VCE Specialist 3/4 Question Thread!

[brightsky]

oh woops...lol

(too tired xD)

[sin0001]

Anyone mind explaining Q9 (MC) of VCAA Exam 2? Link below:
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2012/2012specmath2-w.pdf

Thanks in advance XD

[Stevensmay]

Anyone mind explaining Q9 (MC) of VCAA Exam 2? Link below:
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2012/2012specmath2-w.pdf

Thanks in advance XD

Euler's approximation is
where h is our step size.

Substituting in our initial conditions


Our x value is incremented by the h value.




So now it's either option B or C. Just need to determine whether this will be an over or under estimation.
Graph an antiderivative of , .

We can see that sin(x) is getting less steep as it progresses, which means that over time it will diverge away and down from the tangent line. Thus Euler's approximation will be an overestimation of the value.

https://www.desmos.com/calculator/tolqfsfrck
Pretty basic example but if you zoom in enough you can see how it happens.

[sin0001]

Euler's approximation is
where h is our step size.

Substituting in our initial conditions


Our x value is incremented by the h value.




So now it's either option B or C.

Figured that part out
Should've been clearer: only need help with deciding whether it's an over/under approximation
But thanks anyway, quick reply

[Stevensmay]

I have a habit of hitting post and not preview sorry.

[sin0001]

So now it's either option B or C. Just need to determine whether this will be an over or under estimation.
Graph an antiderivative of , .

We can see that sin(x) is getting less steep as it progresses, which means that over time it will diverge away and down from the tangent line. Thus Euler's approximation will be an overestimation of the value.

Ohh yea I had a similiar approach; I found the anti derivative to be sin(x) +1 (Is this even correct?) and then subbed in x=0.2, which I found was >1.1995... for some reason

[Jaswinder]

with this graph why isn't an asymptote?

[lzxnl]

with this graph why isn't an asymptote?

Rewrite as x/2 + 2/x
As x becomes large, the behaviour of x/2 is what matters. 2/x=>0; only the x/2 term has any real relevance.

[Alwin]

with this graph why isn't an asymptote?

The asymptotes are and

By formal definition of an asymptote,


   not 2/x, even though x=0 is an asymptote of 2/x as well. 2/x has no real effect on the graph

EDIT: beaten

[papertowns]

Can someone help me? I never seemed to have learnt this and I feel stupid but how do you expand something like this.. (4rt2 - 2)^2

[Stevensmay]

Can someone help me? I never seemed to have learnt this and I feel stupid but how do you expand something like this.. (4rt2 - 2)^2

Do you mind clarifying what the expression was?
So far I've got but I am not sure.

[Phy124]

Do you mind clarifying what the expression was?
So far I've got but I am not sure.

Ohhhh I think he was going for

In which case;






If that's not what he meant then I dunno...

[duhherro]

Could anyone help me explain Q9) c)ii) of 2011 exam 1?

[Lejn]

Could anyone help me explain Q9) c)ii) of 2011 exam 1?

So since it's a hence, we've gotta use 3c-a.

For vectors to be linearly dependent, then scalars of two must be able to add up to the other. Since we've found 3c-a and we only need to find a single possible value m, then this must add up to a scalar of b. 3c-a's i and j components are twice that of b's, and so the same must be said for their k components. 3c-a has a -5 k component, which will be twice of m, so m is -5/2.

[duhherro]

So since it's a hence, we've gotta use 3c-a.

For vectors to be linearly dependent, then scalars of two must be able to add up to the other. Since we've found 3c-a and we only need to find a single possible value m, then this must add up to a scalar of b. 3c-a's i and j components are twice that of b's, and so the same must be said for their k components. 3c-a has a -5 k component, which will be twice of m, so m is -5/2.

Thanks for the reply, also on the previous question of 8. My method was :
4/3 = (cosec(pix/6))^2
2/root3 = cosec(pi/6), and then continued onwards, but could only get 2 solutions, did I also need to do -2/root3 = ...  ? I thought that the domain of 0 < x < 12 wouldn't need to do the negative part :/