frac{\sin(4x)}{64} + \frac{\sin^3(2x)}{48} + \frac{x}{16} + C\\ \end{align*}$

Combinations of $\sin(2\theta) = 2\sin(\theta)\cos(\theta),\; \cos^2(\theta) = \frac{1}{2}(1+\cos(2\theta)),\; \sin^2 (\theta)=\frac{1}{2}(1-\cos(2\theta))$ and the substitution method were used.
Title: Re: TrueTears question thread
Post by: Flaming_Arrow on June 28, 2009, 11:45:27 pm: Quote from: Mao on June 28, 2009, 11:42:52 pm
Quote from: Flaming_Arrow on June 28, 2009, 01:17:33 pm
Quote from: kevinn on June 28, 2009, 01:08:46 pm
Hi guys, I have a mind blank
Can someone please do this for me :)
It's question 4g from Essential, exercise 7E

$\int sin(x)^2cos(x)^4$

Thanks :)

$\int (1-\cos^2 x) cos ^4 (x)$

let u = cos x

i think u can do it from there

No, that won't do it, you will have a left over sin(x) term which will be difficult (but not impossible) to get rid of.

$\begin{align*} \int \sin^2(x)\cos^4(x)\; dx &= \int (\sin (x)\cos(x))^2 \cdot \cos^2(x)\; dx \\ &= \int \frac{\sin^2(2x)}{4}\cdot \frac{1+\cos(2x)}{2}\; dx \\ &= \frac{1}{8}\int \sin^2 (2x) + \sin^2(2x)\cos(2x)\; dx\\ &= \frac{1}{8}\left(\int \frac{1-\cos(4x)}{2}\; dx + \int \sin^2(2x)\cos(2x)\; dx\\ &= \frac{1}{16}\left(x - \frac{\sin(4x)}{4}\right) + \frac{\sin^3(2x)}{48} + C \\ I & = -\frac{\sin(4x)}{64} + \frac{\sin^3(2x)}{48} + \frac{x}{16} + C\\ \end{align*}$

oops didnt read the question prlly
Title: Re: TrueTears question thread
Post by: TrueTears on July 10, 2009, 02:56:04 am: Consider the 2nd order DE.

$a(\frac{d^2y}{dx^2}) + b(\frac{dy}{dx}) + cy = 0$ ...[A] where a,b,c $\in R$ and $a \neq 0$

$am^2 + bm + c = 0$ ...(B).

$y = e^{mx}$ is a solution to the above DE.

(B) could have 2 distinct real roots, say $m_1$ and $m_2$ . Without differentiating show that $y = C_1e^{m_1x} + C_2e^{m_2x}$ , where $C_1$ and $C_2$ are real numbers, is a general solution of [A] in this case.
Title: Re: TrueTears question thread
Post by: TrueTears on July 10, 2009, 04:26:41 am: Hmmm I did this:

Let $y_{1} = e^{m_{1}x} \mbox{and} y_{2}= e^{m_{2}x}$ both be soutions to the DE.

Then $a y''_{1} + b y'_{1} + cy_{1} = 0 \mbox{and} a y''_{2} + b y'_{2} + cy_{2} = 0$

Let $y = C_{1}y_{1} + C_{2}y_{2}$

$a (C_{1}y_{1} + C_{2}y_{2})'' + b ( C_{1}y_{1} + C_{2}y_{2})' + c (C_{1}y_{1} + C_{2}y_{2} )$

$= a (C_{1}y''_{1} + C_{2}y''_{2}) + b (C_{1}y'_{1} + C_{2}y'_{2}) + c(C_{1}y_{1} + C_{2}y_{2})$

$= C_{1}(ay''_{1} + by'_{1}+ cy_{1}) + C_{2}((ay''_{2} + by'_{2}+ cy_{2})$

$= C_{1}(0) + C_{2}(0) = 0$

Any other way to do it?
Title: Re: TrueTears question thread
Post by: TrueTears on July 10, 2009, 04:53:49 am: Next part of the question says:

(B) could also have 1 repeated root, say m.

Show that 2am+b=0

Do I just simply implicit differentiate both sides of equation (B) ?
Title: Re: TrueTears question thread
Post by: TrueTears on July 10, 2009, 05:06:19 am: Then finally "Hence without differentiating, show that $y = C_1e^{mx} + C_2xe^{mx}$ , where $C_1$ and $C_2$ are real numbers, is a general solution of (A)."

How to go about this?
Title: Re: TrueTears question thread
Post by: zzdfa on July 10, 2009, 10:56:21 am: Quote from: TrueTears on July 10, 2009, 04:53:49 am
Next part of the question says:

(B) could also have 1 repeated root, say m.

Show that 2am+b=0

Do I just simply implicit differentiate both sides of equation (B) ?

You get the correct answer but I don't think that's correct method, the correct reasoning is:
let y=(B)
then find dy/dx=2am+b
For a quadratic, If m has a repeated root, you know it means the turning point must coincide with the intersection of the x-axis.
Therefore solving
0=2am+b will give you the x-coord of the maximum/minimum which is also the root.

OR you could use the fact that the 2 roots of a quadratic equation are:
$\frac{-b\ + \sqrt{b^2-4ac}}{2a} \text{ and }\frac{-b\ - \sqrt{b^2-4ac}}{2a}$
And for them to be equal, the stuff under the square root has to equal 0. so you have
m=-b/2a, rearrange and youre done.

Quote from: TrueTears on July 10, 2009, 05:06:19 am
Then finally "Hence without differentiating, show that $y = C_1e^{mx} + C_2xe^{mx}$ , where $C_1$ and $C_2$ are real numbers, is a general solution of (A)."

How to go about this?

Well we know that when you sub the $C_1e^{mx}$ bit into [A] the equation equals to 0, so we can ignore that, since differentiation/integration is linear. and now im stuck. are you sure you gave us all the information?

coz atm this is what it's like:

$a y'' + b y' + cy = 0 \mbox{when} y=Ce^{mx}$
$2am+b=0 \mbox{when} am^2+bm+c=0 \mbox{has one solution}$
$\mbox{Therefore} a y'' + b y' + cy = 0 \mbox{when} y=Cxe^{mx}$ (where m=-b/2a)

and we're not allowed to differentiate.
Title: Re: TrueTears question thread
Post by: TrueTears on July 10, 2009, 03:01:46 pm: Thanks for your help.

Yeah that is all the information.

btw is my method for first part correct? I'm just wondering if there's any other method.
Title: Re: TrueTears question thread
Post by: zzdfa on July 10, 2009, 04:00:06 pm: yea that's the way i'd do it, basically using the fact that 0+0=0
Title: Re: TrueTears question thread
Post by: TrueTears on July 10, 2009, 04:02:31 pm: Yeah thanks.

Actually I think the question says given that $y = xe^{mx}$ is a solution of (A) [Part of the Q was cut off from the page]

Then show that $y = C_1e^{mx} + C_2xe^{mx}$ is a general solution of (A).

How would you approach it now?
Title: Re: TrueTears question thread
Post by: zzdfa on July 10, 2009, 04:08:39 pm: lol, then it's just the same as the first question, key thing is
http://en.wikipedia.org/wiki/Linearity_of_differentiation (try to ignore all the scary greek letters, you probably know what it is already)

If $xe^{mx}$ is a solution then $C_2xe^{mx}$ is a solution since $C_2*0 = 0$ . Then $y = C_1e^{mx} + C_2xe^{mx}$ is also a solution since $0+0=0$ . you can fill in the details.
Title: Re: TrueTears question thread
Post by: NE2000 on July 10, 2009, 04:10:18 pm: Quote from: TrueTears on July 10, 2009, 04:02:31 pm
Yeah thanks.

Actually I think the question says given that $y = xe^{mx}$ is a solution of (A) [Part of the Q was cut off from the page]

Then show that $y = C_1e^{mx} + C_2xe^{mx}$ is a general solution of (A).

How would you approach it now?

Won't you do it the same way as the first part or am I missing something :S

EDIT: zzdfa comes in before me :)
Title: Re: TrueTears question thread
Post by: TrueTears on July 10, 2009, 04:11:13 pm: Oh yeah thanks, so if you expand the double prime " into the brackets it would be $(xe^{mx})"$ ?
Title: Re: TrueTears question thread
Post by: TrueTears on July 10, 2009, 04:26:38 pm: Ah nvm I got it. Thanks all.
Title: Re: TrueTears question thread
Post by: TrueTears on July 10, 2009, 04:48:46 pm: What about if (B) had no real roots, and since the coefficients are real then 2 complex roots would be $v \pm wi$ [complex conjugate]. Using the fact that v + wi is a root of (B) show that $a(v^2-w^2) + bv + c = 0$

I just subbed m = v+wi into (B) but it doesn't work.
Title: Re: TrueTears question thread
Post by: zzdfa on July 10, 2009, 04:57:38 pm: sub in m=v+wi, and after some algebra you get

$a(v^2-w^2) + bv + c + wi(2av+b) =0$
thus
to show that $a(v^2-w^2) + bv + c =0$
all we need to do is to show that $2av+b=0$
look familiar? remember the real part of v+wi comes from the stuff outside the square root.
Title: Re: TrueTears question thread
Post by: TrueTears on July 10, 2009, 05:04:38 pm: True, 2av+b = 0 , how to show that?
Title: Re: TrueTears question thread
Post by: zzdfa on July 10, 2009, 05:05:58 pm: the 2 roots of a quadratic equation are:
$\frac{-b\ + \sqrt{b^2-4ac}}{2a} \text{ and }\frac{-b\ - \sqrt{b^2-4ac}}{2a}$

in both cases, $v=-\frac{b}{2a}$
Title: Re: TrueTears question thread
Post by: TrueTears on July 10, 2009, 05:07:03 pm: Oh I see, then what is the imaginary part 'w' of the quadratic equation?

Is it the square root itself or under the square root.
Title: Re: TrueTears question thread
Post by: zzdfa on July 10, 2009, 05:21:25 pm: i'm not sure what you're asking:
solution to a quadratic is:
$x=\frac{-b\ \pm \sqrt{b^2-4ac}}{2a}$
set
$v=-\frac{b}{2a}$
and if the stuff under sqrt is negative, you get an imaginary component with:
$wi=\pm \frac{\sqrt{b^2-4ac}}{2a}$
Title: Re: TrueTears question thread
Post by: TrueTears on July 10, 2009, 05:23:24 pm: Quote from: zzdfa on July 10, 2009, 05:21:25 pm
i'm not sure what you're asking:

$v=-\frac{b}{2a}$
and $wi=\pm \frac{\sqrt{b^2-4ac}}{2a}$

if the stuff under the sqrt is negative.
$wi=\pm \frac{\sqrt{b^2-4ac}}{2a}$ < yeah that was what I was asking.

Thanks I get it now. :)
Title: Re: TrueTears question thread
Post by: TrueTears on July 11, 2009, 05:15:29 am: Also just another Q, what if the coefficients in front of the y" and y' are not reals. Hence the quadratic you end up with does not have complex conjugates as the roots, would the general solution to homogeneous 2nd order DE still work?
Title: Re: TrueTears question thread
Post by: TrueTears on July 12, 2009, 04:49:30 am: ^^ I think it still works because even though the co-efficients are complex they still undergo the same algorithm as real numbers. ie i - i = 0 etc.

amirite?
Title: Re: TrueTears question thread
Post by: zzdfa on July 12, 2009, 11:33:39 am: i dont know, but just make up random complex coefficients for a b and c, get m, then differentiate and see if it works.
Title: Re: TrueTears question thread
Post by: TrueTears on July 13, 2009, 12:57:22 am: Thanks, just tried it now, general formula still holds.
Title: Re: TrueTears question thread
Post by: TrueTears on July 26, 2009, 06:42:11 pm: Just a stupid question lol, are we allowed to use say work, energy [physics related formulas] for the dynamics part of spesh?

If I get the right answer would I get penalised?
Title: Re: TrueTears question thread
Post by: mark_alec on July 26, 2009, 07:26:08 pm: Quote from: TrueTears on July 26, 2009, 06:42:11 pm
Just a stupid question lol, are we allowed to use say work, energy [physics related formulas] for the dynamics part of spesh?
If you look at most dynamics questions, you will see that they cannot be solved by energy considerations, but require solving the second order differential equation, a = F/m. You should always use the methods in the course to solve problems given, not others, as you may not receive credit for solutions using them.
Title: Re: TrueTears question thread
Post by: TrueTears on July 26, 2009, 10:30:51 pm: Thanks mark!
Title: Re: TrueTears question thread
Post by: TrueTears on July 26, 2009, 10:39:19 pm: 1. The graph of $y = cosec(a(x-b)) \mbox{for x} \in \left (-\frac{\pi}{4} , \frac{5\pi}{4} \right )$ is shown below:

(http://img170.imageshack.us/img170/8152/cosec.jpg)

The values of a and b could be?

A. $a = 2 , b = \frac{-\pi}{4}$
B. $a = 2 , b = \frac{\pi}{4}$

2. (http://img338.imageshack.us/img338/682/q7speshheffernan.jpg)

Thank you !
Title: Re: TrueTears question thread
Post by: kamil9876 on July 26, 2009, 11:02:09 pm: 1.)

You can see that at x=0, y=1.

meaning

1=cosec(-2b)

only option A solve the above.

2.)

C is a point, D and E are circles. only A and B seem plausible.

subbing in z=-i (which is part of the locus) we see:

B implies:

|-i|=|-i+1-i|
1=|1-2i|

But |1-2i|>1 hence a contradiction.

Leaving only A

==========================================================

Alternative/Proof that A is true (elimination is good so screw this on the exam but meh).

If you look at the origin and the point 1-i we are after the set of z such that the magnitude of the line connecting z to 1-i and z to 0 are equal. It can be justified that this line satisfies this condition by drawing a line from an arbitrary point on our locus going to 1-i and another one going from this arbitrary point all the way to 0. Then draw a line from 1-i to 0. What we have is an iscoseles triangle because our locus is the perpendicular bisector of the line connecting 1-i and 0. Hence the magnitudes of the two other lines are equal as required.
Title: Re: TrueTears question thread
Post by: TrueTears on July 26, 2009, 11:16:18 pm: Thanks kamil, I did Q 2 with elimination from the choices but I wanted to know a proof for it.

Thanks again.
Title: Re: TrueTears question thread
Post by: TrueTears on July 26, 2009, 11:25:47 pm: Ok for Q2, just out of interest sake, how would you go "backwards" from the cartesian equation to the complex locus?

Ie, the equation of the line is y = x - 1, how do you get from that to the complex locus?
Title: Re: TrueTears question thread
Post by: Mao on July 27, 2009, 09:39:26 am: That would usually involve a bit of ingenuity. Though knowing the geometric interpretations would make it seem fairly easy. In this case, the line bisects the line from (0,0) to (1,-1). Hence |z| = |z-1+i|

There are many different possible expressions. An equivalent form is |z+1| = |z-1+2i|
Title: Re: TrueTears question thread
Post by: kamil9876 on July 27, 2009, 12:32:09 pm: A good systematic way of doing it is to write down the line as a vector, in our case it can be:

$\vec{i} + \vec{j}$

Then find any vector perpendicular to this, ie just assign the x component some arbitrary real number, a, and use the dot product:

$\vec{b}=a\vec{i} - a\vec{j}$ where a is some real constant.

Now pick any point on the line in question, and 'add' to it the vector $\vec{b}$ to end up at some point A and the vector $-\vec{b}$ to end up at some point B. Now we can easily see that:

$|z-A|=|z-B|$ (where A and B are the two points but in complex number form)
Title: Re: TrueTears question thread
Post by: TrueTears on July 27, 2009, 08:57:48 pm: Thanks guys.

Just another one:

(http://img39.imageshack.us/img39/2247/q17heff.jpg)

I got the answer fairly easily, by elimination, A-D are all correct so I picked E. But I can not see how E is correct/incorrect? Can someone please explain why?

Thanks!
Title: Re: TrueTears question thread
Post by: kamil9876 on July 27, 2009, 09:08:04 pm: "angle between two vectors" means the angle formed when the two vectors are placed tail to tail. If you place the tail of a at the tail of b then you can see that the angle between the two vectors is in fact 90 degrees, hence cannot be theta.

In fact, another way to explain is this: you know that a.b=0 while $cos\theta \neq 0$ Since a triangle cannot have more than one right angle (it already has one at the point where the lines meet the circle). Therefore.

$|a||b|cos\theta \neq0$

Hence $|a||b|cos\theta \neq a.b$
Title: Re: TrueTears question thread
Post by: TrueTears on July 27, 2009, 09:21:09 pm: omfg, I thought $\theta$ meant the angle b/w vector a and b. FFS time to go suicide now, I can't read english for shit
Title: Re: TrueTears question thread
Post by: TrueTears on July 28, 2009, 04:49:43 pm: Okay now a really really interesting question heh

(http://img35.imageshack.us/img35/4493/weirdquestion.jpg)

Well first part is very easy to proof. Now second part.

Hence would imply using the previous part to solve the next part.

Now $\frac{sinx}{cosx-1} = 1$

$\frac{sinx}{1-cosx} = -1$

$cot(\frac{x}{2}) = -1$

$tan(\frac{x}{2}) = -1$ [I'm gonna leave out my steps for solving for x cause it's trivial]

Solving for x yields $x = \frac{3\pi}{2}$ for $x \in [0, 2\pi]$

Now let's use the "otherwise" method.

$cosx - sinx = 1$

$\frac{\sqrt{2}}{2}cos(x) - \frac{\sqrt{2}}{2}sin(x) = \frac{\sqrt{2}}{2}$

$cos(\frac{\pi}{4})cos(x) - sin(\frac{\pi}{4})sin(x) = \frac{\sqrt{2}}{2}$

$cos(\frac{\pi}{4} + x) = \frac{\sqrt{2}}{2}$

Solving for x yields $x = 0, \frac{3\pi}{2}, 2\pi$

My question is why does one method give only 1 solution and another method gives 3? Which is the right method? When I type to TI-89 calc to just solve $sin(x) = cos(x) - 1$ with restricted domain I get all 3 solutions. However the answer for this question is just $x = \frac{3\pi}{2}$

I know why the "hence" method only gives 1 solution is because when you divide both sides by $cos(x) - 1$ and if $x = 0$ then it is undefined, but it still does not explain why you can not have 0 and $2\pi$ as solutions because if you just have $sin(x) = cos(x) - 1$ then it clearly satisfies.

Ideas anyone?
Title: Re: TrueTears question thread
Post by: TrueTears on July 28, 2009, 05:07:00 pm: Alright and the very last very interesting question:

(http://img208.imageshack.us/img208/7959/weirdquestion2.jpg)

I've sketched everything etc, now from the information given $|\vec OA| = \lambda |\vec CB|$

There is no way you can get $\vec PQ$ in terms of $\lambda$ and $\vec OA$ because there is no relationship that links the $\vec OA$ with $\vec CB$ , the only information you have is the relationship linking the magnitudes of each but not the actual vectors? How do you go about this....?
Title: Re: TrueTears question thread
Post by: kamil9876 on July 28, 2009, 06:18:19 pm: First question:

It seems as though the otherwise method is correct, here is a simple example of how to miss a solution:

Solve this equation:
$ x^2=x$

Dividing both sides by x:

$x=1$

Where's the x=0? It's because by dividing both sides by x comes with the hidden assumption that $x \neq 0$ Hence all you have proven is that "if $x \neq 0$ then $x=1$ is the only solution". In order to complete the solution you have to now investigate the subset of possibilities that you discarded, and in this case it is: what if x=0?

Hence in your situation you have found the solution when $cosx-1 \neq 0$ . And so now you have to consider what happens when cosx-1=0 and see if any of the solutions coincide with sinx=0.

Hint: Good way to check if you've missed a solution in this fashion is to sub in the missed solution and see if you get something of the form $\frac{0}{0}$
Title: Re: TrueTears question thread
Post by: TrueTears on July 28, 2009, 06:33:25 pm: Ahh well explained, thanks kamil.

Any ideas on the vector Q?
Title: Re: TrueTears question thread
Post by: TrueTears on July 28, 2009, 07:39:04 pm: $18\pi b - 18sin(b \pi) = 9(7\pi + 2)$ without calc.

Find b.
Title: Re: TrueTears question thread
Post by: kamil9876 on July 28, 2009, 07:47:51 pm: I think vector question has a lack of info (which two sides are parralel). But assume OA and CB are parralel. Then assume the other two sides are parralel. If the two ans are the same then that is the ans, if they are not then there isn't enough info.

Hint for next question: Try to "equate coefficients".
Title: Re: TrueTears question thread
Post by: TrueTears on July 28, 2009, 08:05:38 pm: Yeah that is what I did but what do you treat as the variable? I mean I treated $\pi$ as the variable and obviously you probs did the same, but isn't \pi just another number? That means you're equating numbers to numbers?
Title: Re: TrueTears question thread
Post by: kamil9876 on July 28, 2009, 08:29:11 pm: All I'm doing is trying to find a solution.

If a+b=c+d and you can find an instance where a=b AND c=d then that is a solution.

You can prove that this is the only solution by drawing the graphs of $y=2sin(b\pi)$ and $y=\pi(2-7b)-2$ . We know that one intersection happens when the trig function is at a max (b=7/2). Because the line is an increasing one there is no other solution to b greater than 7/2. To prove that there is no solution less than 7/2 we just have to argue that the line has a steeper gradient than the sin grpah. This is because the sin graph only has the same gradient as the line at only one point between a zero and maximum. Hence the linear graph falls down quicker.

Meh that was ugly.
Title: Re: TrueTears question thread
Post by: TrueTears on July 28, 2009, 10:21:08 pm: If $|\vec a| = k |\vec b|$ for some constant k and the 2 vectors are parallel then does $\vec a = k \vec b$ also stand true if and only if the 2 vectors are parallel AND in the same direction?
Title: Re: TrueTears question thread
Post by: Mao on July 28, 2009, 10:38:13 pm: A trapezium is defined as a quadrilateral with two parallel (but not necessarily equal in length) sides.

There is one assumption though: OA || CB, in which case if you get the same trapezium, flip it and attach it to the right, you'll see that it forms a parallelogram, hence PQ + QP = OA + CB $\implies |PQ| = \frac{\lambda + 1}{2}\vec{a}$ , and the ratio is $\frac{|PQ|}{|OA|} = \frac{\lambda + 1}{2\lambda}$
Title: Re: TrueTears question thread
Post by: Mao on July 28, 2009, 10:45:02 pm: Quote from: TrueTears on July 28, 2009, 10:21:08 pm
If $|\vec a| = k |\vec b|$ for some constant k and the 2 vectors are parallel then does $\vec a = k \vec b$ also stand true if and only if the 2 vectors are parallel AND in the same direction?

if k can be negative, then they don't need to point in the same direction.

In general, for some constant k element of real, if $\vec{a} = k\vec{b}$ , then a and b are parallel, not necessarily pointing in the same direction.
Title: Re: TrueTears question thread
Post by: TrueTears on July 28, 2009, 11:02:39 pm: Hmm what if you got: a and b pointing in opposite directions.

$\vec a = i$

and $\vec b = -3i$

so $a = \frac{-1}{3} b$ [ $k = \frac{-1}{3}$ ]

but $|a| = \frac{1}{3}|b|$ , $k = \frac{1}{3}$ in this case, meaning they have to be in same direction for it to be true? Or am I completely wrong.
Title: Re: TrueTears question thread
Post by: Mao on July 28, 2009, 11:05:11 pm: oh, right, that's what you mean. if |a| = k|b|, then a = kb can only be satisfied when they are pointing in the same direction, since k is restricted to non-negative numbers.

but you generally wouldn't be dealing with that kind of condition.
Title: Re: TrueTears question thread
Post by: TrueTears on July 28, 2009, 11:06:58 pm: Ahhh ok, thanks Mao!
Title: Re: TrueTears question thread
Post by: TrueTears on July 29, 2009, 03:54:03 pm: How to change $x^3-21x^2+147x$ into cubic "turning point" form, ie $(x-7)^3 +343$ ?

Thanks!
Title: Re: TrueTears question thread
Post by: dcc on July 29, 2009, 04:12:20 pm: Such a form doesn't necessarily exist, however for this 'special case' you can do:
$x^3 - 3a x^2 + 3a^2 x = (x - a)^3 + a^3$ , where $a = 7$ in the above example.

EDIT:

And I must add, I don't really see a purpose to this form for a cubic. Only a subset of the cubics can be expressed in such a way, and it doesn't tell you much about anything at all (quite unlike the turning point form for a parabola, where you can immediately recognise the minimum value of the function and where it occurs).
Title: Re: TrueTears question thread
Post by: TrueTears on July 30, 2009, 07:05:49 pm: thanks dcc
Title: Re: TrueTears question thread
Post by: TrueTears on August 01, 2009, 07:15:20 pm: Just this question from an exam... I think answer is wrong can anyone show me what they get when they do this question?

Shade the required region of:

$y \le x (\sqrt{3} -2)$ n $y > x(-\sqrt{3} - 2)$ n $x^2+y^2 < \frac{1}{3}$

Many thanks!
Title: Re: TrueTears question thread
Post by: Damo17 on August 01, 2009, 07:38:44 pm: Quote from: TrueTears on August 01, 2009, 07:15:20 pm
Just this question from an exam... I think answer is wrong can anyone show me what they get when they do this question?

Shade the required region of:

$y \le x (\sqrt{3} -2)$ n $y > x(-\sqrt{3} - 2)$ n $x^2+y^2 < \frac{1}{3}$

Many thanks!

Haven't done these questions in a long time. I think it is the shaded area between all 3 graphs in the 4th quadrant.

(http://i232.photobucket.com/albums/ee101/stifla_2007/shadedregion.jpg)

The circle and the line with steep gradient are meant to be dotted.
Title: Re: TrueTears question thread
Post by: TrueTears on August 01, 2009, 08:24:33 pm: Yeah I got that as well, ok thanks Damo! Now I know the answers are wrong :)
Title: Re: TrueTears question thread
Post by: TrueTears on August 02, 2009, 02:52:54 pm: I did this question but I dono if my method is acceptable... anyways how would you guys attempt this question?

(http://img25.imageshack.us/img25/9746/speshquestion.jpg)
Title: Re: TrueTears question thread
Post by: kamil9876 on August 02, 2009, 03:15:17 pm: (1) $\frac{dx}{dt}=5$ so use chain rule after finding $\frac{dx}{d\theta}$ from implicit differentiation.

(2) After implicitly differentiating and finding $\frac{dx}{d\theta}$ , notice the equation $5t=x$ . Use the identity:

$\frac{dx}{d\theta}=\frac{d(5t)}{d\theta}=5\frac{dt}{d\theta}$ and reciprocalise etc.

method 2 only worked because it's a linear function, more complicated ones work better with the chain rule (method 1)
Title: Re: TrueTears question thread
Post by: TrueTears on August 02, 2009, 04:42:03 pm: Quote from: kamil9876 on August 02, 2009, 03:15:17 pm
(1) $\frac{dx}{dt}=5$ so use chain rule after finding $\frac{dx}{d\theta}$ from implicit differentiation.

(2) After implicitly differentiating and finding $\frac{dx}{d\theta}$ , notice the equation $5t=x$ . Use the identity:

$\frac{dx}{d\theta}=\frac{d(5t)}{d\theta}=5\frac{dt}{d\theta}$ and reciprocalise etc.

method 2 only worked because it's a linear function, more complicated ones work better with the chain rule (method 1)
True but that's not using "implicit differentiating" is it?

when you work out $\frac{d\theta}{dx}$ from the $tan(\theta) = \frac{40}{x}$ , that is not implicit differentiation, that is just differentiating with respect to x.
Title: Re: TrueTears question thread
Post by: kamil9876 on August 02, 2009, 04:50:06 pm: "Normal Differentiation" is a subset of implicit differentiation :P. Differentiating both sides by x is implicit differentiation in this case, after all you have to use the chain rule for both sides and all that shiz that normally comes with implicit diff.

But yes, if you were to rearrange first then it woudl be 'regular' differentiation lol.

$ \frac{d}{dx}(tan\theta)=\frac{d}{dx}(40x^{-1})$
$\frac{d\theta}{dx} \frac{d(tan\theta)}{d\theta}=\frac{-40}{x^2}$
$\frac{d\theta}{dx} sec^2 \theta=\frac{-40}{x^2}$
$\frac{d\theta}{dx}=cos^2 \theta \frac{-40}{x^2}$
Title: Re: TrueTears question thread
Post by: TrueTears on August 02, 2009, 05:13:16 pm: Thanks kamil!
Title: Re: TrueTears question thread
Post by: TrueTears on August 02, 2009, 05:21:12 pm: How about this innocent looking question but somewhere I get wrong answer...

Solve the differential equation $\frac{dy}{dx}= \frac{-y}{2}$ given that when x = 0 y = -1
Title: Re: TrueTears question thread
Post by: kamil9876 on August 02, 2009, 05:27:40 pm: $\frac{dx}{dy}=\frac{-2}{y}$
$x=-2 ln|y| +c$

Here's a fun way of getting rid of the modulus sign:

$x=-2 ln Ay$

$0=-2ln(-A) $
$\imples A=-1$

You probably forgot the modulus and thought that the function is +y inside the argument but really it has a - as shown by the negative A value.
Title: Re: TrueTears question thread
Post by: TrueTears on August 02, 2009, 05:40:19 pm: lol true true thanks!
Title: Re: TrueTears question thread
Post by: TrueTears on August 05, 2009, 05:39:00 pm: If a polynomial to the power of 5 has all real coefficients, then what can you say about the solutions?

A. 3 real 2 complex solutions.
B. 5 complex solutions.
C. 5 real solutions.
D. 3 complex 2 real solutions.

thanks
Title: Re: TrueTears question thread
Post by: ryley on August 05, 2009, 05:59:36 pm: As it has all real coefficients, the conjugate root theorem can be used. From the theorem, the conjugate of any roots must also be a root, so it will have to have an even number of complex roots, so I don't think it can be B or D. Now don't quote me on this, but I'm sure I've read that a polynomial of any degree will have some multiple of two LESS real roots, ie, degree eight, it will have 8, 6, 4, 2 or 0 real roots, so a fifth degree polynomial should have 5, 3 or 1 real root. From this, I think both A and C are possible.

EDIT:
I think I read that last bit here, http://fym.la.asu.edu/~tturner/MAT_117_online/Polynomials_and_their_Functions/polynomial_functions.htm
Title: Re: TrueTears question thread
Post by: TrueTears on August 05, 2009, 06:19:04 pm: Hmm yeah thanks ryley I　was thinking between A and C as well but answer says it's A, but I don't know why...
Title: Re: TrueTears question thread
Post by: ryley on August 05, 2009, 06:31:11 pm: Off the top of my head, the only reason I can think of it not being C is how the roots need to be evenly spaced in an argand diagram, ~~can you do that for 5 real solutions~~?

EDIT: nvm, forgot that the solutions don't have to be unique
Title: Re: TrueTears question thread
Post by: kamil9876 on August 05, 2009, 06:36:47 pm: Quote
Now don't quote me on this, but I'm sure I've read that a polynomial of any degree will have some multiple of two LESS real roots, ie, degree eight, it will have 8, 6, 4, 2 or 0 real roots, so a fifth degree polynomial should have 5, 3 or 1 real root. From this, I think both A and C are possible.

That is true. let n be the degree. The number of solutions, including multiplicities(ie getting same factor twice(squared)) will be n=R+C where R is the real solution and C is the non-real solution. Because C must be a multiple of two since they come in pairs we can write it as $2k$ where is any non-negative integer:

$n=R+2k$
$R=n-2k for k=0,1,2,3....$ as long as $R \geq 0$ , which is what you claim.

Lulz just realised I quoted against your wishes :P
Title: Re: TrueTears question thread
Post by: kamil9876 on August 05, 2009, 06:46:34 pm: Quote from: ryley on August 05, 2009, 06:31:11 pm
the roots need to be evenly spaced in an argand diagram?

Sorry for quoting again :P

This is only true in the equation of the form $z^n - z_o=0$ where $z_0$ is some complex(possibly real)number. However if the polynomial had terms of other degree in there then the evenly spaced thing need not be true.

The statement concerning evenly spread out on argand diagram is derived from assuming it's of the form $z^n - z_o=0$ (you can derive this by writing it $z_0=rcis\theta$ and notice that:

$z^n=rcis(\theta) , r cis(\theta + 2\pi) , rcis(\theta + 2\pi + 2\pi).....$
$ z=r^{1/n} cis\frac{\theta}{n} , r^{1/n} cis(\frac{\theta}{n} + \frac{2\pi}{n}) , r^{1/n} cis(\frac{\theta}{n} + \frac{2\pi}{n} + \frac{2\pi}{n})$

Notice how the argument increases by $\frac{2\pi}{n}$ each time. This is why the numbers are all seperated by same angle.

The above reasoning cannot be applied to other kinds of polynomials.
Title: Re: TrueTears question thread
Post by: ryley on August 05, 2009, 06:55:25 pm: Thanks for correcting that kamil, I was just playing around on the CAS and I realised how fucked up what I said was, can't believe (and not sure why) I thought that extended to all polynomials.
Title: Re: TrueTears question thread
Post by: TrueTears on August 08, 2009, 01:16:18 am: Hi here's a question which I think the answer is wrong but can someone please confirm? Thanks!

(http://img257.imageshack.us/img257/2650/speshq2.jpg)
(http://img15.imageshack.us/img15/6353/speshq3.jpg)

You don't need to know part a) - c) to do d).

So d) clearly says the FORCE is inversely proportional which means $F \propto \frac{1}{v^2} \implies F = \frac{-k}{v^2}$ where $k \in R^+$

Now using Newton's second law: $F = ma \implies F = \frac{-k}{v^2} = 60(a) \implies a = \frac{-k}{60v^2}$

Here's the answers:

(http://img139.imageshack.us/img139/373/speshq.jpg)

Why do did they leave out the mass Michael? Are they wrong?
Title: Re: TrueTears question thread
Post by: TrueTears on August 08, 2009, 01:40:31 am: Also another question, how do you do it?

(http://img186.imageshack.us/img186/3746/speshq4.jpg)
(http://img41.imageshack.us/img41/9313/speshq5.jpg)

Thankee
Title: Re: TrueTears question thread
Post by: TrueTears on August 08, 2009, 01:56:02 am: Last one:

Part e) is what I'm having trouble with, I can do part e) 2 different ways but they give an answer which is 0.2 off each other but I can't see a flaw in either of the 2 methods so can someone please check?

(http://img136.imageshack.us/img136/6753/speshq6.jpg)

Very simple $r(t) = (15t(t+2)) i + (10t(t-2)) j$

(http://img196.imageshack.us/img196/9703/speshq7.jpg)

d) $\vec{AP} = \vec{OP} - \vec{OA} = (15t(t+2) -500) i + (10t(t-2) - 450) j$

Now part e).
First method: Closest to the plane is when the dot product of $\vec{AP}$ and $\vec{OP} = 0$ [Straight line is the fastest route]

$\therefore (15t(t+2) -500)(15t(t+2)) + (10t(t-2) - 450)(10t(t-2)) = 0$

Solve for t using a calc yields t = 5.31 min (2 dp)

Second method: Closest to the plane is when the magnitude of $\vec{AP}$ is a minimum, thus whatever is under the square root must be a minimum.

$\therefore \sqrt{(15t(t+2) -500)^2 + (10t(t-2) - 450)^2}$ is a minimum when $(15t(t+2) -500)^2 + (10t(t-2) - 450)^2$ is a minimum.

So, $\frac{d}{dt}((15t(t+2) -500)^2 + (10t(t-2) - 450)^2)$ and set to 0 and solve for t yields t = 5.50 min (2 dp)

Now question says to nearest minute, well method one would be 5 min and method 2 would be 6 min. But both ways seem correct to me? Why is there a discrepancy between the 2 methods?
Title: Re: TrueTears question thread
Post by: kamil9876 on August 08, 2009, 12:16:40 pm: Quote
Now part e).
First method: Closest to the plane is when the dot product of $\vec{AP}$ and $\vec{OP}$ = 0 [Straight line is the fastest route]

Not true. When he is at the point closest to A the direction he is facing is perpendicular to AP (you also need to test endpoints of domain because they could be even less(beginning/ end of journey), but I assume you have ruled them out).

Hence you need to find the dot product between AP and his direction vector (velocity vector).
Title: Re: TrueTears question thread
Post by: TrueTears on August 08, 2009, 01:19:25 pm: Quote from: kamil9876 on August 08, 2009, 12:16:40 pm
Quote
Now part e).
First method: Closest to the plane is when the dot product of $\vec{AP}$ and $\vec{OP}$ = 0 [Straight line is the fastest route]

Not true. When he is at the point closest to A the direction he is facing is perpendicular to AP (you also need to test endpoints of domain because they could be even less(beginning/ end of journey), but I assume you have ruled them out).

Hence you need to find the dot product between AP and his direction vector (velocity vector).
I have no idea what you're talking about.
Title: Re: TrueTears question thread
Post by: kamil9876 on August 08, 2009, 01:38:28 pm: Sorry just realised my method is the same as yours(because the cartesian equation is a straightline). My method is just more general for any curve.
Title: Re: TrueTears question thread
Post by: TrueTears on August 08, 2009, 01:42:44 pm: Quote from: kamil9876 on August 08, 2009, 01:38:28 pm
Sorry just realised my method is the same as yours(because the cartesian equation is a straightline). My method is just more general for any curve.
Yes that's what I thought... so... what's the wrong thing in the 2 methods...?
Title: Re: TrueTears question thread
Post by: kamil9876 on August 08, 2009, 01:55:43 pm: on second thought. The cartesian equation of OP is not a straight line:

$\frac{y}{x}=\frac{3}{2}\frac{t+2}{t-2}=\frac{3}{2}\((1+\frac{4}{t+2})$

But the above must be constnat for a striaght line going through origin. Hence the thing is not a straight line so method1 doesn't work. THe more general variation of it should though.
Title: Re: TrueTears question thread
Post by: TrueTears on August 08, 2009, 02:00:14 pm: Quote from: kamil9876 on August 08, 2009, 01:55:43 pm
on second thought. The cartesian equation of OP is not a straight line:

$\frac{y}{x}=\frac{3}{2}\frac{t+2}{t-2}=\frac{3}{2}\((1+\frac{4}{t+2})$

But the above must be constnat for a striaght line going through origin. Hence the thing is not a straight line so method1 doesn't work. THe more general variation of it should though.
I think you did x/y there :P but anyways I think I get it now...

So how about finding the cartesian equation and making sure it's not a line?
Title: Re: TrueTears question thread
Post by: TrueTears on August 08, 2009, 04:24:49 pm: For any $n \ge 3$ , the complex roots of the equation $z^n = 1$ are vertices of a polygon what perimeter?
Title: Re: TrueTears question thread
Post by: /0 on August 08, 2009, 06:16:35 pm: For all $n\geq 3$ you get regular n-gons with circumradius 1.

If we put a point at (0,0) and draw lines connecting this point to each of the edges of the polygon, we can form $n$ isosceles triangles.

The inside angle of any of these isosceles triangles is $\frac{2\pi}{n}$ radians, and the isosceles legs of the triangle both have length 1, so we can use the cosine rule to find the outside edge of these triangles:

$x = \sqrt{1^2+1^2-2(1)(1)\cos{\frac{2\pi}{n}}}=\sqrt{2-2\cos{\frac{2\pi}{n}}}=\sqrt{2\left(1-\cos\frac{2\pi}{n}\right)}=\sqrt{2\left(1-\left(1-2\sin^2{\frac{\pi}{n}}\right)\right)}=\sqrt{4\sin^2{\frac{\pi}{n}}}=2|\sin\frac{\pi}{n}|=2\sin\frac{\pi}{n}$ $(0< \frac{\pi}{n} \leq \pi)$

But we need to multiply this by n to get the perimeter, so $P = 2n\sin{\frac{\pi}{n}}$

Check: If $n \to \infty$ we get a circle

$\lim_{n \to \infty} P = \lim_{n \to \infty} 2\pi \frac{\sin{\frac{\pi}{n}}}{\frac{\pi}{n}}$

Letting $m = \frac{\pi}{n}$ , $\lim_{n \to \infty} P = \lim_{m \to 0 } 2\pi \frac{\sin{m}}{m} = 2\pi$ ( $\lim_{x \to 0}\frac{\sin{x}}{x}=1$ )
Title: Re: TrueTears question thread
Post by: TrueTears on August 08, 2009, 06:28:32 pm: How do we form the isosceles triangle?

And how do you know they lie on a unit circle?
Title: Re: TrueTears question thread
Post by: NE2000 on August 08, 2009, 06:48:54 pm: Quote from: TrueTears on August 08, 2009, 06:28:32 pm
How do we form the isosceles triangle?

And how do you know they lie on a unit circle?

A drawing would help in this case, but basically a regular polygon can be split into n isosceles triangles (where n is the number of sides) with one vertex of the triangle in the centre and two sides leading out to the corners of the polygon. The distance between the centre and any two corners would be the same. So looking at the centre of the polygon, you'll have lines converging into the middle. There are 360 degrees in the middle and you split that into n to find the angle for each of the isosceles triangles.

We know they lie on a unit circle because z^n = 1 because if you consider the general solutions of that you get $r^{n}cis(n \theta ) = 1$ and therefore r = 1 for all of them. The solutions form a polygon because all of the different solutions plotted on the complex plane can be joined with lines that make a polygon, but each of these solutions (which are vertices in the polygon) have a modulus of 1. Is that what you were asking about?

You can plot the polygon on the complex plane and it will make more sense:
centre is the origin
each vertex will be a point a distance of 1 away from the origin
each vertex will be separated from another vertex by an angle that is $\frac{2 \pi}{n}$
Title: Re: TrueTears question thread
Post by: TrueTears on August 08, 2009, 07:00:36 pm: Thanks.
Title: Re: TrueTears question thread
Post by: TrueTears on August 09, 2009, 05:30:15 pm: (http://img20.imageshack.us/img20/4093/speshq9.jpg)

Thanks!
Title: Re: TrueTears question thread
Post by: TonyHem on August 09, 2009, 05:40:14 pm: is it a?
Title: Re: TrueTears question thread
Post by: TrueTears on August 09, 2009, 05:41:37 pm: Quote from: TonyHem on August 09, 2009, 05:40:14 pm
is it a?
Nope answer is C.
Title: Re: TrueTears question thread
Post by: Flaming_Arrow on August 09, 2009, 05:44:40 pm: Quote from: TonyHem on August 09, 2009, 05:40:14 pm
is it a?

thats what i got
Title: Re: TrueTears question thread
Post by: TrueTears on August 09, 2009, 05:45:21 pm: I got C but I left out the information about a = 1.

I just did u = 0, a = -9.8 and t = 8 =S
Title: Re: TrueTears question thread
Post by: TonyHem on August 09, 2009, 05:54:57 pm: Eh i see why mine is wrong =/
Title: Re: TrueTears question thread
Post by: kamil9876 on August 09, 2009, 05:55:41 pm: $h=-0.5g (8^2) + 8u$ and there is nothing that the acceleration of the balloon can tell you about the initial velocity of the balloon, which is also the initial velocity of the stone. I don't think this has a solution.
Title: Re: TrueTears question thread
Post by: TrueTears on August 09, 2009, 05:56:43 pm: Yeah, because if the hot air balloon is accelerating... then the initial velocity of the stone can't be 0... but you can't find out the initial velocity of stone...
Title: Re: TrueTears question thread
Post by: /0 on August 09, 2009, 05:58:05 pm: I get something way off:
Assuming the balloon starts at rest, for the upwards journey,
$v^2 = u^2+2ax \Rightarrow v = \sqrt{(0)+2(1)(h)}=\sqrt{2h}$

And for the fall,
$-h = \sqrt{2h}(8)-\frac{1}{2}(9.8)(8^2)$

$\therefore h= 167$
Title: Re: TrueTears question thread
Post by: TrueTears on August 09, 2009, 06:02:19 pm: So, can we come up with a mutual answer??
Title: Re: TrueTears question thread
Post by: kamil9876 on August 09, 2009, 06:05:23 pm: or maybe, because its hot air balloon(thing constructed by humans ussually at h=0) we can assume that it accelerated to a height of h wiht a constant acceleration of 1m/s^2 up and started at rest(since that's how humans ussually do it).

edit: oh wait that's what /0 did i think.
Title: Re: TrueTears question thread
Post by: TrueTears on August 09, 2009, 06:22:43 pm: hmmm ok.

What about this question:

Simplify: $\frac{b^2 + b \sqrt{a^2+b^2}}{a^2+b^2+b \sqrt{a^2+b^2}}$
Title: Re: TrueTears question thread
Post by: Flaming_Arrow on August 09, 2009, 06:38:56 pm: $\frac{b^2 + b \sqrt{a^2+b^2}}{a^2+b^2+b \sqrt{a^2+b^2}}$

$\frac{b(b + \sqrt{a^2+b^2})}{a^2+b^2+b \sqrt{a^2+b^2}}$

$\frac{b(b + \sqrt{a^2+b^2})}{(b + \sqrt{a^2+b^2})\sqrt{a^2+b^2)}}$

$\implies \frac{b}{\sqrt{a^2+b^2}}$
Title: Re: TrueTears question thread
Post by: TrueTears on August 09, 2009, 06:41:05 pm: Quote from: Flaming_Arrow on August 09, 2009, 06:38:56 pm
$\frac{b^2 + b \sqrt{a^2+b^2}}{a^2+b^2+b \sqrt{a^2+b^2}}$

$\frac{b(b + \sqrt{a^2+b^2})}{a^2+b^2+b \sqrt{a^2+b^2}}$

$\frac{b(b + \sqrt{a^2+b^2})}{(b + \sqrt{a^2+b^2})\sqrt{a^2+b^2)}}$

$\implies \frac{b}{\sqrt{a^2+b^2}}$

PROOOO
Title: Re: TrueTears question thread
Post by: TrueTears on August 10, 2009, 06:07:31 pm: (http://img177.imageshack.us/img177/9858/speshq10.jpg)

Thanks!!
Title: Re: TrueTears question thread
Post by: TonyHem on August 10, 2009, 06:15:22 pm: I guess C
Title: Re: TrueTears question thread
Post by: TrueTears on August 10, 2009, 06:15:55 pm: Quote from: TonyHem on August 10, 2009, 06:15:22 pm
I guess C
Answer says B =S
Title: Re: TrueTears question thread
Post by: /0 on August 10, 2009, 06:19:27 pm: I get C.
The reactants react in equal parts, so each will produce $\frac{x}{2}$ of product.
So the amount remaining for each of them will be $\left(a-\frac{x}{2}\right)$ and $\left(b-\frac{x}{2}\right)$
$\frac{dx}{dt} \propto \left(a-\frac{x}{2}\right)\left(b-\frac{x}{2}\right)$
Also, there can't be a negative sign in front of the k because if you make $x=0$ then that would imply a back reaction, despite there being no product.
Title: Re: TrueTears question thread
Post by: TrueTears on August 10, 2009, 06:21:10 pm: As I suspected...

I hereby declare Kilbaha answers are the shittest, so many are wrong wdf
Title: Re: TrueTears question thread
Post by: /0 on August 10, 2009, 06:22:31 pm: Quote from: TrueTears on August 10, 2009, 06:21:10 pm
As I suspected...

I hereby declare Kilbaha answers are the shittest, so many are wrong wdf

i agree
Title: Re: TrueTears question thread
Post by: kamil9876 on August 10, 2009, 06:22:51 pm: B defies the conservation of mass :P

I go with C too.
Title: Re: TrueTears question thread
Post by: TrueTears on August 10, 2009, 06:24:01 pm: Thanks guys!
Title: Re: TrueTears question thread
Post by: Mao on August 10, 2009, 08:01:56 pm: 'the velocity of the reaction'..... good job Kilbaha.
Title: Re: TrueTears question thread
Post by: TrueTears on August 13, 2009, 09:46:09 pm: (http://img148.imageshack.us/img148/5360/speshassignment.jpg)

For Q2

What kind of scale should I have on my x and y axis? The scale I get I always get a shitty looking graph when I do it on graph paper...

Any ideas?

Thanks!
Title: Re: TrueTears question thread
Post by: TrueTears on August 13, 2009, 11:08:48 pm: Hi can someone check my working for solving for the general solution for the following system of differential equation:

Did I get the right answer?

[I just started learning this so that's I want to check because this has no answers]

$ \begin{cases} \frac{dx}{dt} = 7x-4y \\ \frac{dy}{dt} = -9x+7y \end{cases}$

(http://img141.imageshack.us/img141/9162/speshassignment4.jpg)
Title: Re: TrueTears question thread
Post by: Mao on August 13, 2009, 11:19:00 pm: hey man, can you maybe use a smaller picture next time...? It's so big I can't be bothered scrolling across to read it...
Use Microsoft picture manager or something to resize to 600pixels wide or something.

as for the second post, you can check whether it is right by substituting back into the original differential equations.
Title: Re: TrueTears question thread
Post by: TrueTears on August 13, 2009, 11:21:13 pm: lol sorry im a noob at computers :(
Title: Re: TrueTears question thread
Post by: TrueTears on August 14, 2009, 12:20:43 am: Ok last question.

(http://img44.imageshack.us/img44/5369/speshassignment5.jpg)

Here is my working so far. But I am just stuck on part i). How do you go about it?

(http://img524.imageshack.us/img524/659/speshassignment6.jpg)

Thanks!!!
Title: Re: TrueTears question thread
Post by: kamil9876 on August 14, 2009, 12:31:46 pm: if $x_0>>y_0$ then by looking at the differential equations given, we see that the value of y decreases a lot, but as the value of y becomes small then the rate of decrease of x becomes small, hence y decreases a lot more than x So x dominates.

a and b can play a similair role to x_0 and y_0. If $b>>a$ then this helps x too as can be seen by analysing the differential equations in a similair fashion.
Title: Re: TrueTears question thread
Post by: aa53558 on August 14, 2009, 12:59:21 pm: hey what the hell, this is a application SAC, I've seen people working on it at school, don't post it online!!
Title: Re: TrueTears question thread
Post by: TrueTears on August 14, 2009, 01:10:29 pm: Am I breaking the rules or anything? I am clearly learning, idiot.
Title: Re: TrueTears question thread
Post by: zzdfa on August 14, 2009, 01:35:29 pm: Quote from: Mao on August 13, 2009, 11:19:00 pm
hey man, can you maybe use a smaller picture next time...? It's so big I can't be bothered scrolling across to read it...
Use Microsoft picture manager or something to resize to 600pixels wide or something.

ctrl+scrolldown will make it small enough to read
and ctrl+0 will restore stuff back to normal size
Title: Re: TrueTears question thread
Post by: kamil9876 on August 14, 2009, 05:05:49 pm: Quote from: aa53558 on August 14, 2009, 12:59:21 pm
hey what the hell, this is a application SAC, I've seen people working on it at school, don't post it online!!

Looks like, contrary to popular belief, TrueEars has found a friend.
Title: Re: TrueTears question thread
Post by: TrueTears on August 14, 2009, 05:11:37 pm: It probs is TrueEars.
Title: Re: TrueTears question thread
Post by: kat148 on August 14, 2009, 08:09:10 pm: You mean FakeEars ???
Title: Re: TrueTears question thread
Post by: TrueTears on August 15, 2009, 01:39:00 pm: So how do we find out whether x dominates?
Title: Re: TrueTears question thread
Post by: TrueTears on August 15, 2009, 08:26:00 pm: Quote from: kamil9876 on August 14, 2009, 12:31:46 pm
if $x_0>>y_0$ then by looking at the differential equations given, we see that the value of y decreases a lot, but as the value of y becomes small then the rate of decrease of x becomes small, hence y decreases a lot more than x So x dominates.

a and b can play a similair role to x_0 and y_0. If $b>>a$ then this helps x too as can be seen by analysing the differential equations in a similair fashion.
Ok then looking at the equation $a(y^2 - y_0^2) = b(x^2 - x_0^2)$

Because the questions tells us to examine some values, I found out if you have $x_0>>y_0$ it doesn't work...

Say you got a = 2 b = 3 and $y_0 = 1000$ and $x_0 = 7000$ and x = 6000 [Clearly here $x_0 >> y_0$ ]

Then the equation becomes $y^2 = -18500000$ which is not solvable with regards to this question.

Now if we increase $y_0$ to say 9000 then the equation becomes $y^2 = 61500000$ which is solvable but here clearly $y_0 >> x_0$

?????
?
Title: Re: TrueTears question thread
Post by: kamil9876 on August 15, 2009, 08:45:26 pm: Quote
and y_0 = 1000 and x_0 = 7000 and x = 6000 [Clearly here x_0 >> y_0]

Not clear to me.
Title: Re: TrueTears question thread
Post by: TrueTears on August 16, 2009, 03:15:46 pm: What you mean not clear?
Title: Re: TrueTears question thread
Post by: TrueTears on August 23, 2009, 01:59:52 am: (http://img197.imageshack.us/img197/1087/speshq11.jpg)

any ideas?
Title: Re: TrueTears question thread
Post by: /0 on August 23, 2009, 02:49:54 am: checked with calculator, don't think there's a correct answer.
Title: Re: TrueTears question thread
Post by: TrueTears on September 03, 2009, 11:02:24 pm: Quote from: /0 on August 23, 2009, 02:49:54 am
checked with calculator, don't think there's a correct answer.
Thanks.
Title: Re: TrueTears question thread
Post by: TrueTears on September 03, 2009, 11:10:11 pm: (http://img441.imageshack.us/img441/2637/speshq12.jpg)

I think answer is wrong but can someone confirm?

I put D but apparently answer says B.

But if you think about it, looking at D, x stands for the amount of tablet dissolved, so as x gets larger then $\frac{dx}{dt}$ will get smaller because 15% of the amount UNDISSOLVED is continuing to get smaller.

Where as B is saying is that as x gets larger, then the rate at which the the tablet is dissolving increases. But how can this be when 15% of the amount UNDISSOLVED is continually decreasing[since the amount of tablet left undissolved is decreasing.]
Title: Re: TrueTears question thread
Post by: Over9000 on September 03, 2009, 11:22:56 pm: I would definitely say D as B isn't logical how can rate start negative, can you have a negative rate in this application. D is right, rate starts high and then slows off.
Negative probably means that the tablet undissolves (nice according to tsfx we got some LCP here)
Title: Re: TrueTears question thread
Post by: TrueTears on September 04, 2009, 12:06:40 am: ok thanks over9000
Title: Re: TrueTears question thread
Post by: TrueTears on September 04, 2009, 12:08:08 am: (http://img40.imageshack.us/img40/8375/speshq13.jpg)

Why is C not right? Answer says E.

Yes I know E is right, but why is C also not one of the answers?
Title: Re: TrueTears question thread
Post by: /0 on September 04, 2009, 01:08:12 am: The diagonals of a kite meet at 90 degrees as well. However, a kite is not a rhombus.
http://en.wikipedia.org/wiki/Kite_(geometry)
Title: Re: TrueTears question thread
Post by: Flaming_Arrow on September 04, 2009, 08:24:50 am: Quote from: /0 on September 04, 2009, 01:08:12 am
The diagonals of a kite meet at 90 degrees as well. However, a kite is not a rhombus.
http://en.wikipedia.org/wiki/Kite_(geometry)

and this is also true for a square
Title: Re: TrueTears question thread
Post by: /0 on September 04, 2009, 12:01:16 pm: Quote from: Flaming_Arrow on September 04, 2009, 08:24:50 am
Quote from: /0 on September 04, 2009, 01:08:12 am
The diagonals of a kite meet at 90 degrees as well. However, a kite is not a rhombus.
http://en.wikipedia.org/wiki/Kite_(geometry)

and this is also true for a square

hmm yep but a square is a subset of the rhombus so if you can prove it is a square then it is automatically a rhombus
Title: Re: TrueTears question thread
Post by: TrueTears on September 04, 2009, 07:36:48 pm: (http://img267.imageshack.us/img267/8446/speshq14.jpg)

All of a, b, c and d are vectors.

say b(t) = qt i + kt j

and d(t) = ft^2 i + ot^2 j

and a = w i + e j

where q, k, f, o, w and e are all real constants.

then how come to find t_1 ie when d and b intersect, you must do a+b (which is vector c) = d and equate i and j components to find t_1

ie, (w + qt_1)i + (kt_1 + e)j = ft_1^2 i + ot_1^2 j

so (w + qt_1) = ft_1^2 and (kt_1 + e) = ot_1^2

why cant u just do b = d and then equate the i and j components of those 2 vectors?

ie, qt_1 i + kt_1 j = ft_1^2 i + ot_1^2 j

so qt_1 = ft_1^2 and kt_1 = ot_1^2
Title: Re: TrueTears question thread
Post by: kamil9876 on September 04, 2009, 08:22:10 pm: the source of your mistake is that you probably assumed that position vector is equal to displacement vector. Really: $r(t)=r(t_0) + \Delta r$ where $\Delta r$ is the displacement from the time $t_0$ to $t$

I think this is the mistake you made since, the way you have described, b(t) it should be a striaght line through the origin. But really it has a 'psuedo-origin' (ie $r(t_o)$ ) which is the position vector $\vec{a}$
Title: Re: TrueTears question thread
Post by: TrueTears on September 04, 2009, 08:42:49 pm: ok thanks kamil
Title: Re: TrueTears question thread
Post by: TrueTears on September 05, 2009, 03:14:14 am: Also I noticed something in some old exams I was doing.

When a question says "...find the quadratic EQUATION that has the ROOTS b and a..."

Then do your final answer have to be in the form of (x-b)(x-a) = 0

Because if you do not set it to 0 then it won't be an equation right?

Would marks be deducted if you just had "...the quadratic equation is (x-b)(x-a)..." ?
Title: Re: TrueTears question thread
Post by: /0 on September 05, 2009, 03:37:57 am: I think that technically, a root of $f$ is any value of x such that f(x) = 0.
i.e. The roots of f(x) = (x-a)(x-b) [or even just (x-a)(x-b), since f(x) can be redundant] are the solutions to the equation (x-a)(x-b) = 0.

So if they asked "...find the quadratic equation that has the solutions b and a...", I would write (x-a)(x-b)=0
But if they asked "...find the quadratic equation that has the roots b and a...", I would write h(x) = (x-a)(x-b) (or any function)
And if they asked "...find the quadratic (expression) that has roots b and a...", I would write (x-a)(x-b)
Title: Re: TrueTears question thread
Post by: TrueTears on September 05, 2009, 02:21:15 pm: What's different between a root and a solution? Ain't they the same thing?

"... a quadratic equation that has THE ROOTS..." isn't that the same thing as "... a quadratic equation that has THE SOLUTION..."?

EDIT:
Quote from: /0 on September 05, 2009, 03:37:57 am
But if they asked "...find the quadratic equation that has the roots b and a...", I would write h(x) = (x-a)(x-b) (or any function)

Because in a very old VCAA exam, it had a question that says "...find the quadratic EQUATION that has the ROOTS $2 + \sqrt{5}i$ and $2 - \sqrt{5}i$ ..."

In the examiners report they specifically said a lot of people lost marks because they did not set it to zero.

But I did what you did, I wrote y(x) = (x - $(2 + \sqrt{5}i)$ )(x - $(2 - \sqrt{5}i)$ ).
Title: Re: TrueTears question thread
Post by: Damo17 on September 05, 2009, 02:31:50 pm: Quote from: TrueTears on September 05, 2009, 02:21:15 pm
What's different between a root and a solution? Ain't they the same thing?

"... a quadratic equation that has THE ROOTS..." isn't that the same thing as "... a quadratic equation that has THE SOLUTION..."?

They are very similar to the point where one would usually say they are equivalent. However there is a subtle difference.
A solution is finding the values in an equation that satisfy the given condition, and finding roots generally implies finding the zeros (a type of solution) that satisfy the given conditions. 'Solution' is a bit broader than 'root'.
Title: Re: TrueTears question thread
Post by: TonyHem on September 05, 2009, 02:33:51 pm: sound the same to me.
Title: Re: TrueTears question thread
Post by: TrueTears on September 05, 2009, 02:42:20 pm: Quote from: Damo17 on September 05, 2009, 02:31:50 pm
Quote from: TrueTears on September 05, 2009, 02:21:15 pm
What's different between a root and a solution? Ain't they the same thing?

"... a quadratic equation that has THE ROOTS..." isn't that the same thing as "... a quadratic equation that has THE SOLUTION..."?

They are very similar to the point where one would usually say they are equivalent. However there is a subtle difference.
A solution is finding the values in an equation that satisfy the given condition, and finding roots generally implies finding the zeros (a type of solution) that satisfy the given conditions. 'Solution' is a bit broader than 'root'.
Oh I get it, so find the roots to a quadratic equations implies setting it to 0 and solving it.

However finding the solutions to a quadratic equation implies that it is already set to something (like an 'initial condition').

So for a quadratic equation like y = (x-a)(x-b), if you say find the roots, you would do (x-a)(x-b) = 0 and solve.

However if you said find the solutions to (x-a)(x-b) = y, 'y' must be set to something, so it might be something such as find the solutions to (x-a)(x-b) = 4 (not necessarily set to 0).

But doesn't that mean if a question said find the "roots" to [this] quadratic equation you need to answer it by setting your expression to 0.

Quote from: /0
But if they asked "...find the quadratic equation that has the roots b and a...", I would write h(x) = (x-a)(x-b) (or any function)
??
Title: Re: TrueTears question thread
Post by: Damo17 on September 05, 2009, 02:50:39 pm: Quote from: TrueTears on September 05, 2009, 02:42:20 pm
Oh I get it, so find the roots to a quadratic equations implies setting it to 0 and solving it.

However finding the solutions to a quadratic equation implies that it is already set to something (like an 'initial condition').

So for a quadratic equation like y = (x-a)(x-b), if you say find the roots, you would do (x-a)(x-b) = 0 and solve.

However if you said find the solutions to (x-a)(x-b) = y, 'y' must be set to something, so it might be something such as find the solutions to (x-a)(x-b) = 4 (not necessarily set to 0).

Exactly :)

Quote from: TrueTears on September 05, 2009, 02:42:20 pm
But doesn't that mean if a question said find the "roots" to [this] quadratic equation you need to answer it by setting your expression to 0.

Yes you would have to let it equal zero. In what instances are you doubting this?
Title: Re: TrueTears question thread
Post by: TrueTears on September 05, 2009, 02:51:41 pm: oh just /0's statement went opposite to my thinking lol, but he's a bright lad so I thought I was wrong.
Title: Re: TrueTears question thread
Post by: TrueTears on September 05, 2009, 03:00:34 pm: Also if the question said this instead: "find the quadratic function that has the roots a and b"

I thought that since roots mean zero (according to wiki) that means it implies the function MUST be set to 0

so answer would just be f(x) = (x-a)(x-b)
Title: Re: TrueTears question thread
Post by: Damo17 on September 05, 2009, 03:13:54 pm: Quote from: TrueTears on September 05, 2009, 03:00:34 pm
Also if the question said this instead: "find the quadratic function that has the roots a and b"

I thought that since roots mean zero (according to wiki) that means it implies the function MUST be set to 0

so answer would just be f(x) = (x-a)(x-b)

Yes, that would be the answer.
If you had f(x)=a(x-b)(x-c)=0, then this would be a quadratic equation.
If you had f(x)=a(x-b)(x-c) , then this would be a quadratic function.
Title: Re: TrueTears question thread
Post by: Over9000 on September 05, 2009, 03:19:20 pm: Quote from: TrueTears on September 05, 2009, 02:51:41 pm
oh just /0's statement went opposite to my thinking lol, but he's a bright lad so I thought I was wrong.
yeh I think his wrong there, it should be set to 0 in my opinion
Title: Re: TrueTears question thread
Post by: TrueTears on September 05, 2009, 03:21:26 pm: Thanks all!!!
Title: Re: TrueTears question thread
Post by: /0 on September 05, 2009, 04:00:41 pm: Quote from: TrueTears on September 05, 2009, 02:42:20 pm
So for a quadratic equation like y = (x-a)(x-b), if you say find the roots, you would do (x-a)(x-b) = 0 and solve.

Wait, aren't you saying that $a$ and $b$ are 'roots' of the equation $y=(x-a)(x-b)$ ?
Title: Re: TrueTears question thread
Post by: TrueTears on September 05, 2009, 04:43:27 pm: Quote from: TrueTears
When a question says "...find the quadratic EQUATION that has the ROOTS b and a..."

Quote from: /0
But if they asked "...find the quadratic equation that has the roots b and a...", I would write h(x) = (x-a)(x-b) (or any function)

Quote from: kamil (msn)
if it says find the roots of an equation then it must be 0=...
Title: Re: TrueTears question thread
Post by: /0 on September 06, 2009, 01:54:47 pm: $f(x) = (x-a)(x-b)$

It has an equals sign so it's an equation (tick)
It has roots a, b (tick)
It is a quadratic equation (tick)

???
Title: Re: TrueTears question thread
Post by: kamil9876 on September 06, 2009, 02:34:56 pm: $0=(x-a)(x-b)$ has roots $x=a, x=b$ is a clear way of saying that if i plug in x=a or x=b I get an equation that holds.

i.e: just like we say $x^2+y^2=z^2$ has a integers roots (x,y,z)=(5,12,13) rather than $f(x,y,z)=x^2+y^2-z^2$ has zero's/roots etc.

Likewise, Pell's equation is called Pell's equation, not Pell's function because we care more about solving it than differentiating it.

I don't know, maybe what you said is technically true according to what's on wiki but my reply to TT on msn was done with haste because I was in a rush and quite annoyed by his numerous nudges and trivial questions. But for aesthetic reasons posted above and observation of what's commonly used, I will always chose my alternative when seeing "equations" and "roots".

Also I think the EQUATION f(x)=(x-a)(x-b) has infinitely many roots since we can rearrange it to:

0=(x-a)(x-b)-f(x) and what do we define f(x) as?

Whereas the FUNCTION f(x)=(x-a)(x-b) has roots(I will assume this is synonymous to zeros as wiki says, I myslef would use zero's instead) a and b only. (that I am sure is true).

edit: some links:

http://www.sosmath.com/algebra/quadraticeq/root/root.html

http://www.thefreedictionary.com/Root+of+an+equation (basically the definition I adhere to)
Title: Re: TrueTears question thread
Post by: TrueTears on September 06, 2009, 09:17:00 pm: ok.

Then according to /0, the VCAA examiners were wrong.
Title: Re: TrueTears question thread
Post by: Over9000 on September 06, 2009, 09:17:51 pm: Quote from: TrueTears on September 06, 2009, 09:17:00 pm
ok.

Then according to /0, the VCAA examiners were wrong.
Rofl
Title: Re: TrueTears question thread
Post by: /0 on September 06, 2009, 09:38:15 pm: Quote from: Over9000 on September 06, 2009, 09:17:51 pm
Quote from: TrueTears on September 06, 2009, 09:17:00 pm
ok.

Then according to /0, the VCAA examiners were wrong.
Rofl

forget it
Title: Re: TrueTears question thread
Post by: Mao on September 06, 2009, 10:57:28 pm: I don't think they assess the exam at that level of technicality.
Title: Re: TrueTears question thread
Post by: TrueTears on September 14, 2009, 12:20:45 am: How to find $\lim_{t \to 0} \frac{3-\sqrt{9-4t^2}}{2t}$

and why is $\lim_{t \to 0} \frac{3+\sqrt{9-4t^2}}{2t}$ undefined?
Title: Re: TrueTears question thread
Post by: /0 on September 14, 2009, 12:56:21 am: For the top you can use L'hopital's rule, since the indeterminate form $\frac{0}{0}$ is reached. For the bottom one, you have $\frac{6}{0}$ , which is not an indeterminate form and diverges to infinity.
Title: Re: TrueTears question thread
Post by: TrueTears on September 14, 2009, 01:01:54 am: Oh I get the 2nd one now, yeah what about not using L'hopitals?
Title: Re: TrueTears question thread
Post by: Mao on September 14, 2009, 01:04:21 am: Quote from: TrueTears on September 14, 2009, 12:20:45 am
How to find $\lim_{t \to 0} \frac{3-\sqrt{9-4t^2}}{2t}$

and why is $\lim_{t \to 0} \frac{3+\sqrt{9-4t^2}}{2t}$ undefined?

be aware that for the question you are applying this to, the limit is only the left-hand limit (as specified by the domain $t\in (-\infty,0]$ , y must also be negative, and 't', in this case, carries the values of y), i.e. $t \to 0^-$ , hence the second limit is equivalent to $\lim_{t \to 0^-} \frac{6}{t} = -\infty$
Title: Re: TrueTears question thread
Post by: TrueTears on September 14, 2009, 01:05:07 am: Quote from: Mao on September 14, 2009, 01:04:21 am
Quote from: TrueTears on September 14, 2009, 12:20:45 am
How to find $\lim_{t \to 0} \frac{3-\sqrt{9-4t^2}}{2t}$

and why is $\lim_{t \to 0} \frac{3+\sqrt{9-4t^2}}{2t}$ undefined?

be aware that for the question you are applying this to, the limit is only the left-hand limit (as specified by the domain $t\in (-\infty,0]$ ), i.e. $t \to 0^-$ , hence the second limit is equivalent to $\lim_{t \to 0^-} \frac{6}{t} = -\infty$
Yeap, thanks Mao!
Title: Re: TrueTears question thread
Post by: Mao on September 14, 2009, 01:11:02 am: The first one can be kind of 'squeezed'.

$y = \frac{2t}{t^2 + 1}$ , $t\in (-\infty,0] \implies y \le 0$

$\lim_{y \to 0^-} \frac{3 - \sqrt{9-4y^2}}{2y} = L$

firstly, $\sqrt{9-4y^2} \le \sqrt{9}$ , this implies $L \ge \frac{0}{2y} = 0$

secondly, we note that $L \in t$ since it is an expression for the inverse and $0^- \in y$ . Hence another condition on L is $L \in (-\infty, 0] \equiv L \le 0$

combining the two conditions $L\ge 0$ and $L \le 0$ , we arrive at $L = \lim_{y \to 0^-} \frac{3 - \sqrt{9-4y^2}}{2y} = 0$
Title: Re: TrueTears question thread
Post by: TrueTears on September 28, 2009, 01:28:40 am: Solve this differential equation:

$x^2 \frac{d^2y}{dx^2} + 5x \frac{dy}{dx} + (4 + \pi^2)y = \log_e(x)$

Given $y(1) = 0$ and $y(e) = 0$
Title: Re: TrueTears question thread
Post by: sachinmachin on September 28, 2009, 07:57:42 am: Quote from: TrueTears on September 28, 2009, 01:28:40 am
Solve this differential equation:

$x^2 \frac{d^2y}{dx^2} + 5x \frac{dy}{dx} + (4 + \pi^2)y = \log_e(x)$

Given $y(1) = 0$ and $y(e) = 0$

Shouldn't an equation for "y" be specified, so that we are able to do the derivative/double derivative??
I dont think theres any other way you can do it.
Title: Re: TrueTears question thread
Post by: Mao on September 28, 2009, 10:44:24 am: that is a second order non-homogenous. [absolutely not in VCE]

$x^2 \frac{d^2y}{dx^2} + 5x \frac{dy}{dx} + (4 + \pi^2) y = \log_e (x)\; \; \; (1)$

It can be seen here that a particular solution is in the form $y_p (x) = \frac{\log_e (x)}{4 + \pi^2} + B$ , where the first and second derivative multiplied by 'x' and 'x^2' coefficients will give constants. Solving this gives $B = -\frac{4}{(4+\pi^2)^2}$

Now, to solve the homogenous part, $\frac{d^2y}{dx^2} + \frac{5}{x} \frac{dy}{dx} + \frac{4+\pi^2}{x^2} y = 0$
since the coefficients are not constants (they are functions of x), we cannot simply use the characteristic equation. To derive a solution from this will be very very hard, and at this point most problem solvers will go forth and use power series, or specifically, the Forbenius method. This stuff you learn in 2nd/3rd year at uni.

So yeah... bitch of a question.

Mathematica tells me the general solution is $\frac{-4 + (4 + \pi^2) \log_e x}{(4 + \pi^2)^2} + \frac{ C_2 \cos(\pi \log_e x) + C_1 \sin(\pi \log_ex)}{x^2}$

And for this boundary value problem, there's no solution.
Title: Re: TrueTears question thread
Post by: TrueTears on September 28, 2009, 07:37:53 pm: Thanks Mao, I was just reading some stuff on non-homogenous DE's and tried to fiddle around with change of variables etc. Didn't work out =S

Again thanks!
Title: Re: TrueTears question thread
Post by: TrueTears on October 09, 2009, 12:17:30 am: What's the systematic way of working out $\tan(\theta) = \sqrt{3}-2$
Title: Re: TrueTears question thread
Post by: Damo17 on October 09, 2009, 02:12:16 pm: Quote from: TrueTears on October 09, 2009, 12:17:30 am
What's the systematic way of working out $\tan(\theta) = \sqrt{3}-2$

This requires the use of the trig identity: $tan^2 (\theta)= \frac{1-cos(2 \theta)}{1+cos(2 \theta)}$

$tan^2 (\theta)= \frac{1-cos(2 \theta)}{1+cos(2 \theta)}= 7-4\sqrt{3}$

$1-cos(2 \theta)=7-4\sqrt{3}+7cos(2 \theta)-4\sqrt{3}cos(2 \theta)$

$4\sqrt{3}-6=cos(2 \theta)(8-4\sqrt{3})$

$\frac{4\sqrt{3}-6}{8-4\sqrt{3}} \times \frac{8+4\sqrt{3}}{8+4\sqrt{3}}=cos(2 \theta)$

$\frac{\sqrt{3}}{2}=cos(2 \theta)$

$\therefore$ $\theta =\pm \frac{\pi}{12}$

But we take the negative as $\sqrt{3}-2$ is negative, so:

$\theta= -\frac{\pi}{12}$
Title: Re: TrueTears question thread
Post by: dejan91 on October 09, 2009, 03:18:11 pm: Quote from: Damo17 on October 09, 2009, 02:12:16 pm
Quote from: TrueTears on October 09, 2009, 12:17:30 am
What's the systematic way of working out $\tan(\theta) = \sqrt{3}-2$

This requires the use of the trig identity: $tan^2 (\theta)= \frac{1-cos(2 \theta)}{1+cos(2 \theta)}$

What identity is this? Is it derived from something (as in manipulated) or are we meant to know it?
Title: Re: TrueTears question thread
Post by: TonyHem on October 09, 2009, 03:33:53 pm: Quote from: dejan91 on October 09, 2009, 03:18:11 pm
What identity is this? Is it derived from something (as in manipulated) or are we meant to know it?
$\frac{sin^2(x)}{cos^2(x)}$
$cos(2x) = 1-2sin^2(x)$
$2sin^2(x) = 1-cos(2x)$
$cos(2x) = 2cos^2(x)-1$
$2cos^2(x)= 1+cos(2x)$
$so\; \frac{sin^2(x)}{cos^2(x)} = \frac{1-cos(2x)}{1+cos(2x)}$
Title: Re: TrueTears question thread
Post by: TrueTears on October 09, 2009, 04:20:58 pm: Quote from: Damo17 on October 09, 2009, 02:12:16 pm
Quote from: TrueTears on October 09, 2009, 12:17:30 am
What's the systematic way of working out $\tan(\theta) = \sqrt{3}-2$

This requires the use of the trig identity: $tan^2 (\theta)= \frac{1-cos(2 \theta)}{1+cos(2 \theta)}$

$tan^2 (\theta)= \frac{1-cos(2 \theta)}{1+cos(2 \theta)}= 7-4\sqrt{3}$

$1-cos(2 \theta)=7-4\sqrt{3}+7cos(2 \theta)-4\sqrt{3}cos(2 \theta)$

$4\sqrt{3}-6=cos(2 \theta)(8-4\sqrt{3})$

$\frac{4\sqrt{3}-6}{8-4\sqrt{3}} \times \frac{8+4\sqrt{3}}{8+4\sqrt{3}}=cos(2 \theta)$

$\frac{\sqrt{3}}{2}=cos(2 \theta)$

$\therefore$ $\theta =\pm \frac{\pi}{12}$

But we take the negative as $\sqrt{3}-2$ is negative, so:

$\theta= -\frac{\pi}{12}$

Awesome, thanks, how did you think of $7 - 4 \sqrt{3}$ ?
Title: Re: TrueTears question thread
Post by: TonyHem on October 09, 2009, 04:23:09 pm: tan^2(theta) = 3 - 4sqrt{3} + 4
Title: Re: TrueTears question thread
Post by: TrueTears on October 09, 2009, 04:26:30 pm: Quote from: TonyHem on October 09, 2009, 04:23:09 pm
tan^2(theta) = 3 - 4sqrt{3} + 4
indeed, thanks.

and yeah thanks again Damo!!!!!!!!!!!!!!!!!!!!!!! pr0!!!!!!!!!!!!!!!
Title: Re: TrueTears question thread
Post by: Damo17 on October 09, 2009, 04:38:29 pm: Quote from: TrueTears on October 09, 2009, 04:26:30 pm
and yeah thanks again Damo!!!!!!!!!!!!!!!!!!!!!!! pr0!!!!!!!!!!!!!!!

No problem. It took me a while, but I rather enjoyed it.
Title: Re: TrueTears question thread
Post by: TrueTears on October 23, 2009, 04:48:38 pm: A small object is dropped vertically down from the top of a building which is h metres high. It takes 0.4 seconds to travel the last 8 metres before it hits the ground. If the air resistance is negligible, show that the height of the building is : $h = \frac{(100+g)^2}{50g}$

Pretty simple question IF you assume it starts from rest. So can you assume it starts from rest?
Title: Re: TrueTears question thread
Post by: kurrymuncher on October 23, 2009, 04:56:18 pm: Quote from: TrueTears on October 23, 2009, 04:48:38 pm
A small object is dropped vertically down from the top of a building which is h metres high. It takes 0.4 seconds to travel the last 8 metres before it hits the ground. If the air resistance is negligible, show that the height of the building is : $h = \frac{(100+g)^2}{50g}$

Pretty simple question IF you assume it starts from rest. So can you assume it starts from rest?

Well, it says "dropped", so Im assuming it starts from rest.

maybe? i dont know lol
Title: Re: TrueTears question thread
Post by: TrueTears on October 23, 2009, 04:57:48 pm: Quote from: kurrymuncher on October 23, 2009, 04:56:18 pm
Quote from: TrueTears on October 23, 2009, 04:48:38 pm
A small object is dropped vertically down from the top of a building which is h metres high. It takes 0.4 seconds to travel the last 8 metres before it hits the ground. If the air resistance is negligible, show that the height of the building is : $h = \frac{(100+g)^2}{50g}$

Pretty simple question IF you assume it starts from rest. So can you assume it starts from rest?

Well, it says "dropped", so Im assuming it starts from rest.

maybe? i dont know lol
Yeah, but when something is dropped it would also be dropped with a certain velocity.
Title: Re: TrueTears question thread
Post by: ed_saifa on October 23, 2009, 05:06:49 pm: I'm sure that drop implies from rest. You can't really drop something with a certain velocity; that would be called thrown :P
Title: Re: TrueTears question thread
Post by: TrueTears on October 23, 2009, 05:14:12 pm: Ahhh haha, alright thanks you guys!
Title: Re: TrueTears question thread
Post by: TrueTears on October 23, 2009, 05:58:42 pm: A particle is moving so that its position at time t seconds is given by $r(t) = 2cos\left(\frac{\pi}{10}t\right) i + 3t j$ . Find the cartesian equation of the particle, state the domain and range.

Let $x(t) = 2cos\left(\frac{\pi}{10}t\right)$ and $y(t) = 3t$

So the cartesian equation is $y = \frac{30}{\pi}\cos^{-1}\left(\frac{x}{2}\right)$ .

So I've always done that the range of $x(t)$ = domain of cartesian and range of $y(t)$ = range of cartesian.

Okay, so the range of $x(t)$ for $t \ge 0$ is $[-2,2]$ which is right domain for cartesian, however range of $y(t)$ is $[0, \infty)$ however the cartesian is not defined for $[0, \infty)$ does that mean we take the highest possible range that is defined for the cartesian? ie, $[0, 30]$
Title: Re: TrueTears question thread
Post by: kamil9876 on October 23, 2009, 06:17:17 pm: The crux is in this step:

$x=2cos(\frac{y\pi}{30})$

when "inverting":
$ cos^{-1}(\frac{x}{2})=\frac{y\pi}{30} + 2k\pi$ and that other scenario that I cbf working out.

THis is analogous to $y=x^2 \implies x=\sqrt{y}$ not being neccesary.
Title: Re: TrueTears question thread
Post by: TrueTears on October 23, 2009, 06:19:38 pm: Quote from: TrueTears on October 23, 2009, 05:58:42 pm
Okay, so the range of $x(t)$ for $t \ge 0$ is $[-2,2]$ which is right domain for cartesian, however range of $y(t)$ is $[0, \infty)$ however the cartesian is not defined for $[0, \infty)$ does that mean we take the highest possible range that is defined for the cartesian? ie, $[0, 30]$
So... what range do I take?
Title: Re: TrueTears question thread
Post by: Mao on October 23, 2009, 06:25:17 pm: the cartesian equation is not necessarily a function

The cartesian equation is $x = 2\cos \left( \frac{\pi y}{30}\right)$

Domain is [-2,2], range is [0,infinity)

The graph is a wave travelling vertically upwards.
Title: Re: TrueTears question thread
Post by: kamil9876 on October 23, 2009, 06:28:32 pm: Quote from: TrueTears on October 23, 2009, 06:19:38 pm
Quote from: TrueTears on October 23, 2009, 05:58:42 pm
Okay, so the range of $x(t)$ for $t \ge 0$ is $[-2,2]$ which is right domain for cartesian, however range of $y(t)$ is $[0, \infty)$ however the cartesian is not defined for $[0, \infty)$ does that mean we take the highest possible range that is defined for the cartesian? ie, $[0, 30]$
So... what range do I take?

it all depends on what they defiend as t. e.g if t can go over 10seconds, say t=11, then the cos^{-1} expression is not complete. Domain of t must be provided.
Title: Re: TrueTears question thread
Post by: TrueTears on October 23, 2009, 08:22:57 pm: Yeah thanks guys, the question doesn't specify a domain... quite a stupid Q heh
Title: Re: TrueTears question thread
Post by: TrueTears on October 24, 2009, 11:26:58 pm: If $3\int_0^k \frac{1}{x+2} - \frac{1}{x-2} dx = 6$ and $0<k<2$ , find the exact value of $k$ .

Okay so I did:

$[\log_e|\frac{x+2}{x-2}|]_0^k = 2$

$e^2 = |\frac{k+2}{k-2}|$

How do I know which half to pick?

I did this (not sure if reasoning is right):

Case 1: if $\frac{k+2}{k-2} > 0$

then $k+2 >0$ and $k-2 > 0$ or $k+2<0$ and $k-2<0$

so $k>-2$ and $k>2$ or $k<-2$ and $k<2$

which means overall we have $k>2$ or $k<-2$

However the domain they gave for k is $0<k<2$ so if can't be the positive half of the modulus ie $\frac{k+2}{k-2} > 0$ is false.

Case 2: $\frac{k+2}{k-2} < 0$

Then we must have $k+2 <0$ and $k-2>0$ or $k+2 >0$ and $k-2<0$

so $k<-2$ and $k>2$ or $k>-2$ and $k<2$

Which means overall we have $-2<k<2$ . However the domain they give $0<k<2$ is just a subset of this which means the right equation to solve is:

$e^2 = -\left(\frac{k+2}{k-2}\right)$

Is that how you go about deciding which half of the modulus to use or is it complete wrong?
Title: Re: TrueTears question thread
Post by: kamil9876 on October 24, 2009, 11:35:50 pm: Yes, that is completely correct reasoning :) .

Although a bit convoluted as I would just do this by noticing that all I need to decide is whether $\frac{k+2}{k-2}$ is positive or negative over the domain given. We know it is positive if numerator and denominator are the same sign, and negative when they are opposite signs.

we have $0<k<2$

$\implies -2<k-2<0 (1)$ (subtracting 2 from both sides)
and $2<k+2<4 (2)$ (adding two to both sides).

(1) implies the denominator must be negatve, (2) implies the numerator is positive. Hence opposite signs and the result follows.
Title: Re: TrueTears question thread
Post by: TrueTears on October 25, 2009, 02:19:09 am: Thanks kamil.

Also what is the convention used for naming polygons, say you a parallelogram ABCD, is the 4 vertices named in a clockwise way? Or anti clockwise?
Title: Re: TrueTears question thread
Post by: TrueTears on October 25, 2009, 03:51:00 am: And also for any complex number $z$ , then $|z^x| = |z|^x$ for $x \in R$ .

Is that true? If so, how do you prove it?
Title: Re: TrueTears question thread
Post by: humph on October 25, 2009, 05:39:53 am: Quote from: TrueTears on October 25, 2009, 03:51:00 am
And also for any complex number $z$ , then $|z^x| = |z|^x$ for $x \in R$ .

Is that true? If so, how do you prove it?
Not sure how to prove it using spec techniques. Using the complex-valued exponential and logarithm functions, we have that for $z \in \mathbb{C}$ , $a \in \mathbb{R}$ ,
$z^a = (\exp(\log z))^a = (\exp(\log|z| + i \arg z))^a = (\exp(\log|z|)^a (\exp(i \arg z))^a = |z|^a \exp(i a \arg z)$ ,
and so
$|z^a| = ||z|^a \exp(i a \arg z)| = |z|^a |\exp(i a \arg z)| = |z|^a$ ,
as
$|\exp(i a \arg z)| = |\cos(a \arg z) + i \sin(a \arg z)| = \sqrt{\cos^2(a \arg z) + \sin^2(a \arg z)} = 1$ .
Title: Re: TrueTears question thread
Post by: TrueTears on October 25, 2009, 03:07:29 pm: Thanks humph so it is true for $x \in \mathbb{R}$ what about $x \in \mathbb{C}$ ?
Title: Re: TrueTears question thread
Post by: Mao on October 25, 2009, 11:08:37 pm: Quote from: TrueTears on October 25, 2009, 03:07:29 pm
Thanks humph so it is true for $x \in \mathbb{R}$ what about $x \in \mathbb{C}$ ?

raising a number to a complex power?

$a^{z} = a^{x+yi} = a^x \cdot a^{yi} = a^x \cdot e^{\log_e (a) y i}$

As you can see, it is just another complex number.

Extending $a$ to a complex plane:

$a^{z} = \left(|a|e^{i\arg a}\right)^{z} = |a|^z \cdot e^{i \arg a \cdot z} = |a|^x \cdot e^{y\log_e |a| i} \cdot \left( e^{i x \arg a} \cdot e^{-y\arg a}\right) = |a|^x \cdot e^{-y\arg a} \cdot \left( e^{(y\log_e |a| + x \theta ) i} \right)$

$|a^z| = |a|^x e^{-y\arg a}$

So no, it does not extend over the complex plane.
Title: Re: TrueTears question thread
Post by: TrueTears on October 31, 2009, 08:04:05 pm: Just a few technicalities.

What's the definition of a line, line segment and ray?

Line: Both ends extend out to infinity.

Ray: Has a fixed starting point, other end goes to infinity.

Line segment: Has 2 fixed ends and thus a fixed length. (Does that mean a point would also be a line segment since you can consider it as 0 length lol)

Is that right..?
Title: Re: TrueTears question thread
Post by: bem9 on October 31, 2009, 08:33:32 pm: but a point doesnt have '2 fixed ends' so its cant be a line segment
Title: Re: TrueTears question thread
Post by: TrueTears on October 31, 2009, 09:01:33 pm: Quote from: bem9 on October 31, 2009, 08:33:32 pm
but a point doesnt have '2 fixed ends' so its cant be a line segment
What if you define the starting to be 'A'. The ending point is 'B' (even though it's the same point).

Thus the point can be called AB with a length of 0.

Just like a random line segment can be called AB with a length of 5.

Or is the definition of the length of a line segment with length $x$ such that $x \in R^+$ only? In which case a point would not be a line segment.
Title: Re: TrueTears question thread
Post by: Over9000 on October 31, 2009, 09:17:17 pm: Quote from: TrueTears on October 31, 2009, 09:01:33 pm
Quote from: bem9 on October 31, 2009, 08:33:32 pm
but a point doesnt have '2 fixed ends' so its cant be a line segment
What if you define the starting to be 'A'. The ending point is 'B' (even though it's the same point).

Thus the point can be called AB with a length of 0.

Just like a random line segment can be called AB with a length of 5.

Or is the definition of the length of a line segment with length $x$ such that $x \in R^+$ only? In which case a point would not be a line segment.
By doing that, wouldn't point a = point b since their exactly the same point.
Title: Re: TrueTears question thread
Post by: TrueTears on October 31, 2009, 09:17:45 pm: Quote from: Over9000 on October 31, 2009, 09:17:17 pm
Quote from: TrueTears on October 31, 2009, 09:01:33 pm
Quote from: bem9 on October 31, 2009, 08:33:32 pm
but a point doesnt have '2 fixed ends' so its cant be a line segment
What if you define the starting to be 'A'. The ending point is 'B' (even though it's the same point).

Thus the point can be called AB with a length of 0.

Just like a random line segment can be called AB with a length of 5.

Or is the definition of the length of a line segment with length $x$ such that $x \in R^+$ only? In which case a point would not be a line segment.
By doing that, wouldn't point a = point b since their exactly the same point.
No I'm saying a point is also a line segment does not imply that any point equals any other point.

However I don't know the rigorous definition of a line segment yet (nor a line or a ray), just that according to my current understanding, they can be proved to be the same.
Title: Re: TrueTears question thread
Post by: TrueTears on October 31, 2009, 09:37:15 pm: (http://img233.imageshack.us/img233/900/asdfvk.jpg)

None of the choices are right? WDF
Title: Re: TrueTears question thread
Post by: dejan91 on October 31, 2009, 09:41:35 pm: Lol you're right (at least I can't see any correct options). Usually I just look for an undefined gradient and sub the coordinates in to each one to see which one is undefined...but not working for this one. And btw...wtf kind of a differential is that??? So trippy :P
Title: Re: TrueTears question thread
Post by: TrueTears on October 31, 2009, 09:43:44 pm: Quote from: dejan91 on October 31, 2009, 09:41:35 pm
Lol you're right (at least I can't see any correct options). Usually I just look for an undefined gradient and sub the coordinates in to each one to see which one is undefined...but not working for this one. And btw...wtf kind of a differential is that??? So trippy :P
True, I subbed in (0,0) all undefined cept for E, but E is wrong clearly.

GG
Title: Re: TrueTears question thread
Post by: TonyHem on October 31, 2009, 10:26:13 pm: gradient is 0 at x = 0, y = -1
A)1/-(-1)
B)0-1/-(-1) = 1
C)0-(-1)/(-1) = -1
D)= 0
E) = 0-(-1)^2 = 1

But yeah, that (0,0) thing, leaves E :S

really weird
i saw this question when it was coming out of my printer 20 mins ago lol
now i dont really wanna do it
Title: Re: TrueTears question thread
Post by: TrueTears on October 31, 2009, 10:29:46 pm: Good paper to do, just some stupid questions (like always with trial exams)
Title: Re: TrueTears question thread
Post by: TrueTears on November 01, 2009, 04:03:52 am: (http://img405.imageshack.us/img405/7252/asdfd.jpg)

$g(x)$ can also be written as $-x^{\frac{3}{2}}\sqrt{4-x}$

Implicit differentiating we get:

$2y\frac{dy}{dx} = -4x^2(x-3)$

so $\frac{dy}{dx} = \frac{-4x^2(x-3)}{2y}$

But the domain they give is $x \in (0,4)$ so both $f(x)$ and $g(x)$ satisfies.

Thus subbing in $y = \pm x^{\frac{3}{2}}\sqrt{4-x} \implies \frac{dy}{dx} = \frac{-4x^2(x-3)}{\pm 2(x^{\frac{3}{2}}\sqrt{4-x})}$

How come answer is just $\frac{dy}{dx} = \frac{-4x^2(x-3)}{2(x^{\frac{3}{2}}\sqrt{4-x})}$ Why don't they also include the negative half?
Title: Re: TrueTears question thread
Post by: TrueTears on November 01, 2009, 04:14:46 am: (http://img256.imageshack.us/img256/7518/asdfes.jpg)

a) This I can do:

$\sum F = ma$

$9g - 0.01v^2 = 9a$

$a = -\frac{(v^2-900g)}{900}$

b) My working is as follows:

$\frac{dv}{dx} = -\frac{(v^2-900g)}{900v}$

$\frac{dx}{dx} =-\frac{900v}{(v^2-900g)}$

$x = -900 \int \frac{v}{(v^2-900g)} dv$

$= -450\log_e|v^2-900g| + c$

When $x = 0$ $v = 0$

$\therefore c = 450\log_e(900g)$

$\implies x = 450\log_e(900g) - 450\log_e|v^2-900g|$

$\implies x = 450\log_e|\frac{900g}{v^2-900g}|$

$x = 450\log_e|\frac{8820}{v^2-8820}|$

Now how do I get rid of the mod signs to get their form? I did the following:

$x = 450\log_e|\frac{8820}{-(8820-v^2)}|$

$x = 450\log_e\left(\frac{8820}{|-1||8820-v^2|}\right)$

$x = 450\log_e\left(\frac{8820}{|8820-v^2|}\right)$

But then how to get rid of the mods in the denominator? Since it still could be negative and you need the mods to ensure it's positive?

Next part says:

(http://img145.imageshack.us/img145/5813/asdfwc.jpg)

$x = 450\log_e\left(\frac{8820}{|8820-v^2|}\right)$

When $x = 150$

$\therefore 150 = 450\log_e\left(\frac{8820}{|8820-v^2|}\right)$

Solving these on the TI-89 gives: $v = \pm 50.002, \pm 123.044$

Now since the question asks for the speed we require $|v| = 50.002, 123.044$

However which one do you choose?

Now if you solve their equation which is $x = 450\log_e\left(\frac{8820}{8820-v^2}\right)$

you get $|v| = 50.002$ only and not $123.044$

So which formula do you use and most importantly when you use my formula how do you dismiss $123.044$ as a solution?
Title: Re: TrueTears question thread
Post by: bem9 on November 01, 2009, 09:01:54 am: the bags velocity can only be from [0, sqrt(8820)) as for any value greater than this it would require the bags velocity to be greater than sqrt(8820) and therefore at some stage its velocity would have to be equal to sqrt(8820). This is impossible as this value is also the terminal velocity. Therefore you dont need to mod the bottom

Also in your second question, since 123.044 is greater than the terminal velocity, it is rejected as in order to reach this velocity, the sand bag must also pass and at some stage equal sqrt(8820), which is impossible.

i hope my reasoning is correct here :S
Title: Re: TrueTears question thread
Post by: Over9000 on November 01, 2009, 11:55:54 am: Quote from: bem9 on November 01, 2009, 09:01:54 am
the bags velocity can only be from [0, sqrt(8820)) as for any value greater than this it would require the bags velocity to be greater than sqrt(8820) and therefore at some stage its velocity would have to be equal to sqrt(8820). This is impossible as this value is also the terminal velocity. Therefore you dont need to mod the bottom

Also in your second question, since 123.044 is greater than the terminal velocity, it is rejected as in order to reach this velocity, the sand bag must also pass and at some stage equal sqrt(8820), which is impossible.

i hope my reasoning is correct here :S
How do you know $v = \sqrt{8820}$ is the terminal velocity, what is ur working/reasoning for it?
Title: Re: TrueTears question thread
Post by: NE2000 on November 01, 2009, 12:16:45 pm: I think it's because when $a=0$

$9g = 0.01v^2$
Solving for v gives the value that bem9 gave. This is the velocity of the object when acceleration would be zero and hence is the terminal velocity.

For your second value, the $0.01v^2$ is greater than 9g, which means the acceleration is actually against gravity which we can reject. If I were doing that question I wouldn't work out terminal velocity but I would state that it is necessary for the resistive force to be less than or equal to the weight force as acceleration can only be down. Or something like that.

EDIT: the reason the don't get your solution is, as you probably know, because of the modulus being absent. That should indicate to you that the difference is that you have a negative log which you changed to be positive. Intuitively you should be wary of that, but obviously you can't just reject it because of the fact that it was negative before the modulus
Title: Re: TrueTears question thread
Post by: Over9000 on November 01, 2009, 12:57:08 pm: Quote from: NE2000 on November 01, 2009, 12:16:45 pm
I think it's because when $a=0$

$9g = 0.01v^2$
Solving for v gives the value that bem9 gave. This is the velocity of the object when acceleration would be zero and hence is the terminal velocity.

For your second value, the $0.01v^2$ is greater than 9g, which means the acceleration is actually against gravity which we can reject. If I were doing that question I wouldn't work out terminal velocity but I would state that it is necessary for the resistive force to be less than or equal to the weight force as acceleration can only be down. Or something like that.

EDIT: the reason the don't get your solution is, as you probably know, because of the modulus being absent. That should indicate to you that the difference is that you have a negative log which you changed to be positive. Intuitively you should be wary of that, but obviously you can't just reject it because of the fact that it was negative before the modulus
Thanks for explaining NE2000 and bem9.
Title: Re: TrueTears question thread
Post by: TrueTears on November 01, 2009, 04:08:20 pm: Cool thanks guys!