VCE Specialist 3/4 Question Thread!

[lzxnl]

(x/b)^2+(y/a)^2=1
xE[-b,b], yE[-a,a]

Rotating around x axis:
volume =

Around the y axis:
volume =

Oh wait look, they're not the same.

[SocialRhubarb]

No, you will get different answers.

Let's assume that your ellipse is much longer than it is tall, so the x-intercepts are much larger than the y intercepts.

What happens if you rotate it about the x-axis? You get a long tubular shape.

What happens if you rotate it about the y-axis? You get a flatter, wider, disc-like shape.

You get different shapes so you'll get different volumes.

[brightsky]

upon second thought, the volumes can still be different. consider rotating a really skinny ellipse around the x and y axis. the solid is different.

proof:

x^2/a^2 + y^2/b^2 = 1

if rotate around x-axis:
V = pi * int^(a)_(-a) b^2 (1-x^2/a^2) dx
= 2*pi*b^2/a^2 * int^(a)_(0) (a^2 - x^2) dx
= 2*pi*b^2/a^2 * [a^2*x - x^3/3]^(a)_(0)
= 2*pi*b^2/a^2 * (a^3 - a^3/3)
= 2*pi*b^2/a^2 * (2a^3/3)
= 4pi/3*ab^2

rotating around y-axis:
V = pi*int^(b)_(-b) (a^2(1-y^2/b^2)) dy
= 2*pi*a^2/b^2 int^(b)_(0) (b^2 - y^2) dy
= 2*pi*a^2 / b^2 [b^2y - y^3/3]^b_0
= 2*pi*a^2/b^2 (b^3 - b^3/3)
= 2*pi*a^2/b^2 ( 2b^3/3)
= 4pi/3*a^2b

edit: beaten. xD

[bonappler]

I hope I don't sound like a noob but shouldn't the shape and therefore the volume be the same?

[lzxnl]

They're not the same shape. Draw them out.

[bonappler]

Oh hang on I understand what I was doing thanks guys it makes sense now

[Jaswinder]

Question 12 Multiple Choice 2012 Exam 2 VCAA paper I understand why it is A, however why can't it be E? I know that when you evaluate it, you get different answers however I can't seem to figure out why E doesn't work. thanks

[availn]

Question 12 Multiple Choice 2012 Exam 2 VCAA paper I understand why it is A, however why can't it be E? I know that when you evaluate it, you get different answers however I can't seem to figure out why E doesn't work. thanks

The limits are wrong. It should be from -2 to 4.

[lzxnl]

nliu logic

Heffernan 2010 exam: gets 79/80
Kilbaha 2010 exam: gets 79/80

Conclusion: Heffernan=Kilbaha

[Jaswinder]

The limits are wrong. It should be from -2 to 4.

aha oh yeh thanks
also

Three points A, B and C are given by A(2,2,1), B(3,0,4) and C(5,-4,10)
How do we show that A, B and C are collinear? What exactly does collinear mean anyway?
Also
Say we are given two unit vectors OA and OB. How do we find a vector that bisects the angle between OA and OB?

Thanks

[lzxnl]

aha oh yeh thanks
also

Three points A, B and C are given by A(2,2,1), B(3,0,4) and C(5,-4,10)
How do we show that A, B and C are collinear? What exactly does collinear mean anyway?
Also
Say we are given two unit vectors OA and OB. How do we find a vector that bisects the angle between OA and OB?

Thanks

Collinear means on the same line. AKA vector AB is a multiple of vector BC; if they're parallel vectors running through point B, they must be on the same line.

For your second question, it's actually quite nice. As OA and OB have the same length, OAB forms an isosceles triangle. Therefore, the position vector of the midpoint of AB will bisect the angle between OA and OB.

This idea may come up later; that a line drawn from the apex of an isosceles triangle to the midpoint of the base is a perpendicular bisector of the base and bisects the apex angle. Remember it. It's not hard to prove if prompted, but will you recognise when to use it?

[Conic]

It says "given 2 unit vectors..." in the second question. He isn't referring to OA and OB form the previous question.

[Jaswinder]

thanks nliu.

A graph pi/2 + sin-1(x) and it asks to find the volume generated when the area between the curve y=f(x) and the y-axis is rotated about the y-axis to form a solid revolution.

why don't they put a negative in front of a integral from 0 to pi/2?

[brightsky]

because from y = 0 to y = pi/2, the graph is to the left of the x-axis...meaning that it is 'negative'.

[lzxnl]

thanks nliu.

A graph pi/2 + sin-1(x) and it asks to find the volume generated when the area between the curve y=f(x) and the y-axis is rotated about the y-axis to form a solid revolution.

why don't they put a negative in front of a integral from 0 to pi/2?

Why would you need one? You've squared x in the integral and it's always positive

Brightsky you've missed the point xP