VCE Specialist 3/4 Question Thread!

[BlinkieBill]

Hi,
I was wondering if methods type transformation questions have come up on Spec exams? Not really sure what I should be remembering abt methods for spec in general  (other than probability)
From the vcaa exams I've done it hasnt been too bad, but ive read in exam reports that methods knowledge is tested sometimes.
thanks

[Willba99]

Hi,
I was wondering if methods type transformation questions have come up on Spec exams? Not really sure what I should be remembering abt methods for spec in general  (other than probability)
From the vcaa exams I've done it hasnt been too bad, but ive read in exam reports that methods knowledge is tested sometimes.
thanks

As in transformations with matrices? If so, I've never heard of one

[Syndicate]

As in transformations with matrices? If so, I've never heard of one

It can be tested, as methods is a prerequisite for specialist (but they usually stick to the study design).

[BlinkieBill]

Transformations as in dilation factors etc

[VanillaRice]

Transformations as in dilation factors etc

It tends to come up indirectly. Consider for example, having to graph y = 3arcsin(1-2x) + 1. Not only would you need to know the shape of the function y=arcsin(x), but also the transformations required to go from the 'base' function to the one in the question.

[Rieko Ioane]

Hi
For this question
https://ibb.co/ep5bTa
https://ibb.co/jWxegv

When I take the cosine of this right-angle triangle, I'm not getting what they are getting. I assigned my theta at the 'corner' formed by the tangent and the horizontal from (k, -2) and the centre of the circle which gives me cosine = -k/-k.

I can't get their adjacent side of '2' it seems but I'm not sure what they did to get it?

[Shadowxo]

Hi
For this question
https://ibb.co/ep5bTa
https://ibb.co/jWxegv

When I take the cosine of this right-angle triangle, I'm not getting what they are getting. I assigned my theta at the 'corner' formed by the tangent and the horizontal from (k, -2) and the centre of the circle which gives me cosine = -k/-k.

I can't get their adjacent side of '2' it seems but I'm not sure what they did to get it?

The theta location shouldn't matter, both would be 45° and if they weren't, you'd just have to make sure you assigned the right value to theta. For your theta location you'd probably have to use sin instead of cos as you know the opposite (2, the radius. That's how they got their value of 2, it's just the length from the centre of the circle to the tangent)
Length of hypotenuse = distance between (k,-2) and the centre of the circle (0,-2) =  k as the imaginary parts are equal
Note that they used (0,-(2+k)) for their hypotenuse instead of (k,-2). This also means the radius forms the adjacent side to their triangle, instead of the opposite side for ours.

You'd end up with sin(45°)= O/H = 2/k
So 2/k = sqrt(2)/2 so k = 2sqrt(2)

A diagram is very helpful for a question like this. If you're still confused, let me know and I'll draw up a diagram

[waldo2000]

Hey am i supposed to integrate this implicit derivative? and how?

this is the question,

Solve the derivative dy/dx + y/(100+4x) = 0
in the form x = (A/(b+t))^ (1/c)
Where a,b, and c are Z

I have no idea how to do this!!!

[keltingmeith]

Hey am i supposed to integrate this implicit derivative? and how?

this is the question,

Solve the derivative dy/dx + y/(100+4x) = 0
in the form x = (A/(b+t))^ (1/c)
Where a,b, and c are Z

I have no idea how to do this!!!

Your issue is with assuming you need to do a straight integration. Notice that this is an example of a separable ODE, and so you'll need to use the method of separation of variables.

Let us know if you're having any further issues.

[Sine]

Hey am i supposed to integrate this implicit derivative? and how?

this is the question,

Solve the derivative dy/dx + y/(100+4x) = 0
in the form x = (A/(b+t))^ (1/c)
Where a,b, and c are Z

I have no idea how to do this!!!

In spec if you have a differential in terms of both variables you know there are 4 different things that can happen
1) unsolvable in terms of spec (especially for higher order differentials)
2)eulers methods - they would make it very clear wehther you need to use this
3)use CAS to solve
4) sepearation of variables.

This ones looks like numebr 4!

[peanut]

If a and b are vectors, and AB vector = b - a and OQ vector = 0.5a + 0.5b, then does AB.OQ = (1 * 0.5) + (-1 * 0.5) = 0? Or does this rule only apply to i,j,k vectors (if so, how would I calculate AB.OQ?)

[LifeisaConstantStruggle]

If a and b are vectors, and AB vector = b - a and OQ vector = 0.5a + 0.5b, then does AB.OQ = (1 * 0.5) + (-1 * 0.5) = 0? Or does this rule only apply to i,j,k vectors (if so, how would I calculate AB.OQ?)

This would only work on i j k vectors because vectors, and is the magnitude of vectors specified in the question? (eg. |a|=|b| or something like that?) because a^2 would be |a|^2 and you can substitute |a| with |b| (i don't really know the objective of the question but hope this helps )

[Shadowxo]

If a and b are vectors, and AB vector = b - a and OQ vector = 0.5a + 0.5b, then does AB.OQ = (1 * 0.5) + (-1 * 0.5) = 0? Or does this rule only apply to i,j,k vectors (if so, how would I calculate AB.OQ?)

Yes, it would only apply to i,j,k vectors. This is because, when you expand them out i.j = 0, i.k=0 and j.k = 0 because they're perpendicular.

For this, you'd use the usual formulas for dot product of vectors: a.b = |a||b|cos(theta) and (a+b).c = a.c + b.c
If b and a are both vectors then AB.OQ = (b-a).(0.5a+0.5b) = 0.5a.b + 0.5b.b -0.5a.a -0.5a.b = 0.5|b|² - 0.5|a|²
Bit rusty but hope this helps

[peanut]

This would only work on i j k vectors because vectors, and is the magnitude of vectors specified in the question? (eg. |a|=|b| or something like that?) because a^2 would be |a|^2 and you can substitute |a| with |b| (i don't really know the objective of the question but hope this helps )

Yes, it would only apply to i,j,k vectors. This is because, when you expand them out i.j = 0, i.k=0 and j.k = 0 because they're perpendicular.

For this, you'd use the usual formulas for dot product of vectors: a.b = |a||b|cos(theta) and (a+b).c = a.c + b.c
If b and a are both vectors then AB.OQ = (b-a).(0.5a+0.5b) = 0.5a.b + 0.5b.b -0.5a.a -0.5a.b = 0.5|b|² - 0.5|a|²
Bit rusty but hope this helps

Yes, |a| = |b| so the dot product is zero. Nice to know I used the wrong method, even though I got the right answer.

[gnaf]

Is a force required to push up an object up an inclined plane?

My teacher told us that this was a misconception??