VCE Specialist 3/4 Question Thread!

[Aelru]

Question 1:

Spoiler

This doesn't seem too hard, but it has quite a few steps and quite a bit of algebra.

Our first step is just finding the points of intersection between the line and the circle.







Fast forward:



These values of y correspond to the points .

These two points are both on the positive half of the circle, so we can represent the equation of the circle using .

.

At x=-1,

At x = 2,

So the tangents have the gradients of half and -2.

Given that we know their gradients and know the points which they must go through, we can work out their equations, which come out to be:



The intersection of these two tangents is (1,3).

We now know the points A, B and C, which are  (2,1), (-1,2) and (1,3).

Now all we need is to find a circle which passes through all three of these points.

If we use the standard equation for a circle, , and sub in our points, we'll get a system of three equations which we can solve, probably with a calculator or by hand. We get , which you could sub in and rearrange to obtain your answer, .

Question 2:

I got something close to the answer, but with a few different negatives, so I think my diagram is wrong.

Heehee, wonderful setting out. The explanation was clear, and crisp.
I got around to simultaneous equations as well, but I wasn't quite sure on what generic circle equation I've used.

So, we don't actually have to include dilations into our generic circle equations? Why?
EG:



Oh e^1, regarding your questions on concurrent proof,
this page helps: http://www.regentsprep.org/Regents/math/geometry/GC3/LConcurrence.htm
(Am I allowed to share links on this forum?)

So basically, I think concurrent is defined as: the concept of three or more lines intersecting in a single (common) point, having a single point of intersection.

So, as the proof defined at the end that since O,Y,X are collinear, this means that OX intersects the point of Y.
(We learned that Y is also a point on the lines of BB' , and AA', as defined in the earlier sections of the working out)
ie. So three lines intersect at one point (Y), which fulfills the concurrent definition.

Yet, regarding that specific question, I'm not sure how it's the 'Medians" that are concurrent.

[BasicAcid]

If you dilate a circle the way you're asking (assuming h and k are different) it becomes an ellipse. But the question says it's a circle so know h and k are equal.
As long as h and k in your equation are equal, it'll be a circle and if that's the case, we.can just multiply though by h or k which gives us:

(x - h)^2 + (y - k)^2 = ch or ck = r^2

[SocialRhubarb]

Question: Prove that

Okay, so we're going to be using two double angle formulae here:





As well as the trigonometric identity,





Now we factorise the top using difference of two squares, and the bottom by recognising that it is a perfect square.



, through cancelling top and bottom.

EDIT: added question since original post was deleted ()

[zvezda]

Hey,
When finding the angle between two intersecting paths of moving particles, technically theres no real way of finding the EXACT value of the angle isnt there?
Just looking for clarification, because the methods i know of either use the gradient of the tangents of the two path at their intersection or the dot product method using velocity vectors.
Cheers

[Stevensmay]

Wouldn't you have an exact value in the form
So is our exact angle.

[Alwin]

Hey,
When finding the angle between two intersecting paths of moving particles, technically theres no real way of finding the EXACT value of the angle isnt there?
Just looking for clarification, because the methods i know of either use the gradient of the tangents of the two path at their intersection or the dot product method using velocity vectors.
Cheers

They give the same result... if ofc you mean the 'dot product' method using cosine yea?

That is the way of finding the exact value of the angle you leave it in tan^-1<blah> or use cas and convert to the required number of decimal places

[Homer]

I get 10 but its wrong

[sin0001]

|OC| : dot product of 'OB' and the unit vector of 'OA'
So |OA| = 5
|OC| = 1/5 (OB.OA) = 1/5 (2 + 8 ) = 2

[lzxnl]

How do we anti differentiate xsin(x)without using our CAS?

I'm not sure if we need to know it for spesh or not, but even if we don't and you guys know, could you please run through it for me?



By the way there was a question asking which involved this - "Use calculus to find... the area of the shaded region shown in Figure 2." - 2000 VCAA exam (well they weren't even called VCAA back then... 'Board of Studies' haha).

But we were already asked to find it two questions before through finding the derivative of xcos(x) which gave us (cos(x) - xsin(x)) then used that to find the anti derivative of xsin(x). So would've just used that answer as part of our working out.

Although I'm still curious haha.

Product rule: d/dx (uv) = v*du/dx + u*dv/dx
Integrating both sides
uv = int(v du/dx dx) + int(u dv/dx dx)
int(u dv/dx dx) = uv-int(v du/dx dx)
aka integration by parts
Not how u is differentiated and dv/dx is integrated. This is the key.

So here with int(x sin x dx)
We want to differentiate the x to get rid of it, so let u = x, du/dx = 1. dv/dx = sin x dx => v = -cos x. The constant of integration isn't necessary.

Then int(x sin x dx) = uv - int(v du/dx dx) = -x cos x + int(cos x dx)
= -x cos x + sin x + constant.

[psyxwar]

How do we anti differentiate xsin(x)without using our CAS?

I'm not sure if we need to know it for spesh or not, but even if we don't and you guys know, could you please run through it for me?



By the way there was a question asking which involved this - "Use calculus to find... the area of the shaded region shown in Figure 2." - 2000 VCAA exam (well they weren't even called VCAA back then... 'Board of Studies' haha).

But we were already asked to find it two questions before through finding the derivative of xcos(x) which gave us (cos(x) - xsin(x)) then used that to find the anti derivative of xsin(x). So would've just used that answer as part of our working out.

Although I'm still curious haha.

https://www.khanacademy.org/math/calculus/integral-calculus/integration_by_parts/v/deriving-integration-by-parts-formula

Integration by parts, as nliu said. I recommend you watch this series of videos

[Alwin]

Sorry, I don't mean to be a party pooper by I asked about integration by parts last year and it was a resounding no

Even if you use it in the exam (I was asking for methods last year) and get the answer correct, you only get 1 mark out of 2 or 3 the question is worth.

This is because of the structure and wording of the question! In part a it will ask you to find the derivative of ________ and in part b it will always say hence find the antiderivative of _________ . This means you have to use the longer way and not integration by parts

It's good to know + interesting, but can't be used on the exam (like a lot of vce science )

[psyxwar]

Sorry, I don't mean to be a party pooper by I asked about integration by parts last year and it was a resounding no

Even if you use it in the exam (I was asking for methods last year) and get the answer correct, you only get 1 mark out of 2 or 3 the question is worth.

This is because of the structure and wording of the question! In part a it will ask you to find the derivative of ________ and in part b it will always say hence find the antiderivative of _________ . This means you have to use the longer way and not integration by parts

It's good to know + interesting, but can't be used on the exam (like a lot of vce science )

wait you mean you can't use this in spesh either? wtf. So what do you actually learn in VCE maths in regards to integration?

[brightsky]

Sorry, I don't mean to be a party pooper by I asked about integration by parts last year and it was a resounding no

Even if you use it in the exam (I was asking for methods last year) and get the answer correct, you only get 1 mark out of 2 or 3 the question is worth.

This is because of the structure and wording of the question! In part a it will ask you to find the derivative of ________ and in part b it will always say hence find the antiderivative of _________ . This means you have to use the longer way and not integration by parts

It's good to know + interesting, but can't be used on the exam (like a lot of vce science )

you technically can, if you're smart about it. while integration by parts isn't on the course, integration by recognition is. so if you really can't think of any other way to integrate something, just make up an appropriate function, derive it, integrate both sides, and then manipulate to get the integral you want. for example:

find int x*sinx dx
let f(x) = x*cosx
f'(x) = cosx - x*sinx
f(x) = int cos x dx - int x*sinx dx
x*cosx = sinx - int x*sinx dx
int x*sinx dx = sinx - x*cosx + c

[Alwin]

wait you mean you can't use this in spesh either? wtf. So what do you actually learn in VCE maths in regards to integration?

The basics. (<-- notice that full stop? )

BUT not so basic as to define δx or to prove anti diff of cosine is sin and the fundamental theorem of calculus. Noo, vce will teach you how to use (sometimes wrong or flawed) things, which you will later learn the proofs / correct things (many sciences thankfully) later on in uni..
That's if vce hasn't driven you insane by then

EDIT (to avoid double posting):
@Brightsky, but that method isn't int by parts, well not 'truely'. Like i can't go like u=__ and dv=__
Have to use the vce method of 'integration by parts' which it technically is I guess, but a really watered down version...

[lzxnl]

nek minnit...integrate e^x*sin x