VCE Specialist 3/4 Question Thread!

[ahat]

Could someone please post a solution for question 4, VCAA 2007 (Spec) Exam 2?

I'm having trouble understanding part c and d.
Thanks muchly.

[Professor Polonsky]

2006 vcaa MCQ 20 - why am I getting B?

[ahat]

If you're using Lami's Theorem, remember that the angle is opposite the force. This could be the problem.

[zvezda]

Hey,
With q9 on the 2007 exam 1, the assessors report says that there is a more rigorous way to answer the question than simply assuming that x=cos(t) and y=sin(t), but doesnt explain this method.
I assumed the same thing when doing the q initially and knew that it was wrong to do so, but couldnt think of another way (although i didnt explicitly state this).
So what is this apparent method?
Cheers

[BasicAcid]

How do we antidifferentiate 4/(x^2 + 4x)?

[Alwin]

How do we antidifferentiate 4/(x^2 + 4x)?

Hint: partial fractions, A/x and B/(x+4)

From a neap exam 1 iirc?

[revcose]

That's a really nice question stankovic. I don't think I could do it, but the assessment report does kind of explain it.

r = x i + y j
v = dx/dt i + dy/dt j (Through standard vector diff)
v = -y i + x j (As it told us)
Therefore dx/dt = -y and dy/dt = x (or we can look at this as - dy/dt = -x)
Using those, we can take v to a
v = -y i + x j
a = -x i - y j
And a = -r

[BasicAcid]

Hint: partial fractions, A/x and B/(x+4)

From a neap exam 1 iirc?

Nope, hefferman exam 1 I believe, I just did one under exam conditions at school.

And fuck, I should've known how to do that -_-



Btw cheers alwin I appreciate you not just spitting the answer out and just giving me the hint

[sin0001]

I know we can anti-differentiate something like this using tan^-1 : 1/(x^2+4), but is there a way to expand this using partial fractions?

[b^3]

You have an irreducible factor in the denominator (i.e. we can't factorise over the reals), so you can't, you have to go the way.

[Stevensmay]

I know we can anti-differentiate something like this using tan^-1 : 1/(x^2+4), but is there a way to expand this using partial fractions?

Not that I know of, as the denominator is not reducible, thus we need to use the arctan derivative.

[SocialRhubarb]

With q9 on the 2007 exam 1, the assessors report says that there is a more rigorous way to answer the question than simply assuming that x=cos(t) and y=sin(t),

So what is this apparent method?

[zvezda]

Lol. This is what i did assuming x and y were cos(t) and sin(t)

Thanks

[lzxnl]

I know we can anti-differentiate something like this using tan^-1 : 1/(x^2+4), but is there a way to expand this using partial fractions?

Let's mess around with this a bit. This IS spesh after all.
If we expand 1/(x^2-m^2), we find that this is equal to 1/2m*(1/(x-m)-1/(x+m))
So if we let m=ai, then our fraction becomes 1/(x^2+a^2)=1/2ai*(1/(x-ai)-1/(x+ai))
Integrating, we get -i/2a*ln((x-ai)/(x+aoi)) plus a constant.
On the other hand, we know that we can integrate 1/(x^2+a^2) to give 1/a*arctan(x/a)
When x=0, both sides are meant to equal zero. So I'm going to multiply the log term by a factor of something as that's what adding a constant does.
We eventually get LHS=-i/2a*ln((ai-x)/(ai+x))=1/a*arctan(x/a)
Or arctan(x/a)=-i/2ln((ai-x)/(ai+x))

If you want the meaning of what the natural log of a complex number means, use the identity e^ix=cis x, easily proven by showing that both sides satisfy dy/dx=iy and y(0)=1. AKA rewrite a complex number rcis t in the form re^it. Then if z=re^it, ln z=ln(r)+it. Note that this means the natural log is multivalued; the principal value of the log corresponds to the one with the principal value of the argument.

Just thought I'd say this.

[Alwin]

Nope, hefferman exam 1 I believe, I just did one under exam conditions at school.

And fuck, I should've known how to do that -_-

Btw cheers alwin I appreciate you not just spitting the answer out and just giving me the hint

lol serves me right for doing my exam 1s in chunks rather than like one per day. my mistake

and I'm here for you always

@nliu, Euler's formula isn't in spesh is it? At least I don't remember seeing it in the essential textbook anywhere