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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: TyErd on April 24, 2010, 12:10:13 pm

Title: TyErd's questions
Post by: TyErd on April 24, 2010, 12:10:13 pm

solve: $2^{x+1} \times 3^{x-1} = 144$

Title: Re: TyErd's questions
Post by: Damo17 on April 24, 2010, 12:17:58 pm

Quote from: TyErd on April 24, 2010, 12:10:13 pm

solve: $2^{x+1} \times 3^{x-1} = 144$

simplify:

$2^x \cdot 2^1 \cdot 3^x \cdot 3^{-1} =144$

$\frac{2\cdot 6^x}{3}=144$

$6^x= \frac{144 \cdot 3}{2}=216$

$\therefore \ x=3$

Title: Re: TyErd's questions
Post by: TyErd on April 24, 2010, 12:24:19 pm

oook thanks I get it!

Title: Re: TyErd's questions
Post by: TyErd on April 24, 2010, 12:31:53 pm

How do you sketch the graph of $y=x^\tfrac{2}{3}$

Title: Re: TyErd's questions
Post by: superflya on April 24, 2010, 12:45:27 pm

as f(-x)=f(x) it is an even function. so reflect in y.

Title: Re: TyErd's questions
Post by: TyErd on April 24, 2010, 12:58:51 pm

what graph do you reflect though?

Title: Re: TyErd's questions
Post by: kyzoo on April 24, 2010, 01:37:14 pm

As the numerator of the index (2/3) is an even number, therefore the function will always be a positive number; it will always be above the x-axis. The denominator being an odd number means that the function exists for all values of x, rather than just positive values. So you have a function that exists in the 1st and 2nd quadrants.

It will have a similar appearance to y = x^0.5, just that it exists in both the 1st and 2nd quadrants. So you have a graph that looks like y = x^0.5 combined with the graph of y = x^0.5 reflected about the y-axis.

Title: Re: TyErd's questions
Post by: TyErd on April 24, 2010, 01:46:30 pm

so you have to pretty much memorise the shapes of those type of functions?

Title: Re: TyErd's questions
Post by: TyErd on April 24, 2010, 01:58:07 pm

When $f(x)= x^3-6x^2 + 9x - 4$ , Find the real values of h for which only one of the solutions of the equation $f(x+h)=0$ is positive.

Title: Re: TyErd's questions
Post by: Blakhitman on April 24, 2010, 03:20:04 pm

This is how I'd go about this question, should be right!

factorised: $(x+h-4)(x+h-1)^{2}$

basic shape will be,

(http://img51.imageshack.us/img51/71/asdasdasdq.jpg)

then we need to solve both solutions when they are >0

$-h+4>0$

$h<4$

And

$-h+1>0$

$h<1$

and the question asks for when only one is positive, so we'll use the more positive one, cause it'll always be more positive than the other one!

so the answer is: $1\leq h<4$

Title: Re: TyErd's questions
Post by: TyErd on April 24, 2010, 03:38:40 pm

okay i get it up to the point you drew thegraph lol, but i think its the question that im finding it difficult to understand. Whats the questions even asking. Can someone rephrase it for me?

Title: Re: TyErd's questions
Post by: Blakhitman on April 24, 2010, 03:41:05 pm

It's asking when is only one x intercept positive.

Title: Re: TyErd's questions
Post by: TyErd on April 24, 2010, 03:43:59 pm

OHHHHHHHHHH! i get it! so pretty much were shifting the graph left 'h' units until we have 1 solution, right?

Title: Re: TyErd's questions
Post by: Blakhitman on April 24, 2010, 04:01:18 pm

Well, we're trying to find the values of 'h' so that only one of the x-int. is positive.

and the x-ints, are given by $-h+1$ and $-h+4$

So what we do is solve for when these two intercepts are $>0$ .

we get $h<1$ and $h<4$ .

and since the question is asking when only one is positive, we need to eliminate when one of them is positive, so h will not be <1 so we are left with $1\leq h<4$ .

Hope this makes sense!

Title: Re: TyErd's questions
Post by: TyErd on April 24, 2010, 04:07:32 pm

Yeaah i get it thankyou! a Graph really does help! thnx

Title: Re: TyErd's questions
Post by: Blakhitman on April 24, 2010, 04:10:59 pm

Anytime :)

Title: Re: TyErd's questions
Post by: TyErd on April 24, 2010, 04:14:31 pm

I have another question. When $f(x)= x^3-6x^2 + 9x - 4$ describe the sequence of transformations which maps the graph of $y=f(x)$ on to the graph of $y=f(2x)+4$

okay so the $y=f(x)$ can be factorised into $y=(x-1)^2(x-4)$ and the new graph with transformations is $x(2x-3)^2$

I think I am correct with that part. If it is, how do I describe the transformaitons that have occured.

Title: Re: TyErd's questions
Post by: the.watchman on April 24, 2010, 04:16:06 pm

You won't need to know what f(x) is to do this:

- Dilation by factor 1/2 from y-axis (this is f(x) ----> f(2x))
- Translation of 4 units up (f(2x) -------> f(2x)+4)

:)

Title: Re: TyErd's questions
Post by: Blakhitman on April 24, 2010, 04:18:32 pm

Quote from: TyErd on April 24, 2010, 04:14:31 pm

I have another question. When $f(x)= x^3-6x^2 + 9x - 4$ describe the sequence of transformations which maps the graph of $y=f(x)$ on to the graph of $y=f(2x)+4$

okay so the $y=f(x)$ can be factorised into $y=(x-1)^2(x-4)$ and the new graph with transformations is $x(2x-3)^2$

I think I am correct with that part. If it is, how do I describe the transformaitons that have occured.

All you really need to do is state the transformations from $f(x)$ into $f(2x)+4$

which would be dilation by a factor of $\frac{1}{2}$ parallel to the x-axis and a translation of +4 units along the y-axis.

Beaten: the.watchman is here....no point in me replying anymore :P

Title: Re: TyErd's questions
Post by: the.watchman on April 24, 2010, 04:21:38 pm

Quote from: Blakhitman on April 24, 2010, 04:18:32 pm

Beaten: the.watchman is here....no point in me replying anymore :P

Oi! Don't make me sound like some repressive autocrat... :P

EDIT: I'm doing language analysis homework atm...

Title: Re: TyErd's questions
Post by: TyErd on April 24, 2010, 04:23:08 pm

Okay I understand that, but how come you dont substitue $2x$ into $f(x)+4$ .

Title: Re: TyErd's questions
Post by: Blakhitman on April 24, 2010, 04:24:13 pm

Quote from: the.watchman on April 24, 2010, 04:21:38 pm

Quote from: Blakhitman on April 24, 2010, 04:18:32 pm
Beaten: the.watchman is here....no point in me replying anymore :P

Oi! Don't make me sound like some repressive autocrat... :P

EDIT: I'm doing language analysis homework atm...

LOL well you do give no one a chance ;).

haha nah just saying you're too quick :P

Title: Re: TyErd's questions
Post by: the.watchman on April 24, 2010, 04:24:55 pm

The process of a dilation by factor 'k' from an axis is to replace x with $\frac{x}{k}$

So if x were to be replaced by 2x, then the dilation factor would be 1/2

Does that make sense?

Title: Re: TyErd's questions
Post by: Blakhitman on April 24, 2010, 04:31:58 pm

Remember these:

$f(nx)$ dilation of $\frac{1}{n}$ along the x-axis.

$nf(x)$ dilation of "n" along y-axis.

$-f(x)$ reflection on the x-axis.

$f(-x)$ reflection on the y-axis.

$f(x-a)$ translation of "a" units to the right.

$f(x+a)$ translation of "a" units to the left.

$f(x)+k$ translation of "k" units up.

$f(x)-k$ translation of "k" units down.

Title: Re: TyErd's questions
Post by: TyErd on April 24, 2010, 04:44:06 pm

yeh i get how to state the dilations and all but im use to substituting back into the original questions for example if we know f(x) and the questions says what is f(0), then I would just substitute 0 into x, do you get what i'm saying? In this example why wouldn't you substitute 2x into f(x) considering we know f(x)

Title: Re: TyErd's questions
Post by: Blakhitman on April 24, 2010, 04:49:02 pm

because it's only asking you to state the transformations, if it said state what f(2x) + 4 is, then you would substitute into the equation. For now it is only asking to state transformations from f(x), note that this can be any function, into f(2x) + 4

Title: Re: TyErd's questions
Post by: the.watchman on April 24, 2010, 04:49:51 pm

Yes, but the question is not: What is $f(2x)+4$ ?

It is to state the transformations from f(x) to f(2x)+4, in which case just apply your knowledge of transformations to the question :)

Title: Re: TyErd's questions
Post by: TyErd on April 24, 2010, 04:55:58 pm

omg thankyou , im such an idiot :idiot2:, you have no idea how dumb i feel lol also if it asks to state the x intercepts of y=f(2x)+4, then I would substitue 2x into the equation and the +4, then factorise to get the intercepts yeah?

Title: Re: TyErd's questions
Post by: the.watchman on April 24, 2010, 04:57:29 pm

Quote from: TyErd on April 24, 2010, 04:55:58 pm

omg thankyou , im such an idiot :idiot2:, you have no idea how dumb i feel lol also if it asks to state the x intercepts of y=f(2x)+4, then I would substitue 2x into the equation and the +4, then factorise to get the intercepts yeah?

Yeah, although the y-intercept should be 4 more than previous from intuition (the dilation does not affect the point at x=0)

Title: Re: TyErd's questions
Post by: TyErd on April 24, 2010, 05:07:19 pm

okay thankyou! Another question which I cant seem to find where I went wrong. if $f(x)=3-6cos(\frac{\pi (x-2)}{4})$ and $x \epsilon [0,12]$ , find the values of x if $f(x)=0$ .

Okay when I did this I got three x values 10/3, 26/3 and 34/3 all which are correct but I am missing 2/3 according to the answers.

Title: Re: TyErd's questions
Post by: the.watchman on April 24, 2010, 05:13:12 pm

Well, $cos(\frac{\pi(x-2)}{4}) = \frac{1}{2}$

Noting that $0 \leq x \leq 12$

$-2 \leq x-2 \leq 10$

$-\frac{\pi}{2} \leq \frac{\pi(x-2)}{4} \leq \frac{5\pi}{2}$

So $\frac{\pi(x-2)}{4} = -\frac{\pi}{3}, \frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}$

$x = \frac{2}{3}, \frac{10}{3}, \frac{26}{3}, \frac{34}{3}$

Here you go!

EDIT: Be careful when limiting solutions, I find it safest to write the ineqn process like above, particularly for ones that are more complicated :)

Title: Re: TyErd's questions
Post by: Blakhitman on April 24, 2010, 05:16:10 pm

Yep, except watchman began adding pies unnecessarily. lol

Title: Re: TyErd's questions
Post by: the.watchman on April 24, 2010, 05:18:50 pm

Quote from: Blakhitman on April 24, 2010, 05:16:10 pm

Yep, except watchman began adding pies unnecessarily. lol

Lol, edited ;D
That's what happens when you copy LaTeX code everywhere...

Title: Re: TyErd's questions
Post by: TyErd on April 24, 2010, 05:36:22 pm

thankyou I found where I made my mistake I did 0-2 = 0. LOL stupid me. thanks guys.

Title: Re: TyErd's questions
Post by: the.watchman on April 24, 2010, 05:37:29 pm

Quote from: TyErd on April 24, 2010, 05:36:22 pm

thankyou I found where I made my mistake I did 0-2 = 0. LOL stupid me. thanks guys.

No probs, always happy to help :)

(Oh congrats on becoming a VN contributor :D)

Title: Re: TyErd's questions
Post by: TyErd on April 24, 2010, 05:41:22 pm

yayyyyyyyy its funny though because i've mainly been getting help from people not really contributing :)

Title: Re: TyErd's questions
Post by: TyErd on April 24, 2010, 05:46:53 pm

oh another important question about inequalities, when is it that you switch the inequality. Is it when you times or divide by a negative number?

Title: Re: TyErd's questions
Post by: superflya on April 24, 2010, 05:51:41 pm

^ both.

Title: Re: TyErd's questions
Post by: TyErd on April 24, 2010, 05:54:33 pm

ok thnx

Title: Re: TyErd's questions
Post by: TyErd on April 24, 2010, 06:01:50 pm

$g(x)=5x-1, find x :\frac{1}{g(2x)} =3, g(x)\neq 0)$

do I substitute 2x into x of g(x) and then proceed to solve for x or do I multiple g(x) by 2 and then find x?

Title: Re: TyErd's questions
Post by: Damo17 on April 24, 2010, 06:21:44 pm

Quote from: TyErd on April 24, 2010, 06:01:50 pm

$g(x)=5x-1, find x :\frac{1}{g(2x)} =3, g(x)\neq 0)$

do I substitute 2x into x of g(x) and then proceed to solve for x or do I multiple g(x) by 2 and then find x?

$g(2x)=5(2x)-1=10x-1$

Title: Re: TyErd's questions
Post by: TyErd on April 24, 2010, 07:06:24 pm

okay a worded question. A capsule is formed from a cone, a cylinder and a hemisphere. The height of the capsule is 60cm. The radius of the cylinder is r cm and the height of the cone is r cm.

Find the volume of the capsule in terms of r which I calculated to be $V=-\pi r^2 (r-60)$

State the maximal domain of the function.

Title: Re: TyErd's questions
Post by: brightsky on April 24, 2010, 07:15:23 pm

The restriction is the height of the capsule, or else $r \in R$ (theoretically). So $r + r + h = 60$ , where $h$ cm is the height of the cone. Hence $2r + h = 60$ . The lowest $h$ can get is 0 cm, hence $2r = 60 \implies r = 30 \text{cm}$ .

Not too sure on this one though...

Title: Re: TyErd's questions
Post by: TyErd on April 24, 2010, 07:27:06 pm

why do you have to know the lowest h can get?

Title: Re: TyErd's questions
Post by: the.watchman on April 24, 2010, 07:48:51 pm

The maximal domain is (0,30) (I think your volume equation looks odd, but I haven't checked it)

This is because of two reasons:

1. COMMON SENSE: The volume cannot be equal to zero and must be positive, so by looking at the intercepts of the graph, you SHOULD get (0,30)

2. The limiting information is that the height of the whole capsule is 60cm. This means that you have your eqn $2r+h=60$

So note that, as r increases, h decreases and vice versa
The two extremes of r are when r=0, or when h=0 <=> r=30

Title: Re: TyErd's questions
Post by: TyErd on April 24, 2010, 08:07:51 pm

I dont get how you can use this height limitation to determine the maximal domain. Why are the extremes when r=0

Title: Re: TyErd's questions
Post by: the.watchman on April 24, 2010, 08:13:16 pm

The radius cannot be less than zero, or equal to zero to have a capsule at all, yes?
So r>0 (1)

Also, the height cannot be less than zero, or equal to zero to have a capsule
So h>0

$60-2r>0$

$2r<60$

$r<30$ (2)

So combining (1) and (2), you have $0<r<30$

Does that make more sense?

Title: Re: TyErd's questions
Post by: TyErd on April 24, 2010, 08:22:06 pm

okay thats makes loads more sense. So how is radius used to find maximal domain?

Title: Re: TyErd's questions
Post by: the.watchman on April 24, 2010, 08:23:20 pm

The radius is the main variable of the eqn, so the domain is the possible values of r

Therefore, the domain is (0,30) (as set out above)

Title: Re: TyErd's questions
Post by: TyErd on April 24, 2010, 08:27:49 pm

ohhhhh I get it now thankyou the.watchman!

Title: Re: TyErd's questions
Post by: TyErd on April 24, 2010, 08:32:42 pm

Okay another problem: Find the general solution of the equation $2cos ((\pi - 3x )\div 2)=-1$

Write answers in exact values

Title: Re: TyErd's questions
Post by: superflya on April 24, 2010, 08:39:53 pm

bring the 2 over. u should be able to do it.

Title: Re: TyErd's questions
Post by: the.watchman on April 24, 2010, 08:40:42 pm

$cos(\frac{-3x+\pi}{2}) = -\frac{1}{2}$

$\frac{-3x+\pi}{2} = 2n\pi + \frac{2\pi}{3}, 2n\pi + \frac{4\pi}{3}$

$-3x+\pi=4n\pi + \frac{4\pi}{3}, 4n\pi + \frac{8\pi}{3}$

$-3x=4n\pi + \frac{\pi}{3}, 4n\pi + \frac{5\pi}{3}$

$x=-\frac{4n\pi + \frac{\pi}{3}}{3}, -\frac{4n\pi + \frac{5\pi}{3}}{3}$

$x=-\frac{(12n+1)\pi}{9}, -\frac{(12n+5)\pi}{9}$

Why do I get the feeling I've stuffed it up...? :P

EDIT: Oops, $n \in Z$

Title: Re: TyErd's questions
Post by: TyErd on April 24, 2010, 08:55:20 pm

Hm.. this is what I did:

$\frac{\pi - 3x}{2} = 2n\pi \pm \frac{2\pi}{3}$
$\pi - 3x =4n\pi\pm \frac{4\pi}{3}$
$-3x= 4n\pi - \pi \pm \frac{4\pi}{3}$
$x= \frac{4n\pi - \pi}{-3} + \frac{4\pi}{9}$

but I know thats wrong coz im always wrong

Title: Re: TyErd's questions
Post by: the.watchman on April 24, 2010, 09:00:11 pm

Looks fine to me, just mine slightly different
I'm not entirely sure though
But you need a plus/minus in the last line :P

Title: Re: TyErd's questions
Post by: TyErd on April 24, 2010, 09:02:49 pm

Wouldn't the last line only be a plus because when you take the -3 over it becomes positive?

Title: Re: TyErd's questions
Post by: the.watchman on April 24, 2010, 09:05:16 pm

Nope, there are two cases for the plus/minus

CASE 1: Second last line is Positive

$\frac{Positive}{-3} = Negative$

CASE 2: Second last line is Negative

$\frac{Negative}{-3} = Positive$

So the result could be +ve or -ve, therefore the last line should have a $\pm$

Title: Re: TyErd's questions
Post by: TyErd on April 24, 2010, 09:08:54 pm

Oh yeah your right its should be plus/minus. Anyway how come we have different answers

Title: Re: TyErd's questions
Post by: the.watchman on April 24, 2010, 09:12:10 pm

Quote from: TyErd on April 24, 2010, 09:08:54 pm

Oh yeah your right its should be plus/minus. Anyway how come we have different answers

You did $2n\pi \pm \frac{2\pi}{3}$

I did $2n\pi + \frac{2\pi}{3}, 2n\pi + \frac{4\pi}{3}$

They are equivalent (you 'built around' your n-pi's, I 'built over' my n-pi's if you get that)

Title: Re: TyErd's questions
Post by: TyErd on April 24, 2010, 09:13:47 pm

No but i'd like to know :)

Title: Re: TyErd's questions
Post by: the.watchman on April 24, 2010, 09:15:41 pm

Just think: $2\pi - \frac{2\pi}{3} = \frac{4\pi}{3}$ :)

Title: Re: TyErd's questions
Post by: TyErd on April 24, 2010, 09:19:55 pm

ohh ..so instead of going from the negative side of the unit circle you have gone the positive way to the same point. yeah?

Title: Re: TyErd's questions
Post by: the.watchman on April 24, 2010, 09:36:35 pm

Quote from: TyErd on April 24, 2010, 09:19:55 pm

ohh ..so instead of going from the negative side of the unit circle you have gone the positive way to the same point. yeah?

Yup :)
So it's the same thing

Title: Re: TyErd's questions
Post by: TyErd on April 24, 2010, 09:44:50 pm

i like it :)

Title: Re: TyErd's questions
Post by: TyErd on April 25, 2010, 10:55:02 am

What is $f(x)$ if $f(x+3)=4x^2+6x$

Title: Re: TyErd's questions
Post by: brightsky on April 25, 2010, 02:40:36 pm

Let $f(x) = ax^2 + bx + c$ , where $a, b, c$ are constants.

Hence, $f(x + 3) = a(x+3)^2 + b(x+3) + c = a(x^2 + 6x + 9) + bx + 3b + c = ax^2 + 6ax + 9a + bx + 3b + c = ax^2 + (6a + b)x + (9a + 3b + c)$

We are given that $f(x+3) = 4x^2 + 6x$ .

That would mean $ax^2 + (6a+b)x + (9a + 3b + c) = 4x^2 + 6x$

$\therefore a = 4$ ...(1), $6a + b = 6$ ...(2) and $9a + 3b + c = 0$ ...(3)

Substitute (1) into (2):

$6 \cdot 4 + b = 6$

$24 + b = 6$

$b = -18$ ...(4)

Substitute (1) and (4) into (3):

$9(4) + 3(-18) + c = 0$

$36 - 54 + c = 0$

$-18 + c = 0$

$c = 18$ ...(5)

From (1), (4) and (5):

$\boxed{f(x) = 4x^2 - 18x + 18}$

Title: Re: TyErd's questions
Post by: kamil9876 on April 25, 2010, 02:57:47 pm

This is a lot simpler (and more rigorous, how do you know it should be a quadratic?, it is but its more difficult to show that than to actually do the question):

$f(x)=f((x-3)+3)=f(u+3)=4u^2+6u=4(x-3)^2 + 6(x-3)$

where $u=x-3$

(not neccesary to let u=x+3 but I did for clarity's sake)

Title: Re: TyErd's questions
Post by: TyErd on April 25, 2010, 03:07:25 pm

why do you let u=x-3?

Title: Re: TyErd's questions
Post by: brightsky on April 25, 2010, 03:12:51 pm

$f(x) = f((x-3) + 3)$

$f(u + 3) = f(x)$

$\because f(x+3) = 4x^2 + 6u$

$\therefore f(u+3) = 4u^2 + 6u$ $\text{This is the step that makes letting u = x-3 so important.}$

Substitute $u = x - 3$ back in.

$f(x) = 4(x-3)^2 + 6(x-3)$

Nice method, kamil! ;D

EDIT: I got a question though, how can f(x) not be degree 2 when f(x+3) is degree 2?

Title: Re: TyErd's questions
Post by: TyErd on April 25, 2010, 03:33:45 pm

I dont get your first line though.

Title: Re: TyErd's questions
Post by: brightsky on April 25, 2010, 03:40:59 pm

The aim is to let $f(x) = f(u + 3)$ . This is because you are already given what $f(x+3)$ is and if you let $f(x) = f(u+3)$ , you can easily find out what f(u+3) is hence finding out what f(x) is.

That is why u = x - 3, as $f((x - 3) + 3) = f(x)$ but it still is in the form $f(u + 3)$ .

Title: Re: TyErd's questions
Post by: the.watchman on April 25, 2010, 03:41:46 pm

It is just trying to get the function with (something-in-terms-of-x)+3, in this case, $x=(x-3)+3$

I agree, nice method!

Title: Re: TyErd's questions
Post by: TyErd on April 25, 2010, 03:53:44 pm

ooh I get it thankyou!

Title: Re: TyErd's questions
Post by: kamil9876 on April 25, 2010, 06:00:08 pm

Quote

EDIT: I got a question though, how can f(x) not be degree 2 when f(x+3) is degree 2?

yeah it always is of degree two. Just found it wierd that you knew a fact that seemed more difficult than what the solution needed.

edit:

in general, say i am given $f(g(x))$ , and i want to find $f(x)$ .

Then i just use:

$ f(x)=f(g(g^{-1}( x)))$

Example: $f:\mathbb{R}^+ \to \mathbb{R}, f(2e^x)=x^2$ , what's f(x)?

Title: Re: TyErd's questions
Post by: TyErd on April 25, 2010, 08:09:57 pm

hmm how would you do that kamil?

Title: Re: TyErd's questions
Post by: kamil9876 on April 25, 2010, 09:24:10 pm

just expanding on the ideas of the previous post:

$ f(x)=f(2e^t)=t^2$

ie: we let $x=2e^t$

then that means $t=log_e (x/2)$

so $f(x)=(log_e(x/2))^2$

Title: Re: TyErd's questions
Post by: TyErd on April 25, 2010, 09:35:28 pm

ohk I getcha, thankyou!

Title: Re: TyErd's questions
Post by: TyErd on April 25, 2010, 11:37:26 pm

Given that $q^p = 25$ find $\log_5 q$ in terms of p.

Title: Re: TyErd's questions
Post by: TyErd on April 25, 2010, 11:52:45 pm

Also solve the equation: $\log_5 x = 16\log_x 5$

Title: Re: TyErd's questions
Post by: superflya on April 26, 2010, 12:08:16 am

$16\log _{x}5=\frac{16}{\log _{5}x}$

just a hint.

Title: Re: TyErd's questions
Post by: TyErd on April 26, 2010, 12:13:25 am

how did you get another 16 in there?

Title: Re: TyErd's questions
Post by: kamil9876 on April 26, 2010, 12:26:05 am

Quote from: TyErd on April 25, 2010, 11:37:26 pm

Given that $q^p = 25$ find $\log_5 q$ in terms of p.

$log_5 q=\frac{1}{p} log_5 q^p=\frac{1}{p}log_5 25=\frac{2}{p}$

Title: Re: TyErd's questions
Post by: moekamo on April 26, 2010, 12:27:05 am

change of base rule, $16 \log_x 5 =16 \cdot \frac{\log_5 5}{\log_5 x} = \frac{16}{\log_5 x}$

so for your question, $\log_5 x = \frac{16}{\log_5 x} \implies (\log_5 x)^2= 16$

$\implies \log_5 x = \pm 4, x=5^{-4} \mbox{ or } 5^4$

other question, $q^p = 25 = 5^2$
$\implies \log_5 q^p = \log_5 5^2$
$\implies p \log_5 q = 2$
$\implies \log_5 q = 2/p$

Title: Re: TyErd's questions
Post by: TyErd on April 26, 2010, 12:35:06 am

OHhh! i get it! thanks so much! I completely forgot about changing bases. It can be used for just about every problem. Thats awesome. Thnx

Title: Re: TyErd's questions
Post by: TyErd on April 26, 2010, 12:42:23 am

Another problem thats really confusing me: For $\pi < x < \frac {3\pi} {2} with \cos x = -\sin \frac {\pi} {6}$ .

Find the value of $x$

Title: Re: TyErd's questions
Post by: Blakhitman on April 26, 2010, 12:47:26 am

well it's like anything else, just sub the exact value of -sin(pie/6) and solve for x like any other trig function.

Sorry on iPod else I would have done it.

Title: Re: TyErd's questions
Post by: TyErd on April 26, 2010, 01:00:59 am

ohhh how stupid am I for not realising that! My answer is $\frac {7\pi} {6}$ . Can anyone confirm?

Title: Re: TyErd's questions
Post by: TyErd on April 26, 2010, 01:23:48 am

It can't be $\frac {5\pi} {3}$ , because $x$ is between $\pi and \frac {3\pi} {2}$

Title: Re: TyErd's questions
Post by: mandy on April 26, 2010, 01:24:35 am

Yeah, it's $\frac{4\pi}{3}$ , sorry.

Edit: Here we go.

$cos x = - sin \frac{\pi}{6}$
$cos x = \frac{-1}{2}$
cos is negative in the 3rd quad, so:
$x = \frac{4\pi}{3}$ :)

Title: Re: TyErd's questions
Post by: TyErd on April 26, 2010, 01:25:39 am

hmm. how'd you get that?

Title: Re: TyErd's questions
Post by: TyErd on April 26, 2010, 01:28:02 am

oh crap i just realised my mistake. Your right mandy!

Title: Re: TyErd's questions
Post by: TyErd on April 26, 2010, 03:06:43 am

The rule for the inverse relation of the function with rule $y=x^2 - 4x + 5$ is ?

and also whats the difference between the inverse function and inverse relation?

Title: Re: TyErd's questions
Post by: moekamo on April 26, 2010, 04:55:20 am

$y=x^2-4x+5,\mbox{ swap } x \mbox{ and } y$

$\implies x= y^2 - 4y + 5 = (y^2-4y +4) - 4 + 5 = (y-2)^2 +1$

$\implies x-1= (y-2)^2$

$\implies y= \pm \sqrt{x-1} + 2$ which is the inverse relation, since it contains both positive and negative square roots, if it were a function it could only have either positive or negative square roots

Title: Re: TyErd's questions
Post by: TyErd on April 26, 2010, 09:57:00 am

OH okay! thankyou for clearing that up.

Title: Re: TyErd's questions
Post by: TyErd on April 26, 2010, 10:01:53 am

so basically for an inverse relation, it doesn't have to be a one to one function?

Title: Re: TyErd's questions
Post by: the.watchman on April 26, 2010, 10:09:37 am

Quote from: TyErd on April 26, 2010, 10:01:53 am

so basically for an inverse relation, it doesn't have to be a one to one function?

A relation is basically an eqn linking two variables, it can be a function but it doesn't have to be.
So if you are asked to find the inverse relation, you use the same process, however, be aware that your answer may not be a function

Title: Re: TyErd's questions
Post by: TyErd on April 26, 2010, 10:54:42 am

ohhh so all functions are relations right?...and for this one $y=x^2 - 4x + 5$ if i was told to find the inverse function then first I would have to make it a one to one function by restricting the domain and then proceed to find the inverse but if it asked for the inverse relation i dont have to restrict the domain.

Title: Re: TyErd's questions
Post by: Blakhitman on April 26, 2010, 10:57:31 am

And then the positive or negative square root depends on how you restricted the domain.

Title: Re: TyErd's questions
Post by: TyErd on April 26, 2010, 10:59:57 am

Yes, thankyou! makes so much more sense now :)

Title: Re: TyErd's questions
Post by: the.watchman on April 26, 2010, 11:00:26 am

Note, the choice of which root to take depends on the range restriction of the inverse (remembering that $Dom f = Ran f^{-1}$ )

Title: Re: TyErd's questions
Post by: TyErd on April 26, 2010, 11:03:56 am

yup :)

Title: Re: TyErd's questions
Post by: TyErd on April 26, 2010, 11:32:52 am

if $sin\theta = 0.3 cos\beta = 0.8$ Find: $sin (\frac {\pi} {2} - \beta)$

Title: Re: TyErd's questions
Post by: the.watchman on April 26, 2010, 11:37:20 am

Well, using the complementary angle formulas,

$\sin (\frac{\pi}{2}-\beta)=\cos \beta=0.8$

Title: Re: TyErd's questions
Post by: TyErd on April 26, 2010, 11:52:03 am

can you use identities to solve this?

Title: Re: TyErd's questions
Post by: the.watchman on April 26, 2010, 11:57:03 am

Quote from: TyErd on April 26, 2010, 11:52:03 am

can you use identities to solve this?

Of course, you can use whatever formulas you like, as long as you can get the answer with the information they give you :)

Title: Re: TyErd's questions
Post by: TyErd on April 26, 2010, 12:01:45 pm

following on from the previous question, how would you solve $cos(\frac {\pi} {2} + \theta )$

if you can can you show me how to do it with complementary angles and identites please :)

Title: Re: TyErd's questions
Post by: the.watchman on April 26, 2010, 12:05:31 pm

Quote from: TyErd on April 26, 2010, 12:01:45 pm

following on from the previous question, how would you solve $cos(\frac {\pi} {2} + \theta )$

if you can can you show me how to do it with complementary angles and identites please :)

By definition (and a unit circle diagram), $\cos (\frac{\pi}{2}+\theta) = -\sin x = -0.3$ from you previous post :)

Title: Re: TyErd's questions
Post by: Stroodle on April 26, 2010, 12:08:07 pm

Using the compound angle formula:

$cos(\frac{\pi}{2}+\theta)=cos({\frac{\pi}{2})cos(\theta)-sin(\frac{\pi}{2})sin(\theta)$

$=0\times cos(\theta)-1\times sin(\theta)$

$=-sin(\theta)$

Title: Re: TyErd's questions
Post by: the.watchman on April 26, 2010, 12:09:08 pm

Quote from: Stroodle on April 26, 2010, 12:08:07 pm

Using the compound angle formula:

$cos(\frac{\pi}{2}+\theta)=cos({\frac{\pi}{2})cos(\theta)-sin(\frac{\pi}{2})sin(\theta)$

$=0\times cos(\theta)-1\times sin(\theta)$

$=-sin(\theta)$

Erm ... it's miles easier to use the complementary angle formulas... :P

Title: Re: TyErd's questions
Post by: TyErd on April 26, 2010, 12:11:49 pm

I just realised I had none of these formulas on my cheat sheet. Thanks!

Title: Re: TyErd's questions
Post by: Stroodle on April 26, 2010, 12:14:13 pm

Quote

Erm ... it's miles easier to use the complementary angle formulas... :P

For sure. I just thought he was asking for a different way to do it.

Title: Re: TyErd's questions
Post by: the.watchman on April 26, 2010, 12:16:18 pm

Quote from: Stroodle on April 26, 2010, 12:14:13 pm

Quote

Erm ... it's miles easier to use the complementary angle formulas... :P

For sure. I just thought he was asking for a different way to do it.

Quote from: TyErd on April 26, 2010, 12:01:45 pm

if you can can you show me how to do it with complementary angles and identites please :)

Really? :D
(jks jks)

Title: Re: TyErd's questions
Post by: TyErd on April 26, 2010, 12:24:59 pm

Hm.. the cambridge essentials book doesn't have any double angle formulae questions. Anyone have a tricky double angle problem that will help me revise for the SAC?

Title: Re: TyErd's questions
Post by: TyErd on April 26, 2010, 01:48:28 pm

I'm not 100% sure about this one. I know that its either A or E. When it says $f(2x - 1)$ , is the 1 a vertical translation or is that part of the horizontal?

Title: Re: TyErd's questions
Post by: Blakhitman on April 26, 2010, 02:10:25 pm

f(2x-1): DIlation by factor of 1/2 along the x-axis. horizontal translation of 1/2 units to the right,

Title: Re: TyErd's questions
Post by: TyErd on April 26, 2010, 02:13:52 pm

ok, so for it to be a vertical translation it would be written as: f(2x-1)-1 yeah?

Title: Re: TyErd's questions
Post by: Blakhitman on April 26, 2010, 02:29:03 pm

yep!

I'd recommend remembering this:

Quote from: Blakhitman on April 24, 2010, 04:31:58 pm

Remember these:

$f(nx)$ dilation of $\frac{1}{n}$ along the x-axis.
$nf(x)$ dilation of "n" along y-axis.
$-f(x)$ reflection on the x-axis.
$f(-x)$ reflection on the y-axis.
$f(x-a)$ translation of "a" units to the right.
$f(x+a)$ translation of "a" units to the left.
$f(x)+k$ translation of "k" units up.
$f(x)-k$ translation of "k" units down.

Title: Re: TyErd's questions
Post by: TyErd on April 26, 2010, 02:42:21 pm

thnx!

Title: Re: TyErd's questions
Post by: Blakhitman on April 26, 2010, 02:47:11 pm

Also note in you previous example, f(2x-1) to get it in the form to determine transformations, it's easier to factorise: $f(2(x-1/2))$

Title: Re: TyErd's questions
Post by: TyErd on April 26, 2010, 02:54:26 pm

yup thnx!

Title: Re: TyErd's questions
Post by: TyErd on April 26, 2010, 04:01:25 pm

In logarithms where for example the question is $3\log_3 (3x+2) = -3$ , you would take the 3 onto the other side instead of making it $\log_3 (3x+2)^3$ . Do we have to divide or can we still get the answer by cubing the $3x+2$

Title: Re: TyErd's questions
Post by: the.watchman on April 26, 2010, 04:05:26 pm

Yes, the simplest way to solve this is to divide both sides by three, before converting to exponential format

Title: Re: TyErd's questions
Post by: TyErd on April 26, 2010, 04:07:25 pm

if you want can you still get the answer by cubing it?

Title: Re: TyErd's questions
Post by: m@tty on April 26, 2010, 04:10:56 pm

Quote from: TyErd on April 26, 2010, 04:07:25 pm

if you want can you still get the answer by cubing it?

Yep.

$\log_3(3x+2)^3=-3$

$(3x+2)^3=3^{-3}$

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

Title: Re: TyErd's questions
Post by: TyErd on April 26, 2010, 04:13:09 pm

ok thnx

Title: Re: TyErd's questions
Post by: TyErd on April 26, 2010, 06:58:32 pm

Um. What happens to the sign of the inequality if the question asked to solve for x in the equation $x^2>2$

Title: Re: TyErd's questions
Post by: m@tty on April 26, 2010, 07:05:36 pm

With inequalities it is safer to always sketch. If you don't you can miss cases, or just get it plain wrong.

So when you sketch the parabola and the line y=2, you see that the parabola is greater for values of x less than the lower intercept, and x greater than the greater intercept. Then figure out the intercepts ( $\pm\sqrt{2}$ ).

So,
$x^2>2 \implies x>\sqrt{2} \ OR \ x<-\sqrt{2}$

But square rooting does not change the inequality.

So you could say:

$x^2>2 \implies \sqrt{x^2}>\sqrt{2} \implies |x|>\sqrt{2}$ .

Title: Re: TyErd's questions
Post by: TyErd on April 26, 2010, 07:13:44 pm

thnx so much for that, really good explanation. Also can the base of an exponential be a negative number. Example $a^{2x+6}=1$ , can 'a' take a negative value?

Title: Re: TyErd's questions
Post by: brightsky on April 26, 2010, 07:36:24 pm

Of course.

Example: $(-1)^{2} = 1$

Title: Re: TyErd's questions
Post by: TyErd on April 27, 2010, 05:31:47 pm

In, $\log_a 1 = 0$ , can 'a' take a value of 0 or less? My teacher said it cannot but I wasn't convinced. Im pretty sure it cannot be zero but what about negative values?

Title: Re: TyErd's questions
Post by: brightsky on April 27, 2010, 05:35:20 pm

It can.

And I was told that $0^0 = 1$ as well, or a lot of maths concepts would be flawed otherwise.

Have a read of this: http://mathforum.org/dr.math/faq/faq.0.to.0.power.html

Title: Re: TyErd's questions
Post by: TyErd on April 27, 2010, 05:42:51 pm

thats very interesting I never knew that 'a' could be zero. It's logical to assume it cant but damn thats pretty interesting.

Title: Re: TyErd's questions
Post by: TyErd on April 27, 2010, 05:43:28 pm

but 'a' can defintely take negative values right?

Title: Re: TyErd's questions
Post by: m@tty on April 27, 2010, 05:48:11 pm

Yep. Think about the statement " $\log_a1=0$ " it is the same as saying $a^0=1$ Now what values can 'a' take where this holds true? All real numbers. So yes, it most definitely can be negative.

Title: Re: TyErd's questions
Post by: TyErd on April 27, 2010, 05:49:41 pm

sweet thnx!

Title: Re: TyErd's questions
Post by: TyErd on April 27, 2010, 10:08:27 pm

whats the easiest and fastest way to sketch this graph: $y=|x+4| + |x-4|$

Title: Re: TyErd's questions
Post by: brightsky on April 27, 2010, 10:22:23 pm

Sketch y = |x + 4| and y = |x - 4| and do addition of ordinates.

Title: Re: TyErd's questions
Post by: m@tty on April 27, 2010, 10:41:36 pm

Take the definition of $|x|=\begin{cases}2, x\geq 0 \\ -x, x<0\end{cases}$

So, $|x+4|=\begin{cases} x+4, x\geq -4 \\ -x-4, x<-4\end{cases}$

And, $|x-4|=\begin{cases}x-4, x\geq 4 \\ -x+4, x<4\end{cases}$

So with $y$ there are three different cases: $x\geq4$ , $-4\leq x<4$ and $x<-4$ .

$y=\begin{cases} x+4+x-4 , x \geq 4 \\ x+4-x+4 , -4 \leq x < 4 \\ -x-4-x+4, x<-4 \end{cases} = \begin{cases} 2x, x\geq 4 \\ 8 , -4\leq x <4 \\ -2x , x<-4\end{cases}$ .

This isn't the fastest way, though it doesn't take too much longer.

Title: Re: TyErd's questions
Post by: TyErd on April 27, 2010, 11:14:27 pm

If $f(x) = \sin 2x$ and $g(x)= 2 \sin x$ , find $(f+g)(\frac {3\pi}{2})$

EDIT: Sorry i forgot to add in the $\frac {3\pi}{2}$

Title: Re: TyErd's questions
Post by: m@tty on April 27, 2010, 11:25:32 pm

What do you want help with?

$(f+g)(x)=f(x)+g(x)$ .

$=\sin{2x}+2\sin{x}=2\cos{x}\sin{x}+2\sin{x}=2\sin{x}(\cos{x}+1)$ .

Title: Re: TyErd's questions
Post by: TyErd on April 27, 2010, 11:51:21 pm

what values can 'a' take in $\log_a x$ . I know that the x value cannot be 0 or a negative number. What about 'a'

Title: Re: TyErd's questions
Post by: m@tty on April 28, 2010, 12:09:53 am

Again, take the more familiar exponential form. $y=\log_ax \implies a^y=x$ .

What values can 'a' take here? 'a' can be any real number, including negatives. But in Methods I am fairly sure that you only consider positive bases.

And what you said about x having to be positive. What about, given that there are negative bases, $\log_{-2}-8$ , is it defined?

Title: Re: TyErd's questions
Post by: TyErd on April 28, 2010, 12:25:09 am

no?

Title: Re: TyErd's questions
Post by: m@tty on April 28, 2010, 12:26:40 am

$\log_{-2}-8=\log_{-2}(-2)^3=3\times 1$

Title: Re: TyErd's questions
Post by: TyErd on April 28, 2010, 12:31:06 am

how come i cant sketch the graph on my calculator?

Title: Re: TyErd's questions
Post by: moekamo on April 28, 2010, 02:53:11 am

Quote from: m@tty on April 28, 2010, 12:09:53 am

Again, take the more familiar exponential form. $y=\log_ax \implies a^y=x$ .

What values can 'a' take here? 'a' can be any real number, including negatives. But in Methods I am fairly sure that you only consider positive bases.

And what you said about x having to be positive. What about, given that there are negative bases, $\log_{-2}-8$ , is it defined?

hmm, $y=a^x$ isnt really defined for $a<0$ since for fractional values you will keep getting answers in the complex number system since you cannot take square, 1/4, 1/6 etc roots of a negative number in real numbers so there will be a bunch of points but it is not a continuous function. Thats probably why it doesnt graph on a calculator anyway...

Title: Re: TyErd's questions
Post by: TyErd on May 03, 2010, 04:11:15 pm

Differentiate: $2x- \frac{5}{3x^{-3}} + \frac{1}{3\sqrt{x}}$

Title: Re: TyErd's questions
Post by: Blakhitman on May 03, 2010, 04:26:24 pm

$2x- \frac{5}{3x^{-3}} + \frac{1}{3\sqrt{x}}$

is the same as

$2x- \frac{5x^{3}}{3} + \frac{x^{-\frac{1}{2}}}{3}$

Then it's just simple power rule.

$2-5x^{2}-\frac{x^{-\frac{3}{2}}}{6}$

positive powers, you get:

$2-5x^{2}-\frac{1}{6x^{\frac{3}{2}}}$

Title: Re: TyErd's questions
Post by: m@tty on May 03, 2010, 04:28:44 pm

This is true.

let $y=2x-\frac{5}{3x^{-3}}+\frac{1}{3\sqrt{x}}=2x-\frac{5}{3}x^3+\frac{1}{3}x^{-\frac{1}{2}}$ .

$\frac{dy}{dx}=2-5x^2-\frac{1}{6}x^{-\frac{3}{2}}$

$=2-5x^2-\frac{1}{6\sqrt{x^3}}$ .

Title: Re: TyErd's questions
Post by: TyErd on May 03, 2010, 04:35:35 pm

ohk thnx guys i see where i went wrong. thnks!

Title: Re: TyErd's questions
Post by: TyErd on May 03, 2010, 04:50:24 pm

For the graph of $y= 4x^3 + 27x^2 - 30x +10$ the subset of R for which the gradient is negative is given by the interval........?

Title: Re: TyErd's questions
Post by: TyErd on May 03, 2010, 04:56:03 pm

Differentiate $f(x) = |2x+4|$ . What are the steps to differentiating a modulus function.

Title: Re: TyErd's questions
Post by: the.watchman on May 03, 2010, 05:05:49 pm

Quote from: TyErd on May 03, 2010, 04:56:03 pm

Differentiate $f(x) = |2x+4|$ . What are the steps to differentiating a modulus function.

You can use a chain rule approach, or this:

When the inside of the mod is positive (eg. when $2x+4\geq 0$ ), then the derivative is [2x-4]'

When the inside of the mod is positive (eg. when $2x+4<0$ ), then f(x) becomes -2x-4, so the derivative is [-2x-4]'

Hence the derivative can be written as a hybrid function according to the above

Title: Re: TyErd's questions
Post by: the.watchman on May 03, 2010, 05:08:58 pm

Quote from: TyErd on May 03, 2010, 04:50:24 pm

For the graph of $y= 4x^3 + 27x^2 - 30x +10$ the subset of R for which the gradient is negative is given by the interval........?

Well, first off, $\frac{dy}{dx} = 12x^2 + 54x - 30$

For the question, $\frac{dy}{dx} < 0$

So $12x^2 + 54x - 30<0$

$2x^2 + 9x - 5<0$

$(2x-1)(x+5)<0$

Solve either graphically or algebraically to get:

$-5<x<\frac{1}{2}$

Title: Re: TyErd's questions
Post by: TyErd on May 03, 2010, 05:15:05 pm

Quote from: the.watchman on May 03, 2010, 05:08:58 pm

Quote from: TyErd on May 03, 2010, 04:50:24 pm
For the graph of $y= 4x^3 + 27x^2 - 30x +10$ the subset of R for which the gradient is negative is given by the interval........?

Well, first off, $\frac{dy}{dx} = 12x^2 + 54x - 30$

For the question, $\frac{dy}{dx} < 0$

So $12x^2 + 54x - 30<0$

$2x^2 + 9x - 5<0$

$(2x-1)(x+5)<0$

Solve either graphically or algebraically to get:

$-5<x<\frac{1}{2}$

okay I get that part except how come when i solve algebraicaly i get x<-5. When i do it graphically i get it right.
and the modulus one im still very confused

Title: Re: TyErd's questions
Post by: m@tty on May 03, 2010, 05:35:09 pm

$f(x)=|2x+4|=\sqrt{(2x+4)^2}$

Or you can say

$f(x)=\begin{cases} 2x+4, x\geq -2 \\ -2x-4, x<-2\end{cases}$

The piece wise defined function is much easier to differentiate. As you don't have to use the chain rule twice...

$f'(x)=\begin{cases}\frac{d}{dx}(2x+4), x> -2 \\ \frac{d}{dx}(-2x-4), x<-2\end{cases} = \begin{cases}2, x>- 2 \\ -2, x<-2\end{cases}$

NOTE: There is no value for the derivative at $x=-2$ .

EDIT: The other method goes like this... (I stupidly used this method in the exam last year and lost a mark.. :( )

$f(x)=\sqrt{(2x+4)^2}$

$f(x)=\sqrt{u}$ where $u=(2x+4)^2$

You could 'officially' use the chain rule again as in write it down on paper, or you could just do it quickly in your head.

$f'(x)=\frac{1}{2\sqrt{u}}\cdot u'$

and

$u'=4(2x+4)^1=8x+16$

So $f'(x)=\frac{8x+16}{2\sqrt{(2x+4)^2}}=4\cdot \frac{x+2}{2|x+2|}=2\cdot \frac{x+2}{|x+2|}$ .

And $\frac{x+2}{|x+2|}=\begin{cases}1, x> -2 \\ -1, x<-2\end{cases}$
(This is because they are always both the same magnitude, and the denominator is always positive, so the sign is determined by the sign of the numerator, which is positive where $x>-2$ and negative where $x<-2$ )

Hence $2\cdot \frac{x+2}{|x+2|}=\begin{cases}2\cdot 1 , x>-2 \\ 2\cdot -1 , x<-2\end{cases} = \begin{cases}2,x>-2 \\ -2, x<-2\end{cases}$ .

Title: Re: TyErd's questions
Post by: TyErd on May 03, 2010, 05:37:19 pm

thankyou! i get it now

Title: Re: TyErd's questions
Post by: the.watchman on May 03, 2010, 05:49:37 pm

Algebraic is more complicated than you think, there are two cases to consider

CASE 1:
(2x-1)>0 AND (x+5)<0 (pos. times neg. = neg.)

CASE 2:
(2x-1)<0 AND (x+5)>0 (neg. times pos. = neg.)

Try it from there!

Quote from: m@tty on May 03, 2010, 05:35:09 pm

$f(x)=|2x+4|=\sqrt{(2x+4)^2}$

Or you can say

$f(x)=\begin{cases} 2x+4, x\geq -2 \\ -2x-4, x<-2\end{cases}$

The piece wise defined function is much easier to differentiate. As you don't have to use the chain rule twice...

$f'(x)=\begin{cases}\frac{d}{dx}(2x+4), x\geq -2 \\ \frac{d}{dx}(-2x-4), x<-2\end{cases} = \begin{cases}2, x\geq 2 \\ -2, x<-2\end{cases}$

BUT the function is not differentiable at x=2 (for the derivative, pos and neg limits different)

Title: Re: TyErd's questions
Post by: m@tty on May 03, 2010, 05:50:31 pm

Quote from: the.watchman on May 03, 2010, 05:49:37 pm

BUT the function is not differentiable at x=-2 (for the derivative, pos and neg limits different)

YEAH I fixed that up. Thanks.

Title: Re: TyErd's questions
Post by: TyErd on May 03, 2010, 06:16:22 pm

why is there no value for the derivative at x=-2

Title: Re: TyErd's questions
Post by: ZachCharge on May 03, 2010, 06:25:17 pm

Quote from: TyErd on May 03, 2010, 06:16:22 pm

why is there no value for the derivative at x=-2

At x=-2 the function is at a cusp, which are impossible to get the gradient function of.

Title: Re: TyErd's questions
Post by: TyErd on May 03, 2010, 06:37:40 pm

Quote from: m@tty on May 03, 2010, 05:35:09 pm

$f(x)=|2x+4|=\sqrt{(2x+4)^2}$

Or you can say

$f(x)=\begin{cases} 2x+4, x\geq -2 \\ -2x-4, x<-2\end{cases}$

The piece wise defined function is much easier to differentiate. As you don't have to use the chain rule twice...

$f'(x)=\begin{cases}\frac{d}{dx}(2x+4), x> -2 \\ \frac{d}{dx}(-2x-4), x<-2\end{cases} = \begin{cases}2, x>- 2 \\ -2, x<-2\end{cases}$

NOTE: There is no value for the derivative at $x=-2$ .

What happened to to the greater than or equal too symbol. Why did you make it just greater than -2?

Title: Re: TyErd's questions
Post by: Potter on May 03, 2010, 07:16:44 pm

Hey guys,

Sorry TyErd for crashing your thread, but I didn't really see the point of making a new thread.

Had a sac last week and my friend and I are in a disagreement with what is right.. Just thought you guys might be able to help.
So, here we go.

y=e^(x+2) is translated 3 units to the right

So, y=e^(x-1)

The next question asked us if the graph is reflected in the y axis, what is the new equation?

Is it y=e^(-x-1)

OR

y=e^-(x-1)

Thanks guys.

Title: Re: TyErd's questions
Post by: TyErd on May 03, 2010, 07:20:09 pm

$y=e^{-(x-1)}$

Title: Re: TyErd's questions
Post by: m@tty on May 03, 2010, 07:21:11 pm

As said above, there is no value for the derivative at x=-2, as there is a cusp. At a cusp you don't have one value for the gradient at this point, you can draw more than one tangent. Or, more correctly, the left and right hand limits are not equal.

let $f(x)=|2x+4|$

To find the gradient using first principles you take the limit:

$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

And for this limit to exist, the left and right hand limits must be equal.

Firstly, the left hand limit:

$\lim_{h\to 0^-}\frac{f(-2+h)-f(-2)}{h}=\lim_{h\to 0^-}\frac{|2(-2+h)+4|-|2(-2)+4|}{h}$

$=\lim_{h \to 0^-} \frac{|2h|-|0|}{h}$

$=\lim_{h \to 0^-} \frac{-2h}{h}$ (as $h<0$ )

$=-2$

And the right hand limit:

$\lim_{h \to 0^+} \frac{f(-2+h)-f(-2)}{h}=\lim_{h\to 0^+} \frac{|2h|}{h}$

$=\lim_{h\to 0^+} \frac{2h}{h}=2$

As these limits are not equal, the derivative is not defined at this point, $x=-2$ .

Title: Re: TyErd's questions
Post by: superflya on May 03, 2010, 07:24:17 pm

first option.

x= -x' y= y'

sub em in.

Title: Re: TyErd's questions
Post by: TyErd on May 03, 2010, 07:34:18 pm

okay i get it! i had to draw the graph to completely understand.

Title: Re: TyErd's questions
Post by: Potter on May 03, 2010, 07:38:08 pm

Thanks buddy. That clears up a lot. I was thinking the same.

Also, one more.. I'm kinda curious(not sure if it's true)

If you have a limit, can you dy/dx the numerator and denominator then sub in the limit value

say for like (x+3)(x-2)/x-2 lim-->2

dy/dx the numerator 2x +1
dy/dx the denominator = 1

(2x+1)/1 = 2x +1

sub in 2 = 5

If you do it the textbook way you get the same answer. I asked my teacher and he said it was just coincidence? But I did it with a couple of other questions and got the right answer.

Title: Re: TyErd's questions
Post by: m@tty on May 03, 2010, 07:44:08 pm

That is indeed true, it is called L'hospital's rule. link.

Title: Re: TyErd's questions
Post by: TyErd on May 03, 2010, 07:46:11 pm

Consider the function $f:R\rightarrow R, f(x) = \sqrt {x^3 +4x}$ .

f'(x) may be written as $f'(x) = \frac {ax^2 + b} { c \sqrt {x^3 +4x}}$

Find values a,b and c.

Title: Re: TyErd's questions
Post by: superflya on May 03, 2010, 07:50:03 pm

same as $(x^{3}+4x)^{\frac{1}{2}}$

chain rule.

a=3 b=4 c=2

Title: Re: TyErd's questions
Post by: TyErd on May 03, 2010, 08:08:12 pm

got it, thanks superflya!

Title: Re: TyErd's questions
Post by: TyErd on May 03, 2010, 08:25:59 pm

Differentiating exponential functions the fastest and easiest way?

Title: Re: TyErd's questions
Post by: m@tty on May 03, 2010, 08:31:25 pm

Example?

For $y=e^{f(x)}$

$\frac{dy}{dx}=e^{f(x)} \cdot f'(x)$ ...

Title: Re: TyErd's questions
Post by: Blakhitman on May 03, 2010, 08:32:58 pm

Quote from: TyErd on May 03, 2010, 08:25:59 pm

Differentiating exponential functions the fastest and easiest way?

lol reminds me of a post where I was laughed at :D

m@tty might remember :P

Title: Re: TyErd's questions
Post by: TyErd on May 03, 2010, 08:38:38 pm

why were you laughed at?

Title: Re: TyErd's questions
Post by: superflya on May 03, 2010, 08:41:54 pm

maybe cause when u differentiate e^x it stays the same. im just guessing :P

Title: Re: TyErd's questions
Post by: TyErd on May 03, 2010, 08:43:07 pm

Quote from: superflya on May 03, 2010, 08:41:54 pm

maybe cause when u differentiate e^x it stays the same. im just guessing :P

it does?

Title: Re: TyErd's questions
Post by: Blakhitman on May 03, 2010, 08:43:16 pm

Here comes the main offender ^ :P (superflya)

Title: Re: TyErd's questions
Post by: superflya on May 03, 2010, 08:44:16 pm

lulz ahahah \head-desk

Title: Re: TyErd's questions
Post by: 99.95 on May 09, 2010, 05:46:52 pm

how do i do this question:

y= X / 1-x

write dy/dx in terms of y

Title: Re: TyErd's questions
Post by: brightsky on May 09, 2010, 05:51:10 pm

http://vcenotes.com/forum/index.php/topic,25258.msg259987.html#msg259987

Hopefully it's right. :p

Title: Re: TyErd's questions
Post by: TyErd on May 19, 2010, 04:12:20 pm

Differentiate: $e^{3x} (2x+1)^{3}$

Title: Re: TyErd's questions
Post by: m@tty on May 19, 2010, 04:17:18 pm

Just use the product rule.
$u=e^{3x}$ and $v=(2x+1)^3$
$u'=3e^{3x}$ and $v'=6(2x+1)^2$ .

And you can take a common factor of $3e^{3x}(2x+1)^2$ .

Title: Re: TyErd's questions
Post by: TyErd on May 19, 2010, 04:31:17 pm

its embarassing but taking the common factor is the part im having trouble with.

Title: Re: TyErd's questions
Post by: the.watchman on May 19, 2010, 04:37:15 pm

So $\frac{d}{dx}(e^{3x}(2x+1)^3) = 3e^{3x}(2x+1)^3 + e^{3x} \cdot 6(2x+1)^2$

From there, we can spot three common factors, e^{3x}, 3 and (2x+1)^2 (always take the lowest powers out)

Taking these three out of the two terms, we have (2x+1) + 2 left-over, right?

So $\frac{d}{dx}(e^{3x}(2x+1)^3) = 3e^{3x}(2x+1)^2[(2x+1) + 2]$

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

Title: Re: TyErd's questions
Post by: TyErd on May 19, 2010, 04:42:27 pm

thnx!

Title: Re: TyErd's questions
Post by: TyErd on May 28, 2010, 04:48:39 pm

Differentiate $y= \frac{sin2x}{x^2+4}$

Okay I use the quick method, and this was my working out but im not sure if it is correct:

$=sin2x \times (x^2+4)^{-1}$
$=2cos2x - 1(x^2+4)^{-2} \times 2x$
$=2cos2x - 2x(x^2+4)^{-2}$
$= {2cos2x} - \frac {2x}{(x^2 + 4)^2}$

is this correct? and do I need to simply further?

Title: Re: TyErd's questions
Post by: the.watchman on May 28, 2010, 05:09:56 pm

Just being pedantic, but so are the examiners

sin(2x) x (x^2... is not equal to 2cos(2x)-1(x^2... :P

You can't write this in the exam, ok? :)
Otherwise, looks good, you can take the two out, but you won't have to :)

Title: Re: TyErd's questions
Post by: TyErd on May 28, 2010, 05:22:22 pm

oh shit sorry forgot dy/dx

so correct answer to this to gain full marks is

$y= \frac{sin2x}{x^2 + 4}$
$y=sin2x \times (x^2+4)^{-1}$

$\frac{dy}{dx} =2cos2x - 1(x^2+4)^{-2} \times 2x$
$=2cos2x - 2x(x^2+4)^{-2}$
$={2cos2x} - \frac {2x}{(x^2 + 4)^2}$

?

Title: Re: TyErd's questions
Post by: m@tty on May 28, 2010, 05:27:24 pm

Should do. Unless they ask for you to use quotient rule...

Title: Re: TyErd's questions
Post by: TyErd on May 28, 2010, 05:36:52 pm

ok thnx

Title: Re: TyErd's questions
Post by: TyErd on May 28, 2010, 06:23:40 pm

Using the first principle, differentiate $f(x)= x^{-3}$

Title: Re: TyErd's questions
Post by: m@tty on May 28, 2010, 06:44:05 pm

Sub $f(x)$ into $f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$ .

Title: Re: TyErd's questions
Post by: brightsky on May 28, 2010, 06:47:38 pm

By first principles, $f'(x) = \frac{f(x+h) - f(x)}{h}$

Because $f(x) = x^{-3} = \frac{1}{x^3}$ , $f(x+h) = \frac{1}{(x+h)^3}$

So $f'(x) = \lim_{h \to 0} \frac{\frac{1}{(x+h)^3} - \frac{1}{x^3}}{h}$

$= \lim_{h \to 0} \frac{\frac{x^3 - (x+h)^3}{x^3(x+h)^3}}{h}$

$= \lim_{h \to 0} -\frac{h^2 +3hx+3x^2}{x^3(x+h)^3}$

Substitute $h = 0$ :

$-\frac{3x^2}{x^6} = -3x^{-4}$

Title: Re: TyErd's questions
Post by: TyErd on May 28, 2010, 06:49:59 pm

error in your latex brightsky

Title: Re: TyErd's questions
Post by: brightsky on May 28, 2010, 06:50:37 pm

Quote from: TyErd on May 28, 2010, 06:49:59 pm

error in your latex brightsky

Fixed. :D (Sorry about that :p)

Title: Re: TyErd's questions
Post by: TyErd on May 28, 2010, 07:16:45 pm

awesome, thnx i get it now! :)

Title: Re: TyErd's questions
Post by: moekamo on May 28, 2010, 07:25:30 pm

Quote from: TyErd on May 28, 2010, 05:22:22 pm

oh shit sorry forgot dy/dx

so correct answer to this to gain full marks is

$y= \frac{sin2x}{x^2 + 4}$
$y=sin2x \times (x^2+4)^{-1}$

$\frac{dy}{dx} =2cos2x - 1(x^2+4)^{-2} \times 2x$
$=2cos2x - 2x(x^2+4)^{-2}$
$={2cos2x} - \frac {2x}{(x^2 + 4)^2}$

?

wait, $\begin{align} \frac{d}{dx} \sin 2x \cdot (x^2+4)^{-1} &= 2\cos 2x \cdot (x^2+4)^{-1} + \sin 2x \cdot (x^2+4)^{-2} \cdot 2x \\ & = (x^2+4)^{-1} (2\cos 2x + 2x\sin2x (x^2+4)^{-1}) \end{align} $

i think you forgot part of the product rule in your workings there tyerd :S

Title: Re: TyErd's questions
Post by: TyErd on May 28, 2010, 10:17:45 pm

what i did was apply the chain rule to the 2nd term. I dunno is that wrong?

Title: Re: TyErd's questions
Post by: TyErd on May 28, 2010, 10:38:18 pm

I am soo bloody confused right now. I know I can use product rule or quotient rule if i wanted to but what about the way i did it?

Title: Re: TyErd's questions
Post by: m@tty on May 28, 2010, 11:32:45 pm

Wait, I'm sorry for not seeing it earlier. You forgot to use the product rule. All you did was find the derivative of both parts, but according to the quotient rule you had to multiply each derivative by the other function...

$\frac{d}{dx}(uv)=u'\bold{v}+\bold{u}v'$ .

You omitted the parts I bolded.

Title: Re: TyErd's questions
Post by: TyErd on May 28, 2010, 11:35:00 pm

Ohh okayys thnx!

Also Differentiate:
$sin^{2} 2\theta$

Title: Re: TyErd's questions
Post by: m@tty on May 28, 2010, 11:42:04 pm

$\frac{d}{d\theta}\left(\sin^2 (2\theta)\right)=2\sin(2\theta)\cdot \cos(2\theta)\cdot 2$ .

Which, using a double angle formula, can be simplified to:

$2\sin(4\theta)$ .

But I don't think you need to know this :P Spesh stuff...

Title: Re: TyErd's questions
Post by: TyErd on May 28, 2010, 11:48:50 pm

i dunno, theres a bit of it in the essentials book, The spesh teacher at my school makes the SACs and she's a bitch because she wants her spesh/methods students to get the top scores so she adds spesh shit to the methods SAC.

i dont get how you did the first line, im confused with the squared bit mainly.

Title: Re: TyErd's questions
Post by: m@tty on May 29, 2010, 12:04:57 am

Okay, $\sin^2(2\theta)=\left[\sin(2\theta)\right]^2$ .

So use the chain rule with $u=\sin(2\theta)$ .

$\frac{du}{d\theta}=2\cos(2\theta)$

$\frac{d}{d\theta}\left(\left[\sin(2\theta)\right]^2\right)=\frac{du}{d\theta}\cdot \frac{d}{du}(u^2)=2u\cdot \frac{du}{d\theta}$

Sub back for $u$ and $\frac{du}{d\theta}$ .

$\frac{d}{d\theta}\left(\left[\sin(2\theta)\right]^2\right)=2\sin(2\theta)\cdot 2\cos(2\theta)$ .

Title: Re: TyErd's questions
Post by: TyErd on May 29, 2010, 12:17:50 am

okay thnx i get it :)

Title: Re: TyErd's questions
Post by: TyErd on May 29, 2010, 02:42:00 am

Differentiate:

$y= 1500 e^\frac{\log_e 3}{3}^{x}$

Title: Re: TyErd's questions
Post by: moekamo on May 29, 2010, 06:18:52 am

$y= 1500 \left ( e^{\frac{x \log_e 3}{3}} \right )$

let $u=\frac{x \log_e 3}{3} \implies \frac{du}{dx} = \frac{ \log_e 3}{3}$

also $y=1500 e^u \implies \frac{dy}{du} = 1500 e^u$

so $ \frac{dy}{dx} = 1500 e^u \cdot \frac{ \log_e 3}{3} = 500 \log_e 3 \cdot e^{\frac{x \log_e 3}{3}} $

Title: Re: TyErd's questions
Post by: the.watchman on May 29, 2010, 09:08:26 am

You could do that, or:

Considering that $\frac{log_e 3}{3}$ is a constant

Then $\frac{dy}{dx}=1500e^{\frac{log_e 3}{3} x} \cdot \frac{log_e 3}{3}$

$\frac{dy}{dx}=500log_e(3)e^{\frac{log_e 3}{3} x}$

This employs the derivative "trick" of $[e^{kx}]'=ke^{kx}$

Title: Re: TyErd's questions
Post by: TyErd on May 30, 2010, 07:03:39 pm

Differentiate: $cos (\frac{\pi}{2} -x)$

Title: Re: TyErd's questions
Post by: m@tty on May 30, 2010, 07:04:29 pm

$\cos\left(\frac{\pi}{2}-x\right)=\sin(x)$ .

Title: Re: TyErd's questions
Post by: TyErd on May 30, 2010, 07:29:36 pm

oh yeah so the derivative of that would be cos x

Title: Re: TyErd's questions
Post by: m@tty on May 30, 2010, 08:03:29 pm

Yep.

Title: Re: TyErd's questions
Post by: TyErd on June 09, 2010, 11:03:45 am

Differentiate: $|x^2 - 4x|$ and also $|x|^2 - 4|x|$

Title: Re: TyErd's questions
Post by: the.watchman on June 09, 2010, 11:09:13 am

Quote from: TyErd on June 09, 2010, 11:03:45 am

Differentiate: $|x^2 - 4x|$ and also $|x|^2 - 4|x|$

Well the first function can be written as:

$x^2-4x$ , $x \in (-\infty,0] \cup [4,\infty)$
$-x^2+4x$ , $x \in (0,4)$

So differentiate each part (noting that the cusp points are not differentiable)

You could also split the second function into parts to differentiate

Title: Re: TyErd's questions
Post by: TyErd on June 09, 2010, 11:29:12 am

Quote from: the.watchman on June 09, 2010, 11:09:13 am

Quote from: TyErd on June 09, 2010, 11:03:45 am
Differentiate: $|x^2 - 4x|$ and also $|x|^2 - 4|x|$
Well the first function can be written as:

$x^2-4x$ , $x \in (-\infty,0] \cup [4,\infty)$

isn't the domain $(-\infty, 0) U (0,4) U (4,\infty)$ ?

Title: Re: TyErd's questions
Post by: the.watchman on June 09, 2010, 11:32:12 am

Well no, the modulus splits it up so that:

First part - $x^2-4x$ , when $x^2-4x\geq 0$ , so when $x\leq 0 \cup x\geq 4$

Second part - $-(x^2-4x)$ , when $x^2-4x< 0$ , so when $0<x<4$

Oh, and for the function itself, it is defined when x=0 and x=4, it's only the derived function which does not exist at these points :)

Title: Re: TyErd's questions
Post by: TyErd on June 09, 2010, 11:47:14 am

Quote from: the.watchman on June 09, 2010, 11:32:12 am

Well no, the modulus splits it up so that:

First part - $x^2-4x$ , when $x^2-4x\geq 0$ , so when $x\leq 0 \cup x\geq 4$

okay, graphically I get it but algebraically I get $x\geq 0 \cup x\geq 4$

Title: Re: TyErd's questions
Post by: the.watchman on June 09, 2010, 12:03:36 pm

Be careful when doing it algebraically, this is the proper method:

$x^2-4x \geq 0$

$x(x-4) \geq 0$

CASE 1:

$x \geq 0$ AND $x-4 \geq 0$ (two positives make a positive)

From these, $x \geq 4$ (finding the intersection)

CASE 2:

$x \leq 0$ AND $x-4 \leq 0$ (two negatives make a positive)

From these, $x \leq 0$ (finding the intersection)

So the solutions are $x \leq 0$ or $x \geq 4$ :)

Title: Re: TyErd's questions
Post by: TyErd on June 11, 2010, 12:21:05 am

If $y=x^4$ , prove that $x\frac{dy}{dx} = 4y$ . I dont even understand what the expression $x\frac{dy}{dx}$ is askin for.

Title: Re: TyErd's questions
Post by: qshyrn on June 11, 2010, 12:30:11 am

Quote from: TyErd on June 11, 2010, 12:21:05 am

If $y=x^4$ , prove that $x\frac{dy}{dx} = 4y$ . I dont even understand what the expression $x\frac{dy}{dx}$ is askin for.

its prettty simple i think: dy/dx=4x^3
x*dy/dx= x*(4x^3)=4x^4=4(x^4) =4y=rhs

Title: Re: TyErd's questions
Post by: TyErd on June 11, 2010, 12:42:38 am

ohhh I getcha thanks!

Title: Re: TyErd's questions
Post by: kenhung123 on June 11, 2010, 07:38:10 am

Hmm yea, the book doesn't really explain how to differentiate absolute value functions well...

Title: Re: TyErd's questions
Post by: Juddinator on June 13, 2010, 12:20:54 pm

P and Q lie on the curve $y=x^3$ . The x-co-ords of P and Q and 2 and (2+h) respectively. What is the gradient of PQ?

I tried to derive the equation of the curve ten just simply sub in the value of the x-co-ords but I got stuck when I realised you have2 x values, not one.

Title: Re: TyErd's questions
Post by: kenhung123 on June 13, 2010, 01:36:59 pm

Is there at all many tricks to find the gradient graph only given the original graph? This is extremely tricky!

Title: Re: TyErd's questions
Post by: TyErd on June 13, 2010, 01:41:25 pm

Quote from: kenhung123 on June 13, 2010, 01:36:59 pm

Is there at all many tricks to find the gradient graph only given the original graph? This is extremely tricky!

I agree, it is very tricky at first, but I found practice was the key. I did all the ones in essentials atleast three times even the basic ones. I dont think theres any tricks with them but If there is I'd be eager to know.

Title: Re: TyErd's questions
Post by: kenhung123 on June 13, 2010, 01:43:15 pm

Do you throw these types of questions at us in exams? Definately would take some time to figure out and then try to confirm its the right answer.

Title: Re: TyErd's questions
Post by: brightsky on June 13, 2010, 01:47:51 pm

Because the x-coordinate of P is 2 and P lies on $y = x^3$ , the y-coordinate of P is given by $y = 2^3 = 8$ . So P $(2,8)$ .

Because the x-coordinate of Q is 2+h and Q lies on $y = x^3$ , the y-coordinate of Q is given by $y = (2+h)^3 = h^3 + 6h^2 + 12h + 8$ . So Q $(2+h, h^3 + 6h^2 + 12h + 8)$ .

So the gradient of P,Q is given by $\frac{y_2 - y_1}{x_2 - x_1} = \frac{h^3 + 6h^2 + 12h + 8 - 8}{2+h - 2} = \frac{h^3 + 6h^2 + 12h}{h} = \frac{h(h^2 + 6h+12)}{h} = h^2 + 6h + 12$ .

Title: Re: TyErd's questions
Post by: brightsky on June 13, 2010, 01:54:30 pm

Quote from: kenhung123 on June 13, 2010, 01:36:59 pm

Is there at all many tricks to find the gradient graph only given the original graph? This is extremely tricky!

Just find the equation of the original graph and then derive it to get gradient graph as far as I know.

Title: Re: TyErd's questions
Post by: the.watchman on June 13, 2010, 02:01:57 pm

Quote from: brightsky on June 13, 2010, 01:54:30 pm

Quote from: kenhung123 on June 13, 2010, 01:36:59 pm
Is there at all many tricks to find the gradient graph only given the original graph? This is extremely tricky!
Just find the equation of the original graph and then derive it to get gradient graph as far as I know.

Well, you can and may be asked to sketch gradient graphs, given the graph of any function
To tackle these questions, look for stationary points and be sure to note the sign of the gradient between any of these.
Also, be careful with any endpoints and cusps that may appear, because functions are not differentiable at these.

Title: Re: TyErd's questions
Post by: TyErd on June 13, 2010, 05:06:08 pm

use the chain rule to prove $\frac{dy}{dx} = nx^{n-1} for y=x^n$ , where n is a negative number. I dont get it.

Title: Re: TyErd's questions
Post by: TyErd on June 13, 2010, 05:28:36 pm

Um.. if $y= (x+ \sqrt{x^2+1})^2 , show \frac {dy}{dx} = \frac {2y}{\sqrt{x^2 +1}}$

I can get through about three quarters of it but then I get stuck. I dont like simplifying, its a pain.

Title: Re: TyErd's questions
Post by: kenhung123 on June 13, 2010, 05:32:39 pm

Quote from: TyErd on June 13, 2010, 05:06:08 pm

use the chain rule to prove $\frac{dy}{dx} = nx^{n-1} for y=x^n$ , where n is a negative number. I dont get it.

Hmm not sure about the "where n is a negative number" but I would just do it normally like $dy/dx=nx^{n-1}*1$ lol thats basically it I guess...

Title: Re: TyErd's questions
Post by: kenhung123 on June 13, 2010, 05:38:04 pm

Quote from: TyErd on June 13, 2010, 05:28:36 pm

Um.. if $y= (x+ \sqrt{x^2+1})^2 , show \frac {dy}{dx} = \frac {2y}{\sqrt{x^2 +1}}$

I can get through about three quarters of it but then I get stuck. I dont like simplifying, its a pain.

Use chain rule to get
$dy/dx=2(x+(x^{2}+x)^{1/2})*(1+0.5(x^{2}+1)^{-0.5}*2x)$
=> $[2x+2(x^{2}+1)][1+x(x^{2}+1)^{-1/2}]$
You can see that the first bracket is equal to 2*y from original equation so you can substitute that in and get the ans

Title: Re: TyErd's questions
Post by: TyErd on June 13, 2010, 05:45:01 pm

how'd u get the first half of your first line?

Title: Re: TyErd's questions
Post by: kenhung123 on June 13, 2010, 05:48:25 pm

Sorry missed a bracket

Title: Re: TyErd's questions
Post by: TyErd on June 13, 2010, 06:00:39 pm

$x^2 + x$ or is it suppose to be $x^2 +1$ in that first part?

Title: Re: TyErd's questions
Post by: kenhung123 on June 13, 2010, 06:15:20 pm

Title: Re: TyErd's questions
Post by: TyErd on June 13, 2010, 06:30:32 pm

Quote

Use chain rule to get
$dy/dx=2(x+(x^{2}+x)^{1/2})*(1+0.5(x^{2}+1)^{-0.5}*2x)$
=> $[2x+2(x^{2}+1)][1+x(x^{2}+1)^{-1/2}]$
You can see that the first bracket is equal to 2*y from original equation so you can substitute that in and get the ans

So the first line should look like this? $dy/dx=2(x+(x^{2}+1)^{1/2})*(1+0.5(x^{2}+1)^{-0.5}*2x)$

Title: Re: TyErd's questions
Post by: TyErd on June 13, 2010, 07:36:15 pm

Also help with this: $f(x)= \frac{x^n}{1+ x^n}$ , where n is a positive even integer.

show that $f(x)= 1- \frac{1}{x^n + 1}$

Also show that $0 \leq f(x) < 1$ for all values of x. How do I go about doing this? draw a graph?

Title: Re: TyErd's questions
Post by: TyErd on June 13, 2010, 07:37:20 pm

Quote from: kenhung123 on June 13, 2010, 05:32:39 pm

Quote from: TyErd on June 13, 2010, 05:06:08 pm
use the chain rule to prove $\frac{dy}{dx} = nx^{n-1} for y=x^n$ , where n is a negative number. I dont get it.
Hmm not sure about the "where n is a negative number" but I would just do it normally like $dy/dx=nx^{n-1}*1$ lol thats basically it I guess...

but it says using chain rule

Title: Re: TyErd's questions
Post by: kenhung123 on June 13, 2010, 07:48:41 pm

Yea thats chain rule, it looks like its very simple but you can see that u=x

Title: Re: TyErd's questions
Post by: TyErd on June 13, 2010, 07:57:48 pm

Also what is the equation of the tangent of $f(x)= x^\frac{2}{3} , where x=0.$

Title: Re: TyErd's questions
Post by: kenhung123 on June 13, 2010, 08:35:04 pm

find dy/dx then sub in x=0
This would give the gradient of the point x=0
Now sub the gradient (m) into y=mx+c
You are given a point also (x,y) when x=0, find y using original equation
Sub the point into the linear equation and find c
Then sub c into the y=mx+c equation.

Title: Re: TyErd's questions
Post by: kenhung123 on June 13, 2010, 08:37:21 pm

Quote from: brightsky on June 13, 2010, 01:54:30 pm

Quote from: kenhung123 on June 13, 2010, 01:36:59 pm
Is there at all many tricks to find the gradient graph only given the original graph? This is extremely tricky!
Just find the equation of the original graph and then derive it to get gradient graph as far as I know.

But sometimes they don't even give you 1 point basically they want you to draw an approximate gradient graph but it has to be somewhat reasonable like...positive gradient at a particular point must have f'(x) above x axis at least..

Title: Re: TyErd's questions
Post by: TyErd on June 13, 2010, 08:51:34 pm

Quote from: kenhung123 on June 13, 2010, 08:35:04 pm

find dy/dx then sub in x=0
This would give the gradient of the point x=0
Now sub the gradient (m) into y=mx+c
You are given a point also (x,y) when x=0, find y using original equation
Sub the point into the linear equation and find c
Then sub c into the y=mx+c equation.

Yeah I got that but at x=0 there is a cusp, does that mean there's many tangents?

Title: Re: TyErd's questions
Post by: /0 on June 13, 2010, 09:22:03 pm

Quote from: TyErd on June 13, 2010, 07:37:20 pm

Quote from: kenhung123 on June 13, 2010, 05:32:39 pm
Quote from: TyErd on June 13, 2010, 05:06:08 pm
use the chain rule to prove $\frac{dy}{dx} = nx^{n-1} for y=x^n$ , where n is a negative number. I dont get it.
Hmm not sure about the "where n is a negative number" but I would just do it normally like $dy/dx=nx^{n-1}*1$ lol thats basically it I guess...

but it says using chain rule

I'm not sure how the chain rule plays into it, but here's how I would do it. Since $n$ is a negative number, it's usually convenient to 'bring the negative out'. To do this, set $v = -n$ , then $v$ is a positive number.

Then $y=x^{-v}$ , so $x^vy=1$

Differentiating implicitly,

$vx^{v-1}y + x^v \frac{dy}{dx}=0$

$\frac{dy}{dx} = \frac{-vx^{v-1}y}{x^v}=\frac{-vx^{v-1}x^{-v}}{x^v} = -vx^{-v-1} = nx^{n-1}$

So we have proved differentiation works for negative exponents. Notice that nowhere in the proof did we actually differentiate a negative exponent, since $v$ is positive.

Title: Re: TyErd's questions
Post by: TyErd on June 13, 2010, 09:51:18 pm

Thanks /0!

Title: Re: TyErd's questions
Post by: TyErd on June 13, 2010, 09:54:18 pm

can someone help me out with post number 240.

Title: Re: TyErd's questions
Post by: TyErd on June 13, 2010, 10:30:45 pm

Im stumped on this one: One side of a rectangle is three times the other. If the perimeter inceases by 2%, what is the percentage increase in the area.

Title: Re: TyErd's questions
Post by: the.watchman on June 14, 2010, 09:33:10 am

Let the length of the shorter side of the rectangle be x
Let the perimeter be P(x)
Let the area be A(x)

Then the other side length is 3x

So $P(x)=2(x+3x)=8x$

$P'(x)=8$

From the linear approximation formula, the change in perimeter is $\delta x \cdot P'(x)$

So if the change in perimeter is 2%, then $\delta x \cdot P'(x)=\frac{P(x)}{50}$

$\delta x \cdot 8=\frac{8x}{50}$

$\delta x=\frac{x}{50}$

Because $A(x)=x \cdot 3x = 3x^2$

Therefore the change in area is $\delta A = \delta x \cdot A'(x)$

$\delta A=\frac{x}{50} \cdot 6x$

$\delta A=\frac{3x^2}{25} = \frac{4}{100} \cdot A(x)$

So the change is 4%

I hope that made some sense :)

Title: Re: TyErd's questions
Post by: TyErd on June 14, 2010, 11:05:53 am

Wow you are a legend mate, question though, what did you sub into the x in the last step to get 4?

Title: Re: TyErd's questions
Post by: TyErd on June 14, 2010, 11:25:25 am

I cant seem to get this one either: A 2% error is made in measuring the radius of a sphere. Find the percentage error in the surface area. (The surface area of a sphere is given by $A=4\pi r^2$

Title: Re: TyErd's questions
Post by: the.watchman on June 14, 2010, 12:27:06 pm

Quote from: TyErd on June 14, 2010, 11:05:53 am

Wow you are a legend mate, question though, what did you sub into the x in the last step to get 4?

Sorry, my LaTeX didnt come out how I wanted, it should be there now :)

Quote from: TyErd on June 14, 2010, 11:25:25 am

I cant seem to get this one either: A 2% error is made in measuring the radius of a sphere. Find the percentage error in the surface area. (The surface area of a sphere is given by $A=4\pi r^2$

From the linear approximation formula:

$A(r+h)=A(r)+h\cdot A'(r)$

In this instance, h is 2% of r, so:

$A(r+\frac{2r}{100})=A(r)+\frac{2r}{100}\cdot A'(r)$

$A(\frac{51r}{50})=4\pi r^2+\frac{r}{50}\cdot (8\pi r)$

$A(\frac{51r}{50})=4\pi r^2+\frac{r}{50}\cdot (8\pi r)$

$A(\frac{51r}{50})=4\pi r^2 + \frac{4\pi r^2}{25} = A(r) + A(r) \cdot \frac{4}{100}$

So the percentage error is 4% :)

Title: Re: TyErd's questions
Post by: TyErd on June 14, 2010, 12:33:34 pm

Quote from: the.watchman on June 14, 2010, 09:33:10 am

Let the length of the shorter side of the rectangle be x
Let the perimeter be P(x)
Let the area be A(x)

Then the other side length is 3x

So $P(x)=2(x+3x)=8x$

$P'(x)=8$

From the linear approximation formula, the change in perimeter is $\delta x \cdot P'(x)$

So if the change in perimeter is 2%, then $\delta x \cdot P'(x)=\frac{P(x)}{50}$

$\delta x \cdot 8=\frac{8x}{50}$

$\delta x=\frac{x}{50}$

Because $A(x)=x \cdot 3x = 3x^2$

Therefore the change in area is $\delta A = \delta x \cdot A'(x)$

$\delta A=\frac{x}{50} \cdot 6x$

$\delta A=\frac{3x^2}{25} = \frac{4}{100} \cdot A(x)$

So the change is 4%

I hope that made some sense :)

How did you get that last line to equal 4/100

Title: Re: TyErd's questions
Post by: kamil9876 on June 14, 2010, 12:33:39 pm

Quote from: /0 on June 13, 2010, 09:22:03 pm

Quote from: TyErd on June 13, 2010, 07:37:20 pm
Quote from: kenhung123 on June 13, 2010, 05:32:39 pm
Quote from: TyErd on June 13, 2010, 05:06:08 pm
use the chain rule to prove $\frac{dy}{dx} = nx^{n-1} for y=x^n$ , where n is a negative number. I dont get it.
Hmm not sure about the "where n is a negative number" but I would just do it normally like $dy/dx=nx^{n-1}*1$ lol thats basically it I guess...

but it says using chain rule

I'm not sure how the chain rule plays into it, but here's how I would do it. Since $n$ is a negative number, it's usually convenient to 'bring the negative out'. To do this, set $v = -n$ , then $v$ is a positive number.

Then $y=x^{-v}$ , so $x^vy=1$

Differentiating implicitly,

$vx^{v-1}y + x^v \frac{dy}{dx}=0$

$\frac{dy}{dx} = \frac{-vx^{v-1}y}{x^v}=\frac{-vx^{v-1}x^{-v}}{x^v} = -vx^{-v-1} = nx^{n-1}$

So we have proved differentiation works for negative exponents. Notice that nowhere in the proof did we actually differentiate a negative exponent, since $v$ is positive.

If you want to use the chain rule, you can prove it for n=-1 first (from first principles, quite easy). Then use the chain rule as follows:

$x^{-n}=(\frac{1}{x})^v$

And since v is a positive integer, you can do it using and what you already know for the n=-1 case and the n=positive integer case.

Also: /0, you assumed y is differentiable without proof, this way shows it is (a technical point that I'm sure doesnt matter in VCE, heck i never remember this being required back in the days, but you might benefit from it for Analysis).

Title: Re: TyErd's questions
Post by: the.watchman on June 14, 2010, 12:46:51 pm

Quote from: TyErd on June 14, 2010, 12:33:34 pm

Quote from: the.watchman on June 14, 2010, 09:33:10 am
Let the length of the shorter side of the rectangle be x
Let the perimeter be P(x)
Let the area be A(x)

Then the other side length is 3x

So $P(x)=2(x+3x)=8x$

$P'(x)=8$

From the linear approximation formula, the change in perimeter is $\delta x \cdot P'(x)$

So if the change in perimeter is 2%, then $\delta x \cdot P'(x)=\frac{P(x)}{50}$

$\delta x \cdot 8=\frac{8x}{50}$

$\delta x=\frac{x}{50}$

Because $A(x)=x \cdot 3x = 3x^2$

Therefore the change in area is $\delta A = \delta x \cdot A'(x)$

$\delta A=\frac{x}{50} \cdot 6x$

$\delta A=\frac{3x^2}{25} = \frac{4}{100} \cdot A(x)$

So the change is 4%

I hope that made some sense :)

How did you get that last line to equal 4/100

3x^2/25 = 4% x 3x^2, right?

Title: Re: TyErd's questions
Post by: TyErd on June 14, 2010, 12:52:08 pm

oh yeah thnx! i get it now

Title: Re: TyErd's questions
Post by: kenhung123 on June 14, 2010, 02:49:57 pm

Just wondering can (e^{3x}-e^{2x})/e^{2x} be cancelled to (e^2x-e^{x})/e^{x}

Title: Re: TyErd's questions
Post by: Cthulhu on June 14, 2010, 03:02:07 pm

Quote from: kenhung123 on June 14, 2010, 02:49:57 pm

Just wondering can (e^{3x}-e^{2x})/e^{2x} be cancelled to (e^2x-e^{x})/e^{x}

It can be Simplified to
$e^x - 1$

Title: Re: TyErd's questions
Post by: kenhung123 on June 14, 2010, 03:10:34 pm

Thanks

Also when you have like common denominator, but the numerator has negative power do I multiply the numerator with denominator to bring it down like: $\frac{e^{-x}}{\sqrt{x+1}}+\frac{4e^{-4x}}{\sqrt{x+1}}=\frac{1}{e^{x}\sqrt{x+1}}+\frac{1}{4e^{4x}\sqrt{x+1}}$

Title: Re: TyErd's questions
Post by: the.watchman on June 14, 2010, 03:19:37 pm

Well the definition of $a^{-n}$ is $\frac{1}{a^n}$
So yes, you do that :)

Title: Re: TyErd's questions
Post by: moekamo on June 14, 2010, 03:52:15 pm

Quote from: kenhung123 on June 14, 2010, 03:10:34 pm

Thanks

Also when you have like common denominator, but the numerator has negative power do I multiply the numerator with denominator to bring it down like: $\frac{e^{-x}}{\sqrt{x+1}}+\frac{4e^{-4x}}{\sqrt{x+1}}=\frac{1}{e^{x}\sqrt{x+1}}+\frac{1}{4e^{4x}\sqrt{x+1}}$

except the 4 has to stay on the numerator in that second fraction since it is not being raised to a negative power...

Title: Re: TyErd's questions
Post by: kenhung123 on June 14, 2010, 04:07:03 pm

Got it, thanks

Title: Re: TyErd's questions
Post by: the.watchman on June 14, 2010, 04:12:31 pm

Quote from: moekamo on June 14, 2010, 03:52:15 pm

Quote from: kenhung123 on June 14, 2010, 03:10:34 pm
Thanks

Also when you have like common denominator, but the numerator has negative power do I multiply the numerator with denominator to bring it down like: $\frac{e^{-x}}{\sqrt{x+1}}+\frac{4e^{-4x}}{\sqrt{x+1}}=\frac{1}{e^{x}\sqrt{x+1}}+\frac{1}{4e^{4x}\sqrt{x+1}}$

except the 4 has to stay on the numerator in that second fraction since it is not being raised to a negative power...

Oops, good point :P

Title: Re: TyErd's questions
Post by: TyErd on June 14, 2010, 06:04:39 pm

Find two positive numbers who sum is 4 and such that the sum of the cube of the first and the square of the seconds is small as possible. I have no idea where to start.

These minima and maxima problems are hard to understand! I dont really understand how differentiation works in some problems. Any tips ?

Title: Re: TyErd's questions
Post by: kamil9876 on June 14, 2010, 06:10:45 pm

$x+y=4$

you want to minimize $z=x^3+y^2$

to turn it into a one variable problem: $z=x^3+(x-4)^2$ using the first equation.

Title: Re: TyErd's questions
Post by: Yitzi_K on June 14, 2010, 06:16:46 pm

x+y=4
y=4-x

So you want x^3 + (4-x)^2 to be as small as possible.

Find the derivative, then solve for 0.

Title: Re: TyErd's questions
Post by: TyErd on June 14, 2010, 06:23:04 pm

Okay thanks , Find the point on the parabola $y=x^2$ that is closest to the point (3,0)

Title: Re: TyErd's questions
Post by: kenhung123 on June 14, 2010, 06:41:19 pm

huh?? Not sure what that question is asking

Title: Re: TyErd's questions
Post by: TyErd on June 14, 2010, 06:49:07 pm

Yeah im not quite sure either, its question 2 in essentials 10G

Title: Re: TyErd's questions
Post by: the.watchman on June 14, 2010, 07:17:21 pm

Well, let D be the distance between a point on $y=x^2$ and $(3,0)$

Then, using the distance formula:

$D=\sqrt{(x-3)^2 + (y-0)^2}=\sqrt{(x-3)^2 + (x^2)^2}=\sqrt{x^4+x^2-6x+9}$

The x-value(s) which give the minimum value of D are the x-values of the required points :)

Title: Re: TyErd's questions
Post by: m@tty on June 14, 2010, 07:19:09 pm

It is asking for the point where the connecting line segment is as small as possible.

A point on the line $y=x^2$ can be given as $(x,x^2)$ .

The connecting line will be of length $\sqrt{(x-3)^2+(x^2)^2}=\sqrt{x^4+x^2-6x+9}$

Then find the minimum of this function and find the coordinates of the point on the parabola.

Title: Re: TyErd's questions
Post by: TyErd on June 14, 2010, 08:19:49 pm

So you differentate : $\sqrt{x^4+x^2-6x+9}$ make it equal zero then find the x values?

Title: Re: TyErd's questions
Post by: the.watchman on June 14, 2010, 08:40:14 pm

Quote from: TyErd on June 14, 2010, 08:19:49 pm

So you differentate : $\sqrt{x^4+x^2-6x+9}$ make it equal zero then find the x values?

Yes, and make sure you determine/prove local mins using a gradient table :)

EDIT: By chance, I noticed that this is my 999th methods post... :P

Title: Re: TyErd's questions
Post by: TyErd on June 14, 2010, 08:44:57 pm

Okay so far I have this:

$\frac{dy}{dx} = \frac{4x^3 + 2x - 6}{ 2 \sqrt{x^4+x^2 - 6x + 9}} = 0$

How do I simplify it further to find x?

Title: Re: TyErd's questions
Post by: Blakhitman on June 14, 2010, 08:46:47 pm

Well the numerator must equal 0.

Title: Re: TyErd's questions
Post by: the.watchman on June 14, 2010, 08:47:25 pm

Quote from: TyErd on June 14, 2010, 08:44:57 pm

Okay so far I have this:

$\frac{dy}{dx} = \frac{4x^3 + 2x - 6}{ 2 \sqrt{x^4+x^2 - 6x + 9}} = 0$

How do I simplify it further to find x?

In this scenario, you can just multiply both sides by the denominator, so it's $4x^3+2x-6=0$

(Don't be worried about removing solutions, the denominator can't be equal to zero anyway :P)

EDIT: Lol, 1000 MM posts :D

Title: Re: TyErd's questions
Post by: TyErd on June 14, 2010, 08:53:13 pm

So you just disregard the denominator altogether when you got something like that equal to zero?

Congrats man 1000 posts, damn im still on 151 haha

Title: Re: TyErd's questions
Post by: the.watchman on June 14, 2010, 08:55:28 pm

Quote from: TyErd on June 14, 2010, 08:53:13 pm

So you just disregard the denominator altogether when you got something like that equal to zero?

Congrats man 1000 posts, damn im still on 151 haha

Yep, if the denominator equaled zero, then the world would have exploded, but it hasn't :P

Title: Re: TyErd's questions
Post by: Juddinator on June 14, 2010, 10:38:43 pm

I've pretty much finished the MC in Chapter 13 of Essentials but there are a few questions in the MC that I don't get..

The graph of $y=bx^2-cx$ crosses the x-axis at the point (4, 0). The gradient at the point
is 1. What is the value of c?

I understand that I have to sub in points (4, 0) but I get stuck finding b. I tried finding the gradient but I was off....

If $f(x)=\frac{3}{x}$ , then what is $\frac{[f(x+h)-f(x)]}{h}$ equal to?

The straight line $y=x+3$ cuts the parabola $y=x^2-2x+3$ at the point (3, 0). The angle
between the line and the parabola at this point is u. What is tan u?

Title: Re: TyErd's questions
Post by: kenhung123 on June 14, 2010, 10:51:56 pm

What does this actually mean?

Title: Re: TyErd's questions
Post by: moekamo on June 14, 2010, 10:59:56 pm

Quote from: kenhung123 on June 14, 2010, 10:51:56 pm

What does this actually mean?

its an addition/subtraction formula, say you wanted to find $\sin\left ( \frac{\pi}{12} \right )$ then you would use the fact that $\frac{\pi}{12} = \frac{\pi}{3} - \frac{\pi}{4}$ so $\sin\left ( \frac{\pi}{12} \right ) = \sin \left (\frac{\pi}{3} - \frac{\pi}{4} \right ) = \sin\frac{\pi}{3} \cos\frac{\pi}{4} - \cos\frac{\pi}{3} \sin\frac{\pi}{4}$ which you can then simplify with exact values...

Title: Re: TyErd's questions
Post by: Blakhitman on June 14, 2010, 11:00:21 pm

Quote from: Juddinator on June 14, 2010, 10:38:43 pm

I've pretty much finished the MC in Chapter 13 of Essentials but there are a few questions in the MC that I don't get..

The graph of $y=bx^2-cx$ crosses the x-axis at the point (4, 0). The gradient at the point
is 1. What is the value of c?

We can sub in the given point: $16b-4c=0$
we also know what the gradient is at this point so differentiate: $\frac{\mathrm{d} y}{\mathrm{d} x}=2bx-c$
then substitute x=4 and the gradient equals 1 at this point: $8b-c=1$

So we have two equations: $16b-4c=0$ ----(1) and
$8b-c=1$ ----(2)

solve simultaneously and you should get $b=1/4$ and $c=-1$

Quote from: Juddinator on June 14, 2010, 10:38:43 pm

If $f(x)=\frac{3}{x}$ , then what is $\frac{[f(x+h)-f(x)]}{h}$ equal to?

$\frac{[f(x+h)-f(x)]}{h}=\frac{\frac{3}{x+h}-\frac{3}{x}}{h}$

Simplify numerator: $\frac{\frac{3x-3(x+h)}{x(x+h)}}{h}=\frac{\frac{3x-3x-3h}{x^{2}+xh}}{h}$

Then simplify the whole thing: $\frac{-3h}{x^{2}+xh}\div h=\frac{-3}{x^{2}+xh}$

I think that's what it's asking for :S, if it wanted the derivative it would be ${f}'(x)=\lim_{h\rightarrow 0}\frac{-3}{x^{2}+xh}$

Title: Re: TyErd's questions
Post by: Juddinator on June 15, 2010, 11:06:36 am

ok sweet thanks blakhitman!

Title: Re: TyErd's questions
Post by: kenhung123 on June 15, 2010, 07:37:31 pm

Quote from: moekamo on June 14, 2010, 10:59:56 pm

Quote from: kenhung123 on June 14, 2010, 10:51:56 pm
What does this actually mean?

its an addition/subtraction formula, say you wanted to find $\sin\left ( \frac{\pi}{12} \right )$ then you would use the fact that $\frac{\pi}{12} = \frac{\pi}{3} - \frac{\pi}{4}$ so $\sin\left ( \frac{\pi}{12} \right ) = \sin \left (\frac{\pi}{3} - \frac{\pi}{4} \right ) = \sin\frac{\pi}{3} \cos\frac{\pi}{4} - \cos\frac{\pi}{3} \sin\frac{\pi}{4}$ which you can then simplify with exact values...

I'm not sure if the text book is using that formula (I thought it was) but they did this:
$2 sin 2(2\theta+1)=2sin(2\theta+1)cos(2\theta+1)$

Title: Re: TyErd's questions
Post by: stonecold on June 15, 2010, 07:43:07 pm

double/compound angle formulae are so good. :)

Title: Re: TyErd's questions
Post by: kenhung123 on June 15, 2010, 07:52:25 pm

Hmm, do we need to know that provided CAS?

Title: Re: TyErd's questions
Post by: stonecold on June 15, 2010, 07:53:24 pm

Yeah, if you read through my question thread there is a massive discussion about this somewhere in there. Best to know them. They come in very useful at times...

Title: Re: TyErd's questions
Post by: kenhung123 on June 15, 2010, 08:38:43 pm

Is it wrong if I leave it in the form I get without applying double angle? I actually don't really get the point of double angle because it doesn't seem to simplify the expression?

Title: Re: TyErd's questions
Post by: brightsky on June 16, 2010, 05:27:06 pm

Quote from: kenhung123 on June 15, 2010, 07:37:31 pm

Quote from: moekamo on June 14, 2010, 10:59:56 pm
Quote from: kenhung123 on June 14, 2010, 10:51:56 pm
What does this actually mean?

its an addition/subtraction formula, say you wanted to find $\sin\left ( \frac{\pi}{12} \right )$ then you would use the fact that $\frac{\pi}{12} = \frac{\pi}{3} - \frac{\pi}{4}$ so $\sin\left ( \frac{\pi}{12} \right ) = \sin \left (\frac{\pi}{3} - \frac{\pi}{4} \right ) = \sin\frac{\pi}{3} \cos\frac{\pi}{4} - \cos\frac{\pi}{3} \sin\frac{\pi}{4}$ which you can then simplify with exact values...
I'm not sure if the text book is using that formula (I thought it was) but they did this:
$2 sin 2(2\theta+1)=2sin(2\theta+1)cos(2\theta+1)$

Is it me or should the RHS be multiplied by two? Like:

$2\sin2(2\theta+1) = 2(2\sin(2\theta+1)\cos(2\theta+1))$

Edited: xD

Title: Re: TyErd's questions
Post by: the.watchman on June 16, 2010, 06:15:34 pm

Yup, but it should be $2(2\sin (2\theta +1)\cos (2\theta +1))$ :)

Oh and congrats for 1000 :)

Title: Re: TyErd's questions
Post by: TyErd on June 16, 2010, 06:40:47 pm

Let $f$ be differentiable for all values of x in $[0,6]$ . The graph with equation $y=f(x)$ has a local minimum at $(2,4)$ . What is the equation of the tangent at the point with the coordinates $(2,4)$ . I cant seem to get it.

Title: Re: TyErd's questions
Post by: brightsky on June 16, 2010, 06:48:17 pm

(2,4) is the local minimum, so the tangent would just be a horizontal straight line (gradient = 0). Hence, equation of the tangent would be y = 4.

Title: Re: TyErd's questions
Post by: TyErd on June 16, 2010, 06:51:25 pm

Ofcourse! cant believe I didn't see that before. Thankyou!

Title: Re: TyErd's questions
Post by: TyErd on June 16, 2010, 07:07:42 pm

A spherical bubble, initially of radius length 1cm, expands steadily, its radius increaes by 1cm/s and it bursts after 5 seconds.

a) Find the rate of increase of volume with respect to the change in radius when the radius is 4cm.

b) Find the rate of increase of volume with respect to time when the radius is 4cm.

Title: Re: TyErd's questions
Post by: TyErd on June 16, 2010, 07:10:30 pm

An open tank is to be constructed with a square base and vertical sides to contain $500m^3$ of water. What must be the dimensions of the area of sheet metal used in its construction if this area is to be a minimum.

Title: Re: TyErd's questions
Post by: TyErd on June 16, 2010, 07:12:16 pm

At noon the captain of a ship sees two fishing boats approaching. One of them is 10km due east and travelling west at 8km/h. The other is 6km due north, travelling south at 6km/h. At what time will the fishing boats be closest together and how far apart will they be?

Title: Re: TyErd's questions
Post by: TyErd on June 16, 2010, 11:47:00 pm

Title: Re: TyErd's questions
Post by: kenhung123 on June 17, 2010, 12:17:40 am

hey soul sister, ain't that mister mister on the radio stereo

Title: Re: TyErd's questions
Post by: naved_s9994 on June 17, 2010, 12:18:17 am

lol!

Title: Re: TyErd's questions
Post by: /0 on June 17, 2010, 01:01:45 am

If you take your ship to be the origin, you can model the position of the boats with vectors. For the first boat, you have

$\mathbf{r}_1(t) = (10-8t)\mathbf{i}$

$\mathbf{r}_2(t)= (6-6t)\mathbf{j}$

The displacement between the boats is then

$\mathbf{r}_1(t)-\mathbf{r}_2(t) = (10-8t)\mathbf{i}+(6t-6)\mathbf{j}$

And the distance is

$|\mathbf{r}_1(t)-\mathbf{r}_2(t)| = \sqrt{(10-8t)^2+(6t-6)^2} = 2\sqrt{25t^2-58t+34}$

By minimizing this you can find the time where distance is minimum.

Quote from: TyErd on June 16, 2010, 07:10:30 pm

An open tank is to be constructed with a square base and vertical sides to contain $500m^3$ of water. What must be the dimensions of the area of sheet metal used in its construction if this area is to be a minimum.

Let the square base have side length $l$ , and let the height of the tank be $h$ .

Then you want to minimize $l^2+4lh$ subject to the constraint $l^2h = 500$ .

Solve the second equation for $h$ , plug into the first, and minimize.

Title: Re: TyErd's questions
Post by: TyErd on June 17, 2010, 05:40:15 pm

Hmm..I dont think we do vectors in Methods. Any other way to do the problem?

Title: Re: TyErd's questions
Post by: TyErd on June 20, 2010, 12:27:50 pm

Quote from: /0 on June 17, 2010, 01:01:45 am

Quote from: TyErd on June 16, 2010, 07:10:30 pm
An open tank is to be constructed with a square base and vertical sides to contain $500m^3$ of water. What must be the dimensions of the area of sheet metal used in its construction if this area is to be a minimum.

Let the square base have side length $l$ , and let the height of the tank be $h$ .

Then you want to minimize $l^2+4lh$ subject to the constraint $l^2h = 500$ .

Solve the second equation for $h$ , plug into the first, and minimize.

I keep getting the wrong answer. Can someone show me how to do it

Title: Re: TyErd's questions
Post by: the.watchman on June 20, 2010, 02:04:30 pm

Okay, first define some variables (I'll just use the ones decided on earlier)

Then make an equation for the area:

$A=l^2 + 4lh$

Make an expression for the volume, equate it to 500, and rearrange for h in terms of l:

$l^2 h=500$

$h=\frac{500}{l^2}$

Sub this into the area equation:

$A=l^2 + 4l \cdot \frac{500}{l^2} = l^2 + \frac{2000}{l}$

So $\frac{dA}{dl}= 2l-\frac{2000}{l^2} =0$ (for st. points)

$2l^3=2000$

$l=10$ (this is a local minimum, but i cbf proving it :P)

sub this into volume equation to get corresponding value for h

Title: Re: TyErd's questions
Post by: Yitzi_K on June 20, 2010, 02:05:35 pm

Quote from: TyErd on June 20, 2010, 12:27:50 pm

Quote from: /0 on June 17, 2010, 01:01:45 am
Quote from: TyErd on June 16, 2010, 07:10:30 pm
An open tank is to be constructed with a square base and vertical sides to contain $500m^3$ of water. What must be the dimensions of the area of sheet metal used in its construction if this area is to be a minimum.

Let the square base have side length $l$ , and let the height of the tank be $h$ .

Then you want to minimize $l^2+4lh$ subject to the constraint $l^2h = 500$ .

Solve the second equation for $h$ , plug into the first, and minimize.

I keep getting the wrong answer. Can someone show me how to do it

Just working on what /0 said, $l^2h = 500$ , therefore $h= \frac {500}{l^2}$

Subbing that into $l^2+4lh$ , we get $l^2+4l \frac {500}{l^2}$ , which simplifies to $l^2+ \frac{2000}{l}$

The derivative of that is $2l - \frac{2000}{l^2} $

Solving that for zero gives: $2l^3 - 2000 = 0$

$l^3 = 1000$

$ l = 10$

So the base of the sheet metal is 10 by 10.

Subbing $l=10$ back into $l^2h = 500$ gives $h=5,$ . So the sides are 10 by 5.

Title: Re: TyErd's questions
Post by: TyErd on June 20, 2010, 03:26:15 pm

Thankyou very much !

Title: Re: TyErd's questions
Post by: TyErd on June 28, 2010, 11:22:53 am

For the function $f(x) = 3- 2sin\frac{x}{2} , -\frac{\pi}{2} \leq x \leq \frac{3\pi}{2}$ has a minimum value at x equals?

Title: Re: TyErd's questions
Post by: TyErd on June 28, 2010, 11:26:21 am

Find the approximation of $\sqrt [4]{6203}$

Title: Re: TyErd's questions
Post by: TyErd on June 28, 2010, 11:36:46 am

Given that $f(x) = e^\frac{x}{10}$ find in terms of $h$ the approximate value of $f(10+h)$ , given that h is small.

Title: Re: TyErd's questions
Post by: TyErd on June 28, 2010, 11:57:33 am

Find the values of $x$ for which $100e^{-x^2 +2x - 5}$ increases as $x$ increases and hence find the maximum value of $100e^{-x^2 +2x - 5}.$ .

I dont even know where to start on this one. Help?

Title: Re: TyErd's questions
Post by: brightsky on June 28, 2010, 11:58:57 am

Quote from: TyErd on June 28, 2010, 11:26:21 am

Find the approximation of $\sqrt [4]{6203}$

Let $f(x) = \sqrt [4] {x} = x^{\frac{1}{4}$

So now our job is to approximate $f(6203)$ .

When you consider the graph, and draw a tangent to the graph at $x = 8^4 = 4096$ (this won't be a really good approximation as 6203 - 4096 is quite big), but the tangent should be very close to the actual point of 4096 + 2107 = 6203.

By linear approximation, $f(x+h) \approx hf'(x) + f(x)$ (derived from derivatives by first principles).

$f'(x) = \frac{1}{4} x^{-\frac{3}{4}}$

So $f(4096+2107) \approx 2107f'(4096) + f(4096)$ is the approximation (just sub in the values in use a calculator to work out the value).

Title: Re: TyErd's questions
Post by: brightsky on June 28, 2010, 12:01:09 pm

Quote from: TyErd on June 28, 2010, 11:36:46 am

Given that $f(x) = e^\frac{x}{10}$ find in terms of $h$ the approximate value of $f(10+h)$ , given that h is small.

Linear approximation formula:

$f(x+h) \approx hf'(x) + f(x)$

Hence $f(10+h) \approx hf'(10) + f(10)$ (just sub values in again to find answer).

Title: Re: TyErd's questions
Post by: the.watchman on June 28, 2010, 12:07:16 pm

Quote from: TyErd on June 28, 2010, 11:57:33 am

Find the values of $x$ for which $100e^{-x^2 +2x - 5}$ increases as $x$ increases and hence find the maximum value of $100e^{-x^2 +2x - 5}.$ .

I dont even know where to start on this one. Help?

This means basically to find when the expression is increasing (eg. derivative > 0), try finding that, then use the right endpoint (when derivative = 0) as the local maximum :)

Title: Re: TyErd's questions
Post by: brightsky on June 28, 2010, 12:08:42 pm

Quote from: TyErd on June 28, 2010, 11:57:33 am

Find the values of $x$ for which $100e^{-x^2 +2x - 5}$ increases as $x$ increases and hence find the maximum value of $100e^{-x^2 +2x - 5}.$ .

I dont even know where to start on this one. Help?

I don't really understand the first part of the question but the second part:

Maxima and minima exists when gradient = dy/dx = 0.

Let $y = 100e^{-x^2 + 2x - 5}$ ...(1)

$\frac{dy}{dx} = 100e^{-x^2 + 2x - 5} (-2x + 2)$

To find maxima, let that equal 0.

$100e^{-x^2 + 2x - 5} (-2x+2) = 0$

By null factor law, either $e^{-x^2 + 2x - 5} = 0$ or $-2x + 2 = 0$

The former has no solutions, the second yields $x = 1$

Sub in $x = 1$ into (1):

$y = \frac{100}{e^4}$

Title: Re: TyErd's questions
Post by: brightsky on June 28, 2010, 12:19:04 pm

Quote from: TyErd on June 28, 2010, 11:22:53 am

For the function $f(x) = 3- 2sin\frac{x}{2} , -\frac{\pi}{2} \leq x \leq \frac{3\pi}{2}$ has a minimum value at x equals?

Minimum for f(x) occurs when $\sin\left(\frac{x}{2}\right) = 1$ . So $\frac{x}{2} = \frac{\pi}{2}$ (only one within the domain) so $x = \pi$ .

Title: Re: TyErd's questions
Post by: TyErd on June 28, 2010, 12:27:47 pm

If im allowed to use the calculator why would I use the approximation formula in the first place?

Quote from: brightsky on June 28, 2010, 12:19:04 pm

Quote from: TyErd on June 28, 2010, 11:22:53 am
For the function $f(x) = 3- 2sin\frac{x}{2} , -\frac{\pi}{2} \leq x \leq \frac{3\pi}{2}$ has a minimum value at x equals?

Minimum for f(x) occurs when $\sin\left(\frac{x}{2}\right) = 1$ . So $\frac{x}{2} = \frac{\pi}{2}$ (only one within the domain) so $x = \pi$ .

how do you know the minimum is when $\sin\left(\frac{x}{2}\right) = 1$ ?

Title: Re: TyErd's questions
Post by: TrueTears on June 28, 2010, 12:30:20 pm

3+/-amplitude

Title: Re: TyErd's questions
Post by: brightsky on June 28, 2010, 12:33:06 pm

Quote from: TyErd on June 28, 2010, 12:27:47 pm

If im allowed to use the calculator why would I use the approximation formula in the first place?

It's just part of the methods course. Of course, you can work it out by hand and give an "exact solution" to the approximation if you want.

Quote from: TyErd on June 28, 2010, 11:22:53 am

how do you know the minimum is when $\sin\left(\frac{x}{2}\right) = 1$ ?

The minimum occurs when $\sin\left(\frac{x}{2}\right)$ is the largest positive number as 3 - (largest positive number) = smallest number. If you consider the graph $y = \sin(x)$ , the graph can only go as far up as 1. Hence this largest positive number is 1.

Title: Re: TyErd's questions
Post by: TyErd on June 28, 2010, 12:45:40 pm

Ohhh okay! thankyou so much

Title: Re: TyErd's questions
Post by: TyErd on June 28, 2010, 08:32:46 pm

Given that $y=e^{2x}$ , find in terms of q the approximate increase in y as x increase from 0 to q.

I get the answer of 2q+1 but the answer is just 2q

I used the linear approximation formula

Title: Re: TyErd's questions
Post by: TyErd on June 28, 2010, 09:10:53 pm

Same trouble with this one:

Fidn the approximate change in y when x changes from a to a+p, where p is small.
$ y=3-2e^x$

Title: Re: TyErd's questions
Post by: TyErd on June 28, 2010, 09:12:35 pm

help please

Title: Re: TyErd's questions
Post by: brightsky on June 28, 2010, 09:27:34 pm

Quote from: TyErd on June 28, 2010, 09:10:53 pm

Same trouble with this one:

Fidn the approximate change in y when x changes from a to a+p, where p is small.
$ y=3-2e^x$

$\delta y \approx \frac{dy}{dx} \delta x$

As $y = 3 - 2e^x$

$\frac{dy}{dx} = -2e^x$

And because x changes from a to a + p, $\delta x = p$

So $\delta y = -2e^ap$

Title: Re: TyErd's questions
Post by: brightsky on June 28, 2010, 09:29:54 pm

Quote from: TyErd on June 28, 2010, 08:32:46 pm

Given that $y=e^{2x}$ , find in terms of q the approximate increase in y as x increase from 0 to q.

I get the answer of 2q+1 but the answer is just 2q

I used the linear approximation formula

$\delta y \approx \frac{dy}{dx} \delta x$

As $y = e^{2x}$

$\frac{dy}{dx} = 2e^{2x}$

As x changes from 0 to q, $\delta x = q$

So $\delta y = 2e^{2 \times 0} q = 2q$

Title: Re: TyErd's questions
Post by: TyErd on June 28, 2010, 09:36:55 pm

Can you solve it using the formula $f(x+h) = f(x) +h f '(x)$ ?

Title: Re: TyErd's questions
Post by: TyErd on June 28, 2010, 09:37:18 pm

Im never sure about which one to use

Title: Re: TyErd's questions
Post by: brightsky on June 28, 2010, 10:04:54 pm

They mean the same thing, you can use either, the change in x in the non-Leibniz notation would be h and the change in y would be f(x+h) - f(x).

Title: Re: TyErd's questions
Post by: TyErd on June 28, 2010, 10:14:10 pm

I tried it on the above questions but it didn't work out for me :(

Title: Re: TyErd's questions
Post by: brightsky on June 28, 2010, 10:36:45 pm

Using your first one for an example, it would be:

$f(a+p) \approx f(a) + pf'(a)$

$f(a+p) - f(a) \approx -2e^{a}p$

Or you can work the left hand side to find the actual:

$f(a+p) - f(a) = 3 - 2e^{a + p} - (3 - 2e^a} = 3 - 2e^{a+p} - 3 + 2e^a = 2e^a - 2e^{a+p} = 2e^a(1 - 2e^p)$

As you can see, the approximation is pretty close to the actual.

Title: Re: TyErd's questions
Post by: TyErd on June 28, 2010, 10:47:11 pm

OHHH i see where i went wrong, Thankyou!!!

Title: Re: TyErd's questions
Post by: TyErd on June 28, 2010, 11:20:14 pm

Let $f(x) = sin(2x)$ . If $x$ is increased by a small amount $h$ , find:

An expression for the percentage change for $x= \frac {\pi}{6}$

Title: Re: TyErd's questions
Post by: brightsky on June 28, 2010, 11:55:28 pm

Percentage change = $100 \left(\frac{f\left(\frac{\pi}{6}+ h\right) - f(h)}{f(h)}\right)$

Just plug in the values and solve.

Or if you want a linear approximation:

$100 \left(\frac{hf'\left(\frac{\pi}{6}\right)}{f\left(\frac{\pi}{6}\right)\right)}$

Title: Re: TyErd's questions
Post by: TyErd on June 29, 2010, 12:13:45 am

your latex doesnt work

Title: Re: TyErd's questions
Post by: brightsky on June 29, 2010, 12:14:46 am

Oh, sorry, should work now. :)

Title: Re: TyErd's questions
Post by: TyErd on June 29, 2010, 12:26:17 am

Thanks!

Title: Re: TyErd's questions
Post by: TyErd on July 01, 2010, 11:56:28 am

$\int (4-3x)^{-1} dx$

What I did was:

$= \frac {(4-3x)^{0}}{-3 \times 0} + c$ which is clearly wrong but I dont know why this working out is wrong.

Title: Re: TyErd's questions
Post by: brightsky on July 01, 2010, 12:04:14 pm

$\int \frac{1}{4-3x} dx$

Substitute $u = 4 - 3x$ , then $du = -3 dx$

So the integrand becomes:

$-\frac{1}{3} \int \frac{1}{u} dx$

$= -\frac{1}{3} \log_e|u| + C$

$= -\frac{1}{3} \log_e|4-3x| + C$

EDIT: Absolute values.

Title: Re: TyErd's questions
Post by: the.watchman on July 01, 2010, 12:05:39 pm

Remember that the derivative of a log is a fraction with a linear polynomial in its denominator

So your answer should be $-\frac{1}{3}log_e |4-3x| + c$ (i think :P)

(@brightsky: absolute values buddy :P)

Title: Re: TyErd's questions
Post by: brightsky on July 01, 2010, 12:07:08 pm

Oh crap! Thanks the.watchman! :p

Title: Re: TyErd's questions
Post by: TyErd on July 01, 2010, 12:09:46 pm

Ohh so that means when there's a question like $\int \frac{5}{x} dx$ , you can't bring it up so that the equations is $\int 5x^{-1} dx$ ?

Title: Re: TyErd's questions
Post by: brightsky on July 01, 2010, 12:19:47 pm

Nope because $\int \frac{1}{x} dx = \log_e|x|$ .

Title: Re: TyErd's questions
Post by: TyErd on July 01, 2010, 12:30:48 pm

Okay thanks for clearing that up :)

Title: Re: TyErd's questions
Post by: superflya on July 01, 2010, 12:59:53 pm

just a tip, taking out the constant has no effect and will make things simpler.

Title: Re: TyErd's questions
Post by: TyErd on July 01, 2010, 01:06:17 pm

anti-derivative of tan(kx)?

Title: Re: TyErd's questions
Post by: TyErd on July 01, 2010, 01:06:58 pm

Quote from: superflya on July 01, 2010, 12:59:53 pm

just a tip, taking out the constant has no effect and will make things simpler.

What do you mean? can you give an example please :)

Title: Re: TyErd's questions
Post by: brightsky on July 01, 2010, 01:08:43 pm

Quote from: TyErd on July 01, 2010, 01:06:58 pm

Quote from: superflya on July 01, 2010, 12:59:53 pm
just a tip, taking out the constant has no effect and will make things simpler.

What do you mean? can you give an example please :)

Just as in your previous example, $\int \frac{5}{x} dx$ , we can simplify this down to $5 \int \frac{1}{x} dx = 5 \log_e(x)$ .

Title: Re: TyErd's questions
Post by: TyErd on July 01, 2010, 01:10:05 pm

OH yeas Thanks!

Title: Re: TyErd's questions
Post by: superflya on July 01, 2010, 01:13:30 pm

Quote from: TyErd on July 01, 2010, 01:06:17 pm

anti-derivative of tan(kx)?

ln|sec(x)|

Title: Re: TyErd's questions
Post by: the.watchman on July 01, 2010, 01:14:36 pm

Quote from: TyErd on July 01, 2010, 01:06:17 pm

anti-derivative of tan(kx)?

There is no 'methods' anti-derivative of tan, so don't worry about it (unless you do spesh, in which case, you should know it)

Quote from: brightsky on July 01, 2010, 01:08:43 pm

Quote from: TyErd on July 01, 2010, 01:06:58 pm
Quote from: superflya on July 01, 2010, 12:59:53 pm
just a tip, taking out the constant has no effect and will make things simpler.

What do you mean? can you give an example please :)

Just as in your previous example, $\int \frac{5}{x} dx$ , we can simplify this down to $5 \int \frac{1}{x} dx = 5 \log_e(x)$ .

This brings up another important point:

If a question is: Find the antiderivative of $\frac{1}{3x}$

There are many answers, one of which is $\frac{1}{3}log_e |3x|+c$ , another is $\frac{1}{3}log_e |x| +c$ , depending on what method you choose

Now most of you will already know this, but because $log_e |x| + log_e (3)=log_e |3x|$
They are all equivalent, as $log_e (3)$ is a constant
So if the textbook has/lacks a constant inside the log, it doesn't matter :)

EDIT: Sorry to overlap your post superflya, but I'm a very slow typer... :D

Title: Re: TyErd's questions
Post by: brightsky on July 01, 2010, 01:22:25 pm

Quote from: TyErd on July 01, 2010, 01:06:17 pm

anti-derivative of tan(kx)?

First find $\int \tan(x) dx$ :

We know that $\int \tan(x) dx = \int \frac{\sin(x)}{\cos(x)} dx$

Make substitution $u = \cos(x)$ , then $du = -\sin(x) dx$

So the original integrand becomes:

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

$= -1 \cdot \log_e|u|$

$= -\log_e|\cos(x)|$

To find $\int \tan(kx) dx$ , use u-susbtitution again.

Title: Re: TyErd's questions
Post by: superflya on July 01, 2010, 01:26:24 pm

isnt too hard to hard to get the integral of tan(x), bet watchman knows how to do it :P

rewrite tanx as sinx/cosx $\int \frac{\sin (x)}{\cos (x)} dx$

let $u=\cos (x) \rightarrow \frac{\mathrm{d} }{\mathrm{d} x}=-\sin (x)$

now it should look something like this $-\int \frac{1}{u} du$ its in the same form as the 5/x so u know theres gotta be -ln|u|

as u=cosx $\Rightarrow -\ln |\cos (x)| +c$

u can leave it like that or notice the neg infront of the natural log, u can bring it up to the cos(x)

$-\ln |\cos (x)| +c = \ln |(\cos (x))^{-1}|+c = \ln |\sec (x)| +c$

EDIT: brightskyy :P

Title: Re: TyErd's questions
Post by: the.watchman on July 01, 2010, 01:29:26 pm

I didn't know, but now I do :)

Title: Re: TyErd's questions
Post by: TyErd on July 01, 2010, 02:00:50 pm

So it doesn't matter if we don't know the integral of tan(x)? Will it come up on the exams?

Title: Re: TyErd's questions
Post by: brightsky on July 01, 2010, 02:50:02 pm

Nah, not on Methods exams, but it's always handy to know just in case. :p

Title: Re: TyErd's questions
Post by: TyErd on July 01, 2010, 03:07:23 pm

okay thanks

Title: Re: TyErd's questions
Post by: TyErd on July 01, 2010, 03:17:08 pm

Find $\frac{d}{dx} (x sin3x)$ , and hence evaluate $\int^\frac{\pi}{6}_0 x cos3x dx$

Not really sure how to anti differentiate when two terms are multiplied. Btw thats $\frac {\pi}{6}$ if you can't read it.

Title: Re: TyErd's questions
Post by: brightsky on July 01, 2010, 03:26:25 pm

$\frac{d}{dx} (x \sin(3x)) = 3x\cos(3x) + \sin(3x)$

Integrate both sides:

$x \sin(3x) = \int 3x\cos(3x) dx + \int \sin(3x) dx$

$x\sin(3x) - \int \sin(3x) dx = 3 \int x \cos(3x) dx$

$\int x \cos(3x) dx = \frac{1}{3} x\sin(3x) + \frac{1}{9} \cos(3x) + C$

The rest should be straightforward. :)

EDIT: Forgot the 3 on the RHS :p.

Title: Re: TyErd's questions
Post by: brightsky on July 01, 2010, 03:34:06 pm

For a lot of integrals with two terms multiplied together, you can work it out through integration by parts, which the first step (finding the derivative of $x \sin(3x)$ ) does in a way.

Integration by parts comes from the product rule, just manipulated around. By the product rule:

$[f(x)g(x)]' = f'(x)g(x) + f(x)g'(x)$

Integrate both sides:

$f(x)g(x) = \int f'(x)g(x) dx + \int f(x)g'(x) dx$

Manipulate around:

$\int f'(x)g(x) dx = f(x)g(x) - \int f(x)g'(x) dx$

In this case, we let $f'(x) = \cos(3x)$ and $g(x) = x$ .

Title: Re: TyErd's questions
Post by: TyErd on July 01, 2010, 03:42:43 pm

Quote from: brightsky on July 01, 2010, 03:26:25 pm

$\frac{d}{dx} (x \sin(3x)) = 3x\cos(3x) + \sin(3x)$

Integrate both sides:

$x \sin(3x) = \int 3x\cos(3x) dx + \int \sin(3x) dx$

$x\sin(3x) - \int \sin(3x) dx = 3 \int x \cos(3x) dx$

$\int x \cos(3x) dx = \frac{1}{3} x\sin(3x) + \frac{1}{9} \cos(3x) + C$

The rest should be straightforward. :)

EDIT: Forgot the 3 on the RHS :p.

Would you be able to clear up the last line, not really sure how you got that, in particular what you did with the 3

Title: Re: TyErd's questions
Post by: brightsky on July 01, 2010, 04:28:03 pm

We have:

$x \sin(3x) - \int \sin(3x) dx = 3 \int x \cos(3x) dx$

$\because \int \sin(3x) dx = -\frac{1}{3} \cos(3x) + C$

$\therefore x \sin(3x) + \frac{1}{3} \cos(3x) + C= 3 \int x \cos(3x) dx$

Bringing the 3 over to the LHS:

$\frac{1}{3} x \sin(3x) + \frac{1}{9} \cos(3x) + C = \int x \cos(3x) dx$

Title: Re: TyErd's questions
Post by: TyErd on July 01, 2010, 04:35:08 pm

Ohhh I get it!! very clear as well thankyou!

Title: Re: TyErd's questions
Post by: brightsky on July 01, 2010, 04:39:47 pm

Quote from: TyErd on July 01, 2010, 04:35:08 pm

Ohhh I get it!! very clear as well thankyou!

No problem! :D

Title: Re: TyErd's questions
Post by: TyErd on July 02, 2010, 12:56:48 pm

Ok In the essentials book it says in one of the questions that one revolution every 10 seconds is equivalent to a rate of $\frac{\pi}{5}$ radians per seconds. How do they get that?

Title: Re: TyErd's questions
Post by: Whatlol on July 02, 2010, 12:58:51 pm

Quote from: TyErd on July 02, 2010, 12:56:48 pm

Ok In the essentials book it says in one of the questions that one revolution every 10 seconds is equivalent to a rate of $\frac{\pi}{5}$ radians per seconds. How do they get that?

Since a full cirlce is 2 pi then one revolution is 2pi / 10 = pi/5

Title: Re: TyErd's questions
Post by: TyErd on July 02, 2010, 01:01:29 pm

ohh ofcourse, thnx for the quick response

Title: Re: TyErd's questions
Post by: Whatlol on July 02, 2010, 01:02:32 pm

no problems!

Title: Re: TyErd's questions
Post by: TyErd on July 02, 2010, 01:27:00 pm

An aeroplane is flying. horizontally at a constant height of 1000m. At a certain instant the angle of elevation is $30^\circ$ and decreasing and the speed of the aeroplane is $480km/h.$

a) How fast is $\theta$ decreasing at this instant?
b) How fast is the distance between the aeroplane and the observation point changing at this instant?

Title: Re: TyErd's questions
Post by: Whatlol on July 02, 2010, 01:40:10 pm

Quote from: TyErd on July 02, 2010, 01:27:00 pm

An aeroplane is flying. horizontally at a constant height of 1000m. At a certain instant the angle of elevation is $30^\circ$ and decreasing and the speed of the aeroplane is $480km/h.$

a) How fast is $\theta$ decreasing at this instant?
b) How fast is the distance between the aeroplane and the observation point changing at this instant?

Hmm... what are the answers?

Title: Re: TyErd's questions
Post by: TyErd on July 02, 2010, 01:57:25 pm

a) $\frac{1}{30} rad/s$

b) $\frac{200 \sqrt 3 }{3} m/s$

Title: Re: TyErd's questions
Post by: Whatlol on July 02, 2010, 03:15:35 pm

Quote from: TyErd on July 02, 2010, 01:57:25 pm

a) $\frac{1}{30} rad/s$

b) $\frac{200 \sqrt 3 }{3} m/s$

Hmm im not sure about part a, but i figured how to do part b.

First step is convert 480km/h to m/s so just do (480x1000) / 3600 = 133m/s
then you need to do cos30 x 133 = 115.5

Title: Re: TyErd's questions
Post by: TyErd on July 02, 2010, 03:18:21 pm

I'm pretty sure you have to use differentiation to solve it.

Title: Re: TyErd's questions
Post by: Whatlol on July 02, 2010, 03:23:43 pm

Quote from: TyErd on July 02, 2010, 03:18:21 pm

I'm pretty sure you have to use differentiation to solve it.

Well that way gives you the right answer :p just using a different way. or do you mean about part a ?
im trying to figure out what you have to differentiate tho.. )= i mean i guess you need to write an expression for theta ? not sure to be honest.

Title: Re: TyErd's questions
Post by: TyErd on July 02, 2010, 03:42:15 pm

its for both im pretty sure, its the chain rule I think.

Title: Re: TyErd's questions
Post by: Whatlol on July 02, 2010, 03:46:53 pm

Hmm sorry i cant help i honestly dont know how to do it )=

Title: Re: TyErd's questions
Post by: moekamo on July 02, 2010, 04:01:29 pm

$\frac{d\theta}{dt} = \frac{dx}{dt} \frac{d\theta}{dx}$ where x is the horizontal distance from the pbservation point to under the plane.

since $\tan \theta = 1000 /x \implies x = 1000/\tan\theta \implies \frac{dx}{d\theta} = \frac{\tan \theta \cdot 0 - 1000\sec^2 \theta}{\tan ^2 \theta} = -1000/\sin ^2 \theta$

when theta is 30 degrees, $\frac{dx}{d \theta} = -4000 \implies \frac{d \theta}{dx} = -\frac{1}{4000} \implies \frac{d \theta}{dt} = -\frac{1}{4000}\times \frac{480}{3.6} = -1/30$ so decreasing at 1/30 rad/sec

the next part is to find dh/dt where h is the hypotenuse, can you try it?

Title: Re: TyErd's questions
Post by: TyErd on July 02, 2010, 04:25:43 pm

Okays

$\frac{dh}{dt} = \frac {dh}{d\theta} \frac {d\theta}{dt}$

$h= \frac{1000}{sin\theta}, \frac{dh}{d\theta} = \frac{-1000cos\theta}{sin^2\theta} $

$\frac{d\theta}{dt}$ was found to be $-\frac{1}{30}$

$So \frac{dh}{dt} = \frac{-1000cos30}{sin^2(30)} \times -\frac{1}{30}$

For some reason i think thats wrong

Title: Re: TyErd's questions
Post by: brightsky on July 02, 2010, 04:26:39 pm

Let me continue on...

$\frac{dh}{dt} = \frac{dh}{d\theta} \times \frac{d\theta}{dt}$ , where h is the length of the hypotenuse.

We know that $\frac{d\theta}{dt} = -\frac{1}{30}$

$\because \sin(\theta) = \frac{1000}{h}$

$\therefore h = \frac{1000}{\sin(\theta)}$

$\frac{dh}{d\theta} = -1000 \cot(\theta) \csc(\theta)$

When $x = \frac{\pi}{6}$ , $\frac{dh}{d\theta} = -2000 \sqrt{3}$

Hence:

$\frac{dh}{dt} = -2000\sqrt{3} \cdot -\frac{1}{30} = \frac{2000\sqrt{3}}{30} = \frac{200\sqrt{3}}{3}$

Title: Re: TyErd's questions
Post by: brightsky on July 02, 2010, 04:30:36 pm

Quote from: TyErd on July 02, 2010, 04:25:43 pm

Okays

$\frac{dh}{dt} = \frac {dh}{d\theta} \frac {d\theta}{dt}$

$h= \frac{1000}{sin\theta}, \frac{dh}{d\theta} = \frac{-1000cos\theta}{sin^2\theta} $

$\frac{d\theta}{dt}$ was found to be $-\frac{1}{30}$

$So \frac{dh}{dt} = \frac{-1000cos30}{sin^2(30)} \times -\frac{1}{30}$

For some reason i think thats wrong

Change 30 degrees into radians. :p

Title: Re: TyErd's questions
Post by: TyErd on July 02, 2010, 04:30:54 pm

Yay I got it right, I just realised instead of 30 i have to sub in $\frac {\pi}{6}$

Title: Re: TyErd's questions
Post by: TyErd on July 02, 2010, 04:31:18 pm

damn Beaten lol

Title: Re: TyErd's questions
Post by: TyErd on July 02, 2010, 04:31:56 pm

Thanks everyone!

Title: Re: TyErd's questions
Post by: TyErd on July 02, 2010, 04:36:53 pm

I have another problem: ABCD is a trapezium with AB=CD, with vertices on the circle and with the centre O. AD is a diameter of the circle. The radius of the circle is 4units.

a) Find BC in terms of $\theta$ .
b) Find the area of the trapezium in terms of $\theta$ and hence find the maximum area.

image attached

Title: Re: TyErd's questions
Post by: brightsky on July 02, 2010, 04:42:42 pm

$\angle COD = \angle BOA$ (They subtend the same size arc).

Hence $\angle BOA = \theta$

So $\angle BOC = 180 - 2\theta$

$\therefore \angle OBC = \angle OCB = \theta$

By the sine rule:

$\frac{4}{\sin\theta} = \frac{BC}{\sin(180 - 2\theta)}$

Solve that to get your answer. :P

Title: Re: TyErd's questions
Post by: TyErd on July 02, 2010, 05:15:34 pm

error in latex btw

umm why is $\angle OBC$ and $\angle OCB$ equal to $\theta .$

Title: Re: TyErd's questions
Post by: brightsky on July 02, 2010, 05:24:48 pm

Because $\triangle OBC$ is an isosceles triangle. Hence its base angles are equal, hence $\angle OBC = \angle OCB$ .

So because $\angle BOC = 180 - 2\theta$

Then $\angle OBC = \angle OCB = \frac{180 - (180 - 2\theta)}{2} = \frac{2\theta}{2} = \theta$

Continuing on from $\frac{4}{\sin(\theta)} = \frac{BC}{\sin(180 - 2\theta)}$

$BC = \frac{4 \sin(180 - 2\theta)}{\sin(\theta)}$

$BC = \frac{4 \sin(2\theta)}{\sin (\theta)}$

$BC = 8\cos(\theta)$

Title: Re: TyErd's questions
Post by: TyErd on July 02, 2010, 05:26:45 pm

okaays thankyou very much I get that part now, what about the second part?

Title: Re: TyErd's questions
Post by: brightsky on July 02, 2010, 05:34:48 pm

Let $A$ be the area of the trapezium.

Then $A = \frac{1}{2} (BC + AD) h$ , where h is the height of the trapezium.

We already know BC and AD in terms of $\theta$ . $h$ is the height of $\triangle OBC$ which can be found by Pythagoras' theorem.

Once we have the equation, finding max area would be straightforward.

Title: Re: TyErd's questions
Post by: TyErd on July 02, 2010, 05:46:00 pm

Okay I got $A= \frac{1}{2} \times (\frac{4sin(2\theta)}{sin(\theta)} + 8 ) \times 4cos(90-\theta)$

not sure if that is correct though....

but then I have to differentiate it make it equal zero find $\theta$ and then sub back into the equation correct?

Title: Re: TyErd's questions
Post by: TyErd on July 02, 2010, 10:13:01 pm

Another question, if z= sin x and sin 1=a then using linear approximation find the value of sin(1.1). I kinda get stuck half way through, can someone show me how to do it please?

Title: Re: TyErd's questions
Post by: brightsky on July 02, 2010, 10:28:15 pm

By linear approximation:

$f(x+h) \approx hf'(x) + f(x)$

We have $z = f(x)$ , so $f(x) = \sin(x)$ .

$\because \sin(1) = a$ ...(1)

$\therefore f(1 +0.1) \approx 0.1 \cdot \cos(1) + \sin(1)$

$f(1.1) = 0.1\cos(1) + \sin(1)$ ...(2)

From (1):

$\sin^2(1) = a^2$

$1 - \cos^2(1) = a^2$

$\cos^2(1) = 1 - a^2$

$\cos(1) = \sqrt{1 - a^2}$ (reject the negative solution)

Substitute that into (2),

$f(1.1) = 0.1 \sqrt{1-a^2} + a$

Hope this is right. :p

Title: Re: TyErd's questions
Post by: TyErd on July 02, 2010, 10:33:55 pm

Thankyou very much! and yes you are correct :) arent you always? lol

Title: Re: TyErd's questions
Post by: TyErd on July 02, 2010, 10:36:37 pm

oh btw why do you have to reject the negative solution?

Title: Re: TyErd's questions
Post by: TyErd on July 02, 2010, 11:06:20 pm

Also how do you do this: A section of a rollercoaster is given by the equation $y= 18 cos (\frac{\pi x}{80}) + 12 , 0 \leq x \leq 80$

State the coordinates of the point of the track for which the magnitude of the gradient is maximum.

Title: Re: TyErd's questions
Post by: TyErd on July 02, 2010, 11:12:52 pm

Also having trouble with this one, A rocket rises vertically from level ground at a point A. It is observed from another point B on the ground where B is 10km from point A. When the angle of elevation ABR has the value $\frac {\pi}{4}$ radians, this angle is increasing at the rate of 0.005 radians per second. Find in km/s the velocity of the rocket at that instant.

I know that you have to use the chain rule but the velocity part confuses me. Any help?

Title: Re: TyErd's questions
Post by: brightsky on July 02, 2010, 11:24:59 pm

$\frac{dy}{dx} = -\frac{18\pi}{80} \sin\left(\frac{\pi x}{80}\right) = -\frac{9\pi}{40} \sin\left(\frac{\pi x}{80}\right)$

Treat this as another function. We want to find when the dy/dx is at its maximum. That is when $\frac{d^2y}{dx^2} = 0$ .

So $\frac{d^2y}{dx^2} = -\frac{9\pi^2}{3200} \cos \left(\frac{\pi x}{80}\right) = 0$

Simplifying this gives:

$\cos\left(\frac{\pi x}{80}\right) = 0$

$\because 0 \le x \le 80$

$\therefore 0 \le \frac{\pi x}{80} \le \pi$

$\frac{\pi x}{80} = \frac{\pi}{2}$

$x = 40$

Substituting that back into the original function:

$y = 18 \cos \left(\frac{40\pi}{80}\right) + 12 = 12$

So the point on the track for which the mangitude of the gradient is maximum is (40,12).

Title: Re: TyErd's questions
Post by: TyErd on July 02, 2010, 11:34:50 pm

I dont get the $\frac{d^2 y}{dx^2}$ part, how did you get that?

Title: Re: TyErd's questions
Post by: brightsky on July 02, 2010, 11:42:52 pm

$\frac{dx}{dt} = \frac{dx}{d\theta} \cdot \frac{d\theta}{dt}$ where x is the distance from A to R.

We know that:

$\tan \theta = \frac{x}{10}$

$x = 10 \tan (\theta)$

$\frac{dx}{d\theta} = 10\sec^2(\theta)$ ...(1)

When $\theta = \frac{\pi}{4}$ , $\frac{dx}{d\theta} = 20$

We also know that:

$\frac{d\theta}{dt} = 0.005$ when $\theta = \frac{\pi}{4}$

Hence $\frac{dx}{dt} = 20 \times 0.005 = 0.1$ km/s.

Title: Re: TyErd's questions
Post by: brightsky on July 02, 2010, 11:44:40 pm

Quote from: TyErd on July 02, 2010, 11:34:50 pm

I dont get the $\frac{d^2 y}{dx^2}$ part, how did you get that?

We treat $\frac{dy}{dx} = ...$ as a normal function. To find this function is at its maximum - that is when $\frac{dy}{dx}$ is a maximum, we need to derive it and then let it = 0.

Deriving $\frac{dy}{dx}$ would yield $\frac{d}{dx} \left(\frac{dy}{dx}\right) = \frac{d^2y}{dx^2}$

Title: Re: TyErd's questions
Post by: brightsky on July 02, 2010, 11:51:57 pm

Quote from: TyErd on July 02, 2010, 10:36:37 pm

oh btw why do you have to reject the negative solution?

Hehe, took a shortcut but we know $\cos(1)$ is positive given its graph. It's only between $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ that the graph goes negative.

Title: Re: TyErd's questions
Post by: TyErd on July 02, 2010, 11:59:36 pm

Quote from: brightsky on July 02, 2010, 11:44:40 pm

Quote from: TyErd on July 02, 2010, 11:34:50 pm
I dont get the $\frac{d^2 y}{dx^2}$ part, how did you get that?

We treat $\frac{dy}{dx} = ...$ as a normal function. To find this function is at its maximum - that is when $\frac{dy}{dx}$ is a maximum, we need to derive it and then let it = 0.

Deriving $\frac{dy}{dx}$ would yield $\frac{d}{dx} \left(\frac{dy}{dx}\right) = \frac{d^2y}{dx^2}$

Oh okay so because gradient is dy/dx, where finding when that is maximum so as usual we differentiate and make it equal zero. Okay I get it thanks

Title: Re: TyErd's questions
Post by: brightsky on July 03, 2010, 12:00:42 am

Quote from: TyErd on July 02, 2010, 11:59:36 pm

Oh okay so because gradient is dy/dx, where finding when that is maximum so as usual we differentiate and make it equal zero.

Yep.

Title: Re: TyErd's questions
Post by: TyErd on July 03, 2010, 12:06:24 am

Thanks so much brightsky! you've helped me so much today really appreciate it!

Title: Re: TyErd's questions
Post by: TyErd on July 03, 2010, 12:08:54 am

Um what do you need to know about the fundamental theorem of calculus? There's like 3 pages explaining it but is it important to know?

Title: Re: TyErd's questions
Post by: brightsky on July 03, 2010, 12:29:47 am

The fundamental theorem of calculus is basically just the basic ideas related to integration, differentiation, indefinite/definite integrals, etc. Don't think you need to know the real details of it and where it came from but it's useful to understand; the actual "statements" of the theorem you would've already known, like integration is the opposite of differentiation, etc.

Title: Re: TyErd's questions
Post by: /0 on July 03, 2010, 02:55:23 pm

All you need to know about the fundamental theorem of calculus is this:

$\int_a^b f(x)\,dx = F(b)-F(a)$ , where $F'(x)=f(x)$

Title: Re: TyErd's questions
Post by: TyErd on July 03, 2010, 03:16:55 pm

okays thnx /0 , how do you simplify this: $cos(\pi + a ) + cos(\pi - a)$

Title: Re: TyErd's questions
Post by: brightsky on July 03, 2010, 03:31:23 pm

$\cos(\pi + a) + \cos(\pi - a) = -\cos(a) - \cos(a) = -2\cos(a)$

Imagine it on a unit circle. For $\cos(\pi + a)$ , we are given the angle $a$ , then rotate that around $\pi$ radians. We end up in the 3rd quadrant (but the angle is still the same by symmetry), which is negative for cosine, hence it becomes $-\cos(a)$ . Same deal for $\cos(\pi-a)$ .

Title: Re: TyErd's questions
Post by: TyErd on July 03, 2010, 03:44:07 pm

Thankyou again brightsky, clear as always :)

Title: Re: TyErd's questions
Post by: TyErd on July 03, 2010, 06:00:40 pm

$\int \frac{1}{2-2x} dx$

Title: Re: TyErd's questions
Post by: the.watchman on July 03, 2010, 06:02:24 pm

$\int \frac{1}{2-2x}dx$

$= -\frac{1}{2}\int \frac{1}{x-1}dx$

$= -\frac{1}{2}log_e |x-1| + c$

Title: Re: TyErd's questions
Post by: TyErd on July 03, 2010, 06:14:26 pm

thanks the.watchman

I have another problem:

the rate of flow of water from a reservoir is given by $\frac{dv}{dt} = 1000 - 30t^2 + 2t^3$ ,

$0 \leq t \leq 15$ where $V$ is measured in millions of litres and $t$ is the number of hours after the sluice gates are opened.

Find

a) The times when the rate of flow is maximum
b) The maximum flow

Title: Re: TyErd's questions
Post by: brightsky on July 03, 2010, 06:36:59 pm

$\frac{d^2V}{dt^2} = -60t + 6t^2$

The maxima or minima exists when the derivative = 0, so:

$6t^2 - 60t = 0$

$6t(t - 10)=0$

$6t = 0$ or $t - 10 = 0$

$t = 0$ or $t = 10$

Substitute the values for t back into the original equation:

$\frac{dV}{dt} = 1000$ or $\frac{dV}{dt} = 0$

Hence clearly a) is t = 0 and b) is 1000.

I think $t = 15$ is also an answer, because by letting the derivative = 0, we are only finding the turning points, but with a restricted domain, t = 15 also yields $\frac{dV}{dt} = 1000$ .

Title: Re: TyErd's questions
Post by: TyErd on July 03, 2010, 06:42:53 pm

hm.. the back of the book says the answer for a are t=0 and t=15. I got the same answers as you as well.

Title: Re: TyErd's questions
Post by: TyErd on July 03, 2010, 06:43:19 pm

lol damn beaten

Title: Re: TyErd's questions
Post by: TyErd on July 03, 2010, 06:44:27 pm

Do you know how to get 15 with working out?

Title: Re: TyErd's questions
Post by: brightsky on July 03, 2010, 06:45:51 pm

Yeah, for the {x:dy/dx = 0} method, we're only finding the local maximum/minimum, but in your case, we need a global maximum, which is really just one step more. We know that the largest "y-value" in this case is 1000, so we substitute dV/dt = 1000 and solve for t to get the other t = 15 answer.

Title: Re: TyErd's questions
Post by: TyErd on July 03, 2010, 06:48:20 pm

OHH okayys thankyou heaps!

Title: Re: TyErd's questions
Post by: TyErd on July 03, 2010, 06:48:46 pm

oh yeah another problem What are the coordinates of the point of intersection of the graphs $y=3 sin(x) and y = 4 cos(x) , (0 \leq x \leq \frac{\pi}{2})$

Title: Re: TyErd's questions
Post by: brightsky on July 03, 2010, 06:52:40 pm

$3\sin(x) = 4\cos(x)$

$\frac{3\sin(x)}{4\cos(x)} = 1$

$\frac{3}{4} \tan(x) = 1$

$\tan(x) = \frac{4}{3}$

$x = \arctan\left(\frac{4}{3}\right)$

Sub back to the original equation to find y coordinate.

Lol, I don't know if this method is legitimate or not...:S

Title: Re: TyErd's questions
Post by: TyErd on July 03, 2010, 06:56:23 pm

hmm never learnt arctan

Title: Re: TyErd's questions
Post by: TyErd on July 03, 2010, 06:57:39 pm

another problem while I quickly go have dinner :)

$\int_0^\frac{\pi}{4} \frac { 1 - cos (2x) } {1 + cos (2x)} dx$

is this even in the methods course?

Title: Re: TyErd's questions
Post by: superflya on July 03, 2010, 06:58:38 pm

Quote from: TyErd on July 03, 2010, 06:56:23 pm

hmm never learnt arctan

inverse tan.

Title: Re: TyErd's questions
Post by: brightsky on July 03, 2010, 07:36:48 pm

First find the indefinite integral $\int \frac{1 - \cos(2x)}{1 +\cos(2x)} dx$

$= \int \frac{1 - \frac{e^{2ix} + e^{-2ix}}{2}}{1 + \frac{e^{2ix} + e^{-2ix}}{2}} dx$

$= \int \frac{\frac{2 - e^{2ix} - e^{-2ix}}{2}}{\frac{2 + e^{2ix} + e^{-2ix}}{2}} dx$

$= \int \frac{\frac{e^{2ix} - 2 + e^{-2ix}}{4i^2}}{\frac{2 + e^{2ix} + e^{-2ix}}{4}} dx$

$= \int \frac{\sin^2(x)}{\cos^2(x)} dx$

$= \int \tan^2(x) dx$

But $\tan^2(x) = \sec^2(x) - 1$

So the original integrand becomes:

$\int \sec^2(x) - 1 dx$

$= \int \sec^2(x) dx - \int 1 dx$

$= \int \sec^2(x) dx - x + C$

$= \tan(x) - x + C$

EDIT: Latex error.

Using this:

$\int^{\frac{\pi}{4}}_{0} \frac{1 - \cos(x)}{1 +\cos(x)} dx$

$= \tan\left(\frac{\pi}{4}\right) - \frac{\pi}{4} - \tan(0) + 0$

$= 1 - \frac{\pi}{4}$

Title: Re: TyErd's questions
Post by: moekamo on July 03, 2010, 07:38:37 pm

Quote from: TyErd on July 03, 2010, 06:57:39 pm

another problem while I quickly go have dinner :)

$\int_0^\frac{\pi}{4} \frac { 1 - cos (2x) } {1 + cos (2x)} dx$

is this even in the methods course?

a bit of simplifying using double angle formulas: since $\cos 2x = 2\cos ^2 x - 1 = 1 - 2 \sin ^2 x$ we can do this:

$\frac{1-\cos 2x}{1 + \cos 2x} = \frac{1-1+2\sin ^2 x}{1+2\cos ^2 x - 1} = \frac{\sin ^2 x}{\cos ^2 x} = \tan ^2 x = \sec ^2 x - 1$

therefore $\int_0^{\frac{\pi}{4}} \frac{1-\cos 2x}{1 + \cos 2x} dx = \int_0^{\frac{\pi}{4}} \sec ^2 x - 1 dx = \left [ \tan x -x \right ]_0^{\frac{\pi}{4}} = 1-\frac{\pi}{4}$

since you have to use double angle formulas and stuff, its not going to appear on a methods exam, but if you can follow it and understand then thats better for you :)

Title: Re: TyErd's questions
Post by: TyErd on July 03, 2010, 08:01:46 pm

so it definitely wont appear in the exam?

Title: Re: TyErd's questions
Post by: TyErd on July 04, 2010, 06:13:08 pm

The straight line $y=2x+1$ cuts the curve $y=e^x + 1 at (0,1)$ . At this point the acute angle between the line and the curve is $\theta$ . What is $tan \theta$

I'm confused with this question because they don't intersect. Question is from essentials pg 500.

Title: Re: TyErd's questions
Post by: brightsky on July 04, 2010, 06:15:55 pm

Quote from: TyErd on July 04, 2010, 06:13:08 pm

The straight line $y=2x+1$ cuts the curve $y=e^x + 1 at (0,1)$ . At this point the acute angle between the line and the curve is $\theta$ . What is $tan \theta$

I'm confused with this question because they don't intersect. Question is from essentials pg 500.

LOL. :p

Title: Re: TyErd's questions
Post by: TyErd on July 04, 2010, 06:20:30 pm

Also $\int_0^{36} \frac{dx}{2x + 9} = log_e k$ , then $k$ is ?

Title: Re: TyErd's questions
Post by: TyErd on July 04, 2010, 06:25:46 pm

and this one too $\int_a^b sin(2x) dx = 0$ . What does a and b equal?

Title: Re: TyErd's questions
Post by: m@tty on July 04, 2010, 06:27:22 pm

$\int \frac{dx}{2x+9}=\frac{1}{2}\log_e(2x+9)+C$

$\int_0^{36}\frac{dx}{2x+9}=\frac{1}{2}\left(\log_e{81}-\log_e{9}\right)=\frac{1}{2}\log_e9$

Therefore $k=\sqrt{9}=3$

Title: Re: TyErd's questions
Post by: m@tty on July 04, 2010, 06:30:55 pm

Quote from: TyErd on July 04, 2010, 06:25:46 pm

and this one too $\int_a^b sin(2x) dx = 0$ . What does a and b equal?

For this there are infinite solutions.

It will equal zero for all $a=b$ .

But I assume that you are meant to use symmetry to find a situation such as $a=0$ and $b=\pi$ . Where the net result is zero.

Title: Re: TyErd's questions
Post by: crappy on July 04, 2010, 06:35:06 pm

Quote from: m@tty on July 04, 2010, 06:27:22 pm

$\int \frac{dx}{2x+9}=\frac{1}{2}\log_e(2x+9)+C$

$\int_0^{36}\frac{dx}{2x+9}=\frac{1}{2}\left(\log_e{81}-\log_e{9}\right)=\frac{1}{2}\log_e9$

Therefore $k=\frac{9}{2}$

Shouldn't k = 3?

Title: Re: TyErd's questions
Post by: m@tty on July 04, 2010, 06:36:53 pm

Ah, yes. How stupid of me.

Title: Re: TyErd's questions
Post by: TyErd on July 04, 2010, 06:40:42 pm

Okay thankyou, um for the $sin 2x$ one the answers are $a = \frac{3\pi}{4} and b = \frac { \pi}{4}$

Title: Re: TyErd's questions
Post by: m@tty on July 04, 2010, 06:42:45 pm

Any terminals equally spaced about a zero will satisfy the equation.

So that answer is satisfactory.

Were there any restrictions with the question?

Title: Re: TyErd's questions
Post by: TyErd on July 04, 2010, 06:52:37 pm

what do you mean by any terminals equally spaced about a zero? and nah there weren't any restrictions

Title: Re: TyErd's questions
Post by: TyErd on July 04, 2010, 07:28:20 pm

also I need some help with this question:

if $\frac{ f(2+h) - f(2) } {h} = h^2 +6h +12, then f '(2)$ is equal to..?

Title: Re: TyErd's questions
Post by: brightsky on July 04, 2010, 07:38:19 pm

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

So in this case:

$f'(2) = \lim_{h\to 0} h^2 + 6h + 12 = 12$

Title: Re: TyErd's questions
Post by: TyErd on July 04, 2010, 07:55:19 pm

thankyou!

Title: Re: TyErd's questions
Post by: m@tty on July 04, 2010, 08:11:00 pm

Quote from: TyErd on July 04, 2010, 06:52:37 pm

what do you mean by any terminals equally spaced about a zero? and nah there weren't any restrictions

Because a sine curve is symmetrical, the area confined from a zero, say x=0, to another value, say x=a, will be equal to the area from x=-a to x=0. But, since sine changes sign at a zero, one of the areas will be negative, meaning that the areas will sum to zero.

That was probably confusing, but consider one period of a sine curve; both 'halves' are of equal area, but one segment is above the x-axis while the other is below. So when you take the integral over the entire period you get zero.

$\int_0^t\sin(x)dx=-\int_{-t}^0\sin(x)dx$ due to symmetry. (You can replace 0 with any other x-intercept.)

Hopefully this is clearer =S

Title: Re: TyErd's questions
Post by: TyErd on July 04, 2010, 08:31:07 pm

Ohhh I getcha, its much easier to understand with a graph. $\pi and 0$ would also be a solution yeah?

Title: Re: TyErd's questions
Post by: m@tty on July 04, 2010, 09:16:53 pm

Yep.

Title: Re: TyErd's questions
Post by: TyErd on July 04, 2010, 09:20:43 pm

ok thnkyou vm

Title: Re: TyErd's questions
Post by: TyErd on July 06, 2010, 12:13:55 am

A computer manufacturer notes that 5% of the computers are returned owing to faulty disk drives, 2% are returned owing to faulty keyboards and 0.3% are returned because both disk drives and keyboards are faulty.

Find the probability that the next computer has a faulty disk drive and a working keyboard.

I did: 0.05+0.98 - 0.03 = 1.

I dont get it

Title: Re: TyErd's questions
Post by: vexx on July 06, 2010, 12:26:39 am

^ TyErd, it's fairly simple in that using the formula
Pr(AuB) since it's it is both faulty disk(A) and working keyboard(B) which = Pr(A) + Pr(B) - Pr(AnB)
That's just the rule used, and so it's 0.05 (5%) + 0.02 (2%) - 0.003 (0.3%) = 0.067
Do you get it?

Title: Re: TyErd's questions
Post by: TyErd on July 06, 2010, 12:32:19 am

the answers actually 0.047.

Title: Re: TyErd's questions
Post by: vexx on July 06, 2010, 12:55:36 am

Quote from: TyErd on July 06, 2010, 12:32:19 am

the answers actually 0.047.

so sorry i thought you said both were faulty for some reason as i did this question as one of my randomly selected ones:)

well
faulty disk drive and a working keyboard.
as i calculated before, pr(either are faulty)=0.067
but we just want pr(faulty disk drive and working keyboard) = pr (either) - pr(faulty keyboard) as the keyboard is working so we cannot account for a faulty keyboard
=0.047 :p

Title: Re: TyErd's questions
Post by: TyErd on July 06, 2010, 01:12:59 am

okay i get it now, thankyou vm vexx

Title: Re: TyErd's questions
Post by: TyErd on July 06, 2010, 01:31:30 am

Nathan knows that his probability of kicking more than 4 goals on a wet day is 0.3, while on a dry day it is 0.6. The probability that it will be wet on the day of the next game is 0.7. Calculate the probability that Nathan will kick more than 4 goals in the next game.

Title: Re: TyErd's questions
Post by: TyErd on July 06, 2010, 01:37:03 am

For a particular petrol station, 30% of customers buy 'super' and 60% will buy unleaded and 10% diesel. When a customer buys super there is a 25% chance they will fill the tank. Customers buying unleaded have a 20% chance they will fill the tank. Of those buying diesel, 70% will fill their tank.

a) What is the probability that when a car leaves a petrol station it will not have a full tank?
b) Given that a car leaving the petrol station station has a full tank, what is the probability that the tank contain unleaded petrol?

Title: Re: TyErd's questions
Post by: TyErd on July 06, 2010, 01:39:49 am

The test used to determine if a person suffers from a particular disease is not perfect. The probability of a person with the disease returning a positive result is 0.95, while the probability of a person without the disease returning a positive result is 0.02. The probability that a randomly selected person has the disease is 0.03. What is the probability that a randomly selected person will return a positive result?

Title: Re: TyErd's questions
Post by: TyErd on July 06, 2010, 01:42:55 am

This year, 70% of the population have been immunised against a certain disease. Records indicate that an immunised person has a 5% chance of contracting the disease whereas for a non-immunised person this figure is 60%. Calculate the overall percentage of the population who are expected to contract the disease.

Title: Re: TyErd's questions
Post by: moekamo on July 06, 2010, 02:19:17 am

Quote from: TyErd on July 06, 2010, 01:31:30 am

Nathan knows that his probability of kicking more than 4 goals on a wet day is 0.3, while on a dry day it is 0.6. The probability that it will be wet on the day of the next game is 0.7. Calculate the probability that Nathan will kick more than 4 goals in the next game.

Draw a tree diagram, W= wet day, W' = not wet, >4 = kick more than 4 goals, >4 ' = $ or less goals

so we want Pr(>4) = Pr(W>4) + Pr(W'>4) = 0.7 * 0.3 + 0.3 * 0.6 = 0.39

Quote from: TyErd on July 06, 2010, 01:37:03 am

For a particular petrol station, 30% of customers buy 'super' and 60% will buy unleaded and 10% diesel. When a customer buys super there is a 25% chance they will fill the tank. Customers buying unleaded have a 20% chance they will fill the tank. Of those buying diesel, 70% will fill their tank.

a) What is the probability that when a car leaves a petrol station it will not have a full tank?
b) Given that a car leaving the petrol station station has a full tank, what is the probability that the tank contain unleaded petrol?

another tree diagram

a) Pr(Not Full Tank) = .3*.75 + .6*.8 + .1*.3 = .735

b) $Pr(U|F) = \frac{Pr(U \cap F)}{Pr(F)}$ since Pr Unleaded and full = .6*.2, and Pr(Full) = 1-.735= .265, Pr(U|F) = 24/53 as exact value

the rest are similar, you should probably try them now ive given you some working, i find tree diagrams are the best way to do these questions

Title: Re: TyErd's questions
Post by: TyErd on July 06, 2010, 11:24:51 am

Thanks so much man, the tree diagrams really did help.

Title: Re: TyErd's questions
Post by: TyErd on July 06, 2010, 12:03:33 pm

Quote from: TyErd on July 06, 2010, 01:39:49 am

The test used to determine if a person suffers from a particular disease is not perfect. The probability of a person with the disease returning a positive result is 0.95, while the probability of a person without the disease returning a positive result is 0.02. The probability that a randomly selected person has the disease is 0.03. What is the probability that a randomly selected person will return a positive result?

Im having a lil trouble with this one still

Title: Re: TyErd's questions
Post by: cipherpol on July 06, 2010, 12:46:52 pm

probability of random person with disease is 0.03
probability of +ve result, in this case, is 0.95

probability of random person without disease is 0.97
probability of +ve result, in this case, is 0.02

total prob= $(0.03*0.95)+(0.97*0.02)$

Title: Re: TyErd's questions
Post by: TyErd on July 06, 2010, 12:55:16 pm

OH yeah thankyou for that

Title: Re: TyErd's questions
Post by: TyErd on July 20, 2010, 06:32:03 pm

$\int \frac{2x+4}{x^{2}+4x+2} dx$

Title: Re: TyErd's questions
Post by: brightsky on July 20, 2010, 06:44:16 pm

Let $u = x^2 + 4x + 2$ , hence $du = 2x + 4 dx \implies dx = 1$

So the integrand becomes:

$\int \frac{du}{u}$

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

$= \log_e|u| + C$

$= \log_e|x^2 + 4x + 2| + C$

Title: Re: TyErd's questions
Post by: TyErd on July 20, 2010, 09:27:12 pm

is that the only way of doing it?

Title: Re: TyErd's questions
Post by: superflya on July 20, 2010, 09:31:57 pm

Quote from: TyErd on July 20, 2010, 09:27:12 pm

is that the only way of doing it?

ummm y would u do it any other way..that took 2 lines :P

Title: Re: TyErd's questions
Post by: m@tty on July 20, 2010, 09:37:40 pm

In methods it is done, in my experience, a little less formally.

You need to only recognise that the numerator is the derivative(or a multiple thereof) of the denominator. Then the next step is skipped, and you write log_e(whatever was in the denominator)... of course bringing a factor out the front if necessary.

Well, that was the way I did it. Seeing as methods students aren't required to know substitution.

But really this is just substitution without actually knowing what you are doing, and it is quite prone to errors, as you do it all in your head, basically.

Title: Re: TyErd's questions
Post by: 98.40_for_sure on July 20, 2010, 09:45:29 pm

Methods don't know substitution?!
So you can't actually use that method in the exams? wtf

Title: Re: TyErd's questions
Post by: TyErd on September 17, 2010, 03:02:11 pm

I cant seem to get this question. Help?

Title: Re: TyErd's questions
Post by: Blakhitman on September 17, 2010, 03:35:38 pm

start you off...

Pr(X<a)=0.8

should be able to do it now.

Title: Re: TyErd's questions
Post by: akira88 on September 17, 2010, 03:45:23 pm

Integrate that function between the values a and 1/2 (make sure the a is on the bottom and 1/2 at the top) where 0<a<1/2 (meant to be equal or less than signs) and make it equal 0.2.
Sorry I don't know how to use Latex :P
Hopefully the answer should be D.

Title: Re: TyErd's questions
Post by: Blakhitman on September 17, 2010, 03:52:20 pm

Quote from: akira88 on September 17, 2010, 03:45:23 pm

Integrate that function between the values a and 1/2 (make sure the a is on the bottom and 1/2 at the top) where 0<a<1/2 (meant to be equal or less than signs) and make it equal 0.2.
Sorry I don't know how to use Latex :P
Hopefully the answer should be D.

Lol thought domain was x>0 cause a similar question our practice exam today haha.

But yep their way is better.

Title: Re: TyErd's questions
Post by: TyErd on September 17, 2010, 04:38:25 pm

i keep getting stuck at $cos(2\pi a ) = 0.6$

Title: Re: TyErd's questions
Post by: TyErd on September 17, 2010, 04:40:09 pm

i've honestly tried everything i can think off but i still cant get it

Title: Re: TyErd's questions
Post by: fady_22 on September 17, 2010, 04:44:49 pm

Quote from: TyErd on September 17, 2010, 04:38:25 pm

i keep getting stuck at $cos(2\pi a ) = 0.6$

Solve for a, and the first positive solution is your answer.

Title: Re: TyErd's questions
Post by: TyErd on September 17, 2010, 04:47:24 pm

i tried using the calculator but get weird answers with n8 , n9 etc in them

Title: Re: TyErd's questions
Post by: fady_22 on September 17, 2010, 04:53:12 pm

Quote from: TyErd on September 17, 2010, 04:47:24 pm

i tried using the calculator but get weird answers with n8 , n9 etc in them

You have to specify the domain (the n8 is a parameter, you have been given a general solution) to get a numerical answer, i.e. type in |0<x<1/2 at the end.

Title: Re: TyErd's questions
Post by: TyErd on September 17, 2010, 04:58:24 pm

still getting a general solution...

Title: Re: TyErd's questions
Post by: akira88 on September 17, 2010, 05:10:27 pm

Quote from: TyErd on September 17, 2010, 04:58:24 pm

still getting a general solution...

You sure? What fady_22 said should fix the problem.
If you have a TI-89, put the | sign in (given that), and then specify the domain 0<a<1/2
So it should look like
solve( ∫ (pi*sin(2*pi*x),x,a,1/2)=0.2,a)|0≤a≤1/2

Title: Re: TyErd's questions
Post by: Blakhitman on September 17, 2010, 05:14:27 pm

Quote from: akira88 on September 17, 2010, 05:10:27 pm

Quote from: TyErd on September 17, 2010, 04:58:24 pm
still getting a general solution...
You sure? What fady_22 said should fix the problem.
If you have a TI-89, put the | sign in (given that), and then specify the domain 0<a<1/2
So it should look like
solve( ∫ (pi*sin(2*pi*x),x,a,1/2)=0.2,a)|0≤a≤1/2

Same with Ti-nspire

Title: Re: TyErd's questions
Post by: TyErd on September 17, 2010, 05:40:10 pm

okay what i typed into the calculator is: $solve ( \int_a^\frac{1}{2} \! (\pi sin (2\pi x ) ) \, dx= 0.2 , a, |0<a<0.5)$ and that gives me a general solution

Title: Re: TyErd's questions
Post by: Blakhitman on September 17, 2010, 06:11:57 pm

Tried it myself and yes you're right.

Dunno what's wrong.

Title: Re: TyErd's questions
Post by: Blakhitman on September 17, 2010, 06:18:39 pm

Is the answer 0.35?

Title: Re: TyErd's questions
Post by: TyErd on September 17, 2010, 06:20:54 pm

yes it is
how'd you get it?

Title: Re: TyErd's questions
Post by: Blakhitman on September 17, 2010, 06:31:25 pm

haha I sketched and used integral tool, but it can't be the only way.

Title: Re: TyErd's questions
Post by: Blakhitman on September 17, 2010, 06:33:52 pm

trial and error works too.

define the function and keep subbing the different choices.

Title: Re: TyErd's questions
Post by: TyErd on September 17, 2010, 06:36:40 pm

ohh lol why didn't i think of that.

i'd probably end up doing trial and error in the exam if it came up but still i don't know why the solve function doesn't get the answer.

Title: Re: TyErd's questions
Post by: fady_22 on September 17, 2010, 06:54:23 pm

Quote from: TyErd on September 17, 2010, 05:40:10 pm

okay what i typed into the calculator is: $solve ( \int_a^\frac{1}{2} \! (\pi sin (2\pi x ) ) \, dx= 0.2 , a, |0<a<0.5)$ and that gives me a general solution

The | sign should be put in AFTER the brackets:
$solve ( \int_a^\frac{1}{2} \! (\pi sin (2\pi x ) ) \, dx= 0.2 , a)|0<a<0.5$

Title: Re: TyErd's questions
Post by: TyErd on September 17, 2010, 06:57:28 pm

ohhhhhhh and yess it works thankyou vm

Title: Re: TyErd's questions
Post by: akira88 on September 17, 2010, 08:12:59 pm

Quote

solve( ∫ (pi*sin(2*pi*x),x,a,1/2)=0.2,a)|0≤a≤1/2

Lol guess you didn't look at what I typed in carefully enough :P

Title: Re: TyErd's questions
Post by: TyErd on September 17, 2010, 10:26:06 pm

lol my bad

Title: Re: TyErd's questions
Post by: 99.95 on September 18, 2010, 03:38:29 pm

how do i find integral:

if ∫(0,2) f(x)dx=-5 and ∫(5,2) 4f(x)=12

find ∫(5,0) f(x)dx

Title: Re: TyErd's questions
Post by: TrueTears on September 18, 2010, 03:39:19 pm

$\int_a^b f(x) dx + \int_b^c f(x) dx = \int_a^c f(x)dx$

Title: Re: TyErd's questions
Post by: 99.95 on September 18, 2010, 03:43:55 pm

Quote from: TrueTears on September 18, 2010, 03:39:19 pm

$\int_a^b f(x) dx + \int_b^c f(x) dx = \int_a^c f(x)dx$

i know that formula but how do i use it in this particular q

Title: Re: TyErd's questions
Post by: Blakhitman on September 18, 2010, 04:24:15 pm

$4\int_{2}^{5}f(x)dx=12$

therefore,

$\int_{2}^{5}f(x)dx=3$

and $\int_{0}^{5}f(x)dx=\int_{0}^{2}f(x)dx+\int_{2}^{5}f(x)dx$

so

$\int_{0}^{5}f(x)dx=-5+3=-2$

therefore

$\int_{5}^{0}f(x)dx=2$

Title: Re: TyErd's questions
Post by: 99.95 on September 18, 2010, 05:19:06 pm

Quote from: Blakhitman on September 18, 2010, 04:24:15 pm

$4\int_{2}^{5}f(x)dx=12$

therefore,

$\int_{2}^{5}f(x)dx=3$

and $\int_{0}^{5}f(x)dx=\int_{0}^{2}f(x)dx+\int_{2}^{5}f(x)dx$

so

$\int_{0}^{5}f(x)dx=-5+3=-2$

therefore

$\int_{5}^{0}f(x)dx=2$

the answer is 8 but i get what you mean

Title: Re: TyErd's questions
Post by: Blakhitman on September 18, 2010, 06:35:43 pm

lol yea I made a mistake with the first integral i typed, lower limit should be 5 so $\int_{2}^{5}f(x)dx=-3$

and then same procedure and you get 8.

Title: Re: TyErd's questions
Post by: 99.95 on September 18, 2010, 11:29:12 pm

how do i do this question:

find the exact area bounded by the line and the parabola:

y=sinx; y=1/2; 0≤X≤2∏

Title: Re: TyErd's questions
Post by: XXIII on September 18, 2010, 11:41:28 pm

find the 2 points of intersection of the line y=1/2 and sin(x). these points are pi/6 and 5pi/6

these become lower and upper limits respectively, then integrate sin(x).

Title: Re: TyErd's questions
Post by: brightsky on September 19, 2010, 12:37:08 pm

To find intersection points, let $\sin(x) = \frac{1}{2}$ , then $x = \frac{\pi}{6}, \frac{5\pi}{6}$ (taking into consideration the domain).

Drawing the graph, we know that the area is given by:

$\int^{0}_{\frac{\pi}{6}} \left(\frac{1}{2} - \sin(x)\right) + \int^{\frac{\pi}{6}}_{\frac{5\pi}{6}} \left(\sin(x) - \frac{1}{2}\right)$

Solve and you have the area. :)

Title: Re: TyErd's questions
Post by: letsride on September 19, 2010, 12:54:58 pm

Quote from: fady_22 on September 17, 2010, 06:54:23 pm

Quote from: TyErd on September 17, 2010, 05:40:10 pm
okay what i typed into the calculator is: $solve ( \int_a^\frac{1}{2} \! (\pi sin (2\pi x ) ) \, dx= 0.2 , a, |0<a<0.5)$ and that gives me a general solution

The | sign should be put in AFTER the brackets:
$solve ( \int_a^\frac{1}{2} \! (\pi sin (2\pi x ) ) \, dx= 0.2 , a)|0<a<0.5$

i type this exactly in my CAS but get incorrect number of arguments :( i've been tryna figure out how to solve bounds with integrals for a while but it wont work

-edit- nvm just found out :D space b/w dx and =

Title: Re: TyErd's questions
Post by: TyErd on September 23, 2010, 02:26:34 pm

what is the dilation factor from the X axis of the equation:
$ y= 3 \pm \sqrt{6(x-3)}$

Title: Re: TyErd's questions
Post by: itolduso on September 23, 2010, 03:28:42 pm

sqrt(6)

Title: Re: TyErd's questions
Post by: TyErd on September 23, 2010, 03:40:23 pm

why is sqrt(6) and not just 6

Title: Re: TyErd's questions
Post by: kamil9876 on September 23, 2010, 04:06:44 pm

please be more specific, what is dilated from what?

Anyway I assume you mean the original function is $\sqrt{x-3}$ then it gets dilated to $\sqrt{6(x-3)}$ and then translated.

In that case the dilation (in bold) is a factor of $\sqrt{6}$ from the X-axis, why? If $f(x)=\sqrt{x-3}$ then

$\sqrt{6(x-3)}=\sqrt{6} \sqrt{x-3}=\sqrt{6} f(x)$