Author Topic: VCE Specialist 3/4 Question Thread! (Read 2578495 times) Tweet Share

IndefatigableLover · « **Reply #3420 on:** August 03, 2014, 09:18:04 pm »

Quote from: EulerFan101 on August 03, 2014, 08:57:42 pm

This is the ideal way to do it - no other nice way. You could use a trig substitution - maybe - but this is a much nicer, cleaner method.

The only thing I'd fault you on is you shouldn't have modulus signs in your answer - since you have an initial condition, you should be able to gauge what that should look like without modulus signs.

EDIT: Did just try a trig substitution for interest's sake - it's not pretty. At all.

Oh really I shouldn't have modulus signs in my answer? Would it still be correct if I did anyway though (despite the initial condition)?

And yeah I did do a trig substitution beforehand but ran into a dead end (or rather I didn't know what to do) so I gave up and tried partial fractions LOL :')

keltingmeith · « **Reply #3421 on:** August 03, 2014, 09:38:49 pm »

Quote from: IndefatigableLover on August 03, 2014, 09:18:04 pm

Oh really I shouldn't have modulus signs in my answer? Would it still be correct if I did anyway though (despite the initial condition)?

And yeah I did do a trig substitution beforehand but ran into a dead end (or rather I didn't know what to do) so I gave up and tried partial fractions LOL :')

You're right, but you could be MORE right, if that makes sense. I'm not entirely sure if one would lose marks for including the modulus signs, though - my teacher never touched on it, rather just always told us about when we could do it (so I moved on to doing it, like a good obedient year 12 sheep).

You wouldn't have been able to solve it - it requires integration by parts.

Ends up being something like ln|(sec(arcsin(x))+tan(arcsin(x))| or similar.

lzxnl · « **Reply #3422 on:** August 03, 2014, 10:56:42 pm »

LOL. Trig sub that? It's not as bad as you'd think, EulerFan101

$\int{\frac{2}{1-x^2}dx} = \int{-\frac{2\sin{\theta} d\theta}{1-\cos^2{\theta}} = \int{\frac{-2}{\sin{\theta}} d\theta$

Above was after letting x = cos theta
Now to use the best substitution in the book
Let $t = \tan{\frac{\theta}{2}}$

$\sin{\theta} = \frac{2t}{1+t^2}$

$d\theta = \frac{2dt}{1+t^2}$

So subbing all of that into our integral

$\int{\frac{-2}{\sin{\theta}} d\theta = \int{\frac{-2(1+t^2)*2 dt}{2t(1+t^2)} = \int{\frac{-2}{t} dt} = -2\ln{|t|} + c = -2\ln{|\tan{\frac{\theta}{2}}}| +c = -2\ln{|\frac{1-\cos{\theta}}{\sin{\theta}}| + c = \ln{|\frac{\sin^2{\theta}}{(1-\cos{\theta})^2}}| + c$

$= \ln{|\frac{(1+\cos{\theta})(1-\cos{\theta})}{(1-\cos{\theta})^2}}| + c = \ln{|\frac{1+\cos{\theta}}{1-\cos{\theta}}}| + c$ = $\ln{|\frac{1+x}{1-x}}| + c$

And remove mod signs if necessary

There. No integration by parts needed at all, only a (standard) set of substitutions needed that no one at VCE level needs to care about

Special At Specialist · « **Reply #3423 on:** August 04, 2014, 12:10:35 am »

But in a real exam, you would want to use this approach to save time:

$y = \int \frac{2}{1-x^{2}}dx$

Let $\frac{2}{1-x^{2}} = \frac{A}{1 - x} + \frac{B}{1 + x}$

$2 = A + Ax + B - Bx$

$A + B = 2$ and $A - B = 0$

$A = B = 1$

Therefore $y = \int \frac{1}{1 - x}dx + \int \frac{1}{1 + x}dx$

$y = - ln|1 - x| + ln|1 + x| + C$

IndefatigableLover · « **Reply #3424 on:** August 04, 2014, 11:45:59 am »

Haha thanks lzxnl and Special at Specialist (though I think you integrated $\frac{1}{1-x}$ incorrectly but that'd be the route I would have taken)!

keltingmeith · « **Reply #3425 on:** August 04, 2014, 08:33:22 pm »

Quote from: lzxnl on August 03, 2014, 10:56:42 pm

LOL. Trig sub that? It's not as bad as you'd think, EulerFan101

$\int{\frac{2}{1-x^2}dx} = \int{-\frac{2\sin{\theta} d\theta}{1-\cos^2{\theta}} = \int{\frac{-2}{\sin{\theta}} d\theta$

Above was after letting x = cos theta
Now to use the best substitution in the book
Let $t = \tan{\frac{\theta}{2}}$

$\sin{\theta} = \frac{2t}{1+t^2}$

$d\theta = \frac{2dt}{1+t^2}$

So subbing all of that into our integral

$\int{\frac{-2}{\sin{\theta}} d\theta = \int{\frac{-2(1+t^2)*2 dt}{2t(1+t^2)} = \int{\frac{-2}{t} dt} = -2\ln{|t|} + c = -2\ln{|\tan{\frac{\theta}{2}}}| +c = -2\ln{|\frac{1-\cos{\theta}}{\sin{\theta}}| + c = \ln{|\frac{\sin^2{\theta}}{(1-\cos{\theta})^2}}| + c$

$= \ln{|\frac{(1+\cos{\theta})(1-\cos{\theta})}{(1-\cos{\theta})^2}}| + c = \ln{|\frac{1+\cos{\theta}}{1-\cos{\theta}}}| + c$ = $\ln{|\frac{1+x}{1-x}}| + c$

And remove mod signs if necessary

There. No integration by parts needed at all, only a (standard) set of substitutions needed that no one at VCE level needs to care about

That's pretty cool, actually. I never learned about trig substitutions, so all I really know is a botched effort of teaching myself just before my last exam.

I'll definitely remember that one for my repertoire!

Mieow · « **Reply #3426 on:** August 04, 2014, 09:57:44 pm »

How would I sketch the path of this position vector?
r(t) = $(sin\pi t)i + [4cos(2\pi t)]j$

yang_dong · « **Reply #3427 on:** August 04, 2014, 10:05:30 pm »

given that f(X) = 12/sqrt(25/x^2)
find integral (with limits 3 and 0) [f(X)]^2 dx in the form p + qloge(r)

what i've done so far:
integral (with limits 3 and 0) 144/(25-x^2) dx
i don't know did you get p = 0, q = 14.4 and r =4?

polar1 · « **Reply #3428 on:** August 06, 2014, 02:38:52 pm »

Haha, I'm so bored that I've signed back up to do fun maths questions again!

Quote from: Mieow on August 04, 2014, 09:57:44 pm

How would I sketch the path of this position vector?
r(t) = $(sin\pi t)i + [4cos(2\pi t)]j$

Hint: $\sin^2(\pi t) + \cos^2(\pi t) = 1$ and $\cos(2\pi t) = 2\cos^2(2\pi t) - 1$

Spoiler

We have $\sin(\pi t) = x$ , so, $\sin^2(\pi t) = x^2$ and,
$y = 4\cos(2\pi t) = 4(2\cos^2(\pi t) -1) \implies \cos^2(\pi t) =\frac{1}{2}\left( \frac{y}{4}+1\right)$
Hence, we have $\sin^2(\pi t) + \cos^2(\pi t) = x^2 + \frac{1}{2}\left(\frac{y}{4}+1\right) = 1$ .

(There'd usually be a condition on $t$ (like, $t \ge 0$ ) and you'd have to work out the domain and range from that.)

Quote from: yang_dong on August 04, 2014, 10:05:30 pm

given that f(X) = 12/sqrt(25/x^2)
find integral (with limits 3 and 0) [f(X)]^2 dx in the form p + qloge(r)

what i've done so far:
integral (with limits 3 and 0) 144/(25-x^2) dx
i don't know did you get p = 0, q = 14.4 and r =4?

You're on the right track. Would it help if I said 'partial fractions'?

polar1 · « **Reply #3429 on:** August 06, 2014, 09:30:26 pm »

Quote from: Zezima. on August 06, 2014, 09:06:17 pm

Sorry if this is a stupid or confusing question.

But are all solutions of a complex number
ie. of form z^n=a where n E Z and a E R equally spaced out when plotted on an Argand diagram?

What if it's of form z^n=a, n E z and a E C?

Cheers!

Hints: How would you usually solve these equations? What does it mean for the solutions to be equally spaced apart?

Spoiler

In general, you could rewrite $z^n = a$ as $z^n = r\mathrm{cis}(\theta + 2\pi)$ for $|a| = r$ and $\mathrm{Arg}(a) = \theta$ . So, $z = r^{\frac{1}{n}}\mathrm{cis}\left(\frac{\theta}{n} + \frac{2\pi}{n} \right)$ . What does this tell you about the solutions? That they're $\frac{2\pi}{n}$ apart from each other.

If you had $z \in \mathbb{R}$ , you'd just have $\theta = 0$ , so it's a special case.

IndefatigableLover · « **Reply #3430 on:** August 09, 2014, 03:30:23 pm »

Alright so I need to find the intersection points of polar curves however I have come across a problem...

$r=\cos(\theta) \text { and } r=1-\cos(\theta)$

I've solved the equation and I've found two points of intersection

Spoiler

$\theta=\frac{\pi}{3} \text{ and } \theta=\frac{5\pi}{3}$
or
$(\frac{1}{2},\frac{\pi}{3}) \text{ and } (\frac{1}{2}, \frac{5\pi}{3})$

but when I graph the question, I can see that it also intersects at a third position (that is, the origin) yet I don't know how to work out that position in an algebraic form... any help as to how I could go about this?

lzxnl · « **Reply #3431 on:** August 10, 2014, 12:00:28 am »

Quote from: IndefatigableLover on August 09, 2014, 03:30:23 pm

Alright so I need to find the intersection points of polar curves however I have come across a problem...

$r=\cos(\theta) \text { and } r=1-\cos(\theta)$

I've solved the equation and I've found two points of intersection
Spoiler
$\theta=\frac{\pi}{3} \text{ and } \theta=\frac{5\pi}{3}$
or
$(\frac{1}{2},\frac{\pi}{3}) \text{ and } (\frac{1}{2}, \frac{5\pi}{3})$

but when I graph the question, I can see that it also intersects at a third position (that is, the origin) yet I don't know how to work out that position in an algebraic form... any help as to how I could go about this?

They shouldn't intersect at the origin. The origin is where r = 0 and these two polar relations are never both simultaneously zero

IndefatigableLover · « **Reply #3432 on:** August 10, 2014, 09:24:20 am »

Quote from: lzxnl on August 10, 2014, 12:00:28 am

They shouldn't intersect at the origin. The origin is where r = 0 and these two polar relations are never both simultaneously zero

Well here's the graph of the two equations:

And they do intersect since I understand what you mean (which was what I was getting at as well) except when I graph it they do appear to intersect..

keltingmeith · « **Reply #3433 on:** August 10, 2014, 09:47:58 am »

They appear to intersect at the origin. The only way to touch the origin in a polar plot is by letting the length of your line, your r, equal 0, as the angle won't affect how far or close a line is to the origin.

So, for both of your curves, we let r=0:

$r=\cos(\theta) \implies \theta =\frac{\pi}{2}<br />\\ r=1-\cos(\theta) \implies \cos(\theta)=1 \implies \theta=0$

As you can see, while they both hit the origin, they do so at different angles. You'd think that this doesn't matter, but it does, and for now the best way to think of it is if you tried to write a point of intersection, you wouldn't be able to, as one has the point $(0,0)$ and the other has the point $(0,\frac{\pi}{2})$ .

Also, look at the graph - you can actually see that they hit the origin at different angles. Notice how one graph rises off at an angle of 90 degrees, and the other comes off of it flat?

IndefatigableLover · « **Reply #3434 on:** August 10, 2014, 10:04:18 am »

Ah that makes much more sense (but damn these polar graphs LOL) T_T
Thanks lzxnl and EulerFan101

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