HSC 2015 MX2 Marathon (archive) (3 Viewers)

InteGrand · Apr 19, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Prove that the only way an ellipse and a circle with the same centre, can be tangent to each other at their intersection, is if the radius of the circle is the same in length as the semi-minor axis of the ellipse$$

$Or, the radius can be the length of the semi-\emph{major} axis.$

Drsoccerball · Apr 19, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Prove that the only way an ellipse and a circle with the same centre, can be tangent to each other at their intersection, is if the radius of the circle is the same in length as the semi-minor or semi-major axis of the ellipse$$

It is because if they share the same center and are tangents then the semi major or semi minor is equal to the radius. The value of the semi minor/semi major axis gives the distance from the center to the point. Therefore if they share the same center and the circle has a radius equal to the semi major or semi minor axis it is a tangent at both sides of the axis. Can't explain it D:

Zlatman · Apr 19, 2015

Re: HSC 2015 4U Marathon

Hopefully, this suffices (still no LATEX):
http://i.imgur.com/QJNbrGy.jpg?1

Couple of fixes:
*r is the length of the semi-major/semi-minor axis (in final answer)
*gradients are equal

Drsoccerball · Apr 19, 2015

Re: HSC 2015 4U Marathon

Zlatman said:
Hopefully, this suffices (still no LATEX):
http://i.imgur.com/QJNbrGy.jpg?1

Couple of fixes:
*r is the length of the semi-major/semi-minor axis (in final answer)
*gradients are equal

Start up a forum where you just post random Latex thats how me and ekman learnt it and i think usinng logic is more simpler than calculations for this

Sy123 · Apr 19, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
It is because if they share the same center and are tangents then the semi major or semi minor is equal to the radius. The value of the semi minor/semi major axis gives the distance from the center to the point. Therefore if they share the same center and the circle has a radius equal to the semi major or semi minor axis it is a tangent at both sides of the axis. Can't explain it D:

That is sufficient to prove that if the radius and the one of the axes of the ellipse were equal, then they are tangent to each other. However the question is asking, given that they are tangent, that therefore the radius and one of the axes are equal.

So while the question asked to prove (P implies Q) you have shown (Q implies P)

glittergal96 · Apr 19, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Prove that the only way an ellipse and a circle with the same centre, can be tangent to each other at their intersection, is if the radius of the circle is the same in length as the semi-minor or semi-major axis of the ellipse$$

Note that both of these shapes have reflective symmetry in the coordinate axes so it suffices to consider points P of tangential intersection in the first quadrant.

The problem then reduces to showing that the only normals to an ellipse that pass through the origin are the axes (which clearly do so I won't prove this).

If P does not lie on the coordinate axes, chucking (0,0) into the parametric normal equation and simplifying gives $(a^2-b^2)\sin(\theta)=0$ which has no solutions.

(The dot-product would provide a cleaner way of doing this because it avoids dealing with gradients, and the semi-minor axis is vertical.)

Kaido · Apr 19, 2015

Re: HSC 2015 4U Marathon

Anyone got an efficient method to solve 2001 Q8 b) i)
I feel that for 2 marks, should be under 1 page =.=

integral95 · Apr 19, 2015

Re: HSC 2015 4U Marathon

Kaido said:
Anyone got an efficient method to solve 2001 Q8 b) i)
I feel that for 2 marks, should be under 1 page =.=

$0<x<1 \Rightarrow e^x < 3 \\ \\ \therefore \int_0^1 x^{\alpha}e^x < \int_0^1 3x^{\alpha} = \frac{3}{\alpha+1}$

Kaido · Apr 19, 2015

Re: HSC 2015 4U Marathon

Cheers pokemon hunter, If I wrote that in the HSC, will they accept it and give the 2 marks?

InteGrand · Apr 19, 2015

Re: HSC 2015 4U Marathon

Kaido said:
Cheers pokemon hunter, If I wrote that in the HSC, will they accept it and give the 2 marks?

They should. Marking criteria for that Q is on Page 15 here: http://arc.boardofstudies.nsw.edu.a.../support_documentation/math_ext2_guide_01.pdf

T-R-O-L-O-L · Apr 20, 2015

Re: HSC 2015 4U Marathon

Far this stuff looks hard, I'm sticking to 2 Unit.

Drsoccerball · Apr 20, 2015

Re: HSC 2015 4U Marathon

T-R-O-L-O-L said:
Far this stuff looks hard, I'm sticking to 2 Unit.

its not hard once you learn it -.- trust me if you are capable to extension 2

Drsoccerball · Apr 21, 2015

Re: HSC 2015 4U Marathon

I dont understand volumes slicing method D: i can't do even the simplest questions... need help
EDIT: for example i dont know what to put as the arbitary value of the x or y slice

InteGrand · Apr 21, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
I dont understand volumes slicing method D: i can't do even the simplest questions... need help
EDIT: for example i dont know what to put as the arbitary value of the x or y slice

The slicing method uses $V=\int _a ^b A(x)\text{ d}x$ . So you need to find the area of a typical slice at a point x (this isA(x)), and integrate this area function over the given interval.

Drsoccerball · Apr 21, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
The slicing method uses $V=\int _a ^b A(x)\text{ d}x$ . So you need to find the area of a typical slice at a point x (this isA(x)), and integrate this area function over the given interval.

a simple question:
$y= 2\sqrt{x} , $ bounded by x axis $, x=4 $ about $ x=4$

Ekman · Apr 21, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
a simple question:
$y= 2\sqrt{x} , $ bounded by x axis $, x=4 $ about $ x=4$

The volume would be given by:
$V=\pi\int_{0}^{4} (4-x)^2 dy$

You substitute: $\frac{y^2}{4} = x$ and integrate accordingly

Drsoccerball · Apr 21, 2015

Re: HSC 2015 4U Marathon

Ekman said:
The volume would be given by:
$V=\pi\int_{0}^{4} (4-x)^2 dy$

You substitute: $\frac{y^2}{4} = x$ and integrate accordingly

solid explanation

Ekman · Apr 21, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
solid explanation

Since the radius of the cylinders are 4-x, and the height/thickness of the cylinders is dy, you use the formula pir^2h to find the volume. The integral is the summation of the cylinders between 0 to 4

InteGrand · Apr 21, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
a simple question:
$y= 2\sqrt{x} , $ bounded by x axis $, x=4 $ about $ x=4$

(It helps to draw a diagram)

If you're using the slicing method for this, you'll be integrating with respect to y, from y = 0 to y = 4 (since these are the y-values of the endpoints in question).

At a given $h\in [0,4]$ , the radius of a typical slice will be equal to the distance of the line x = 4 to the point on the given curve where y = h. Since the curve is $y=2\sqrt{x}$ , at the point where y = h, we have $2\sqrt{x}=h\Rightarrow x = \frac{h^2}{4}$ .

So the required radius at a height h is $r(h)=4-x(h)=4-\frac{h^2}{4}$ . The thickness of the slice is $\text{d}h$ . So the slice has the volume $\text{d}V=\pi (r(h))^2 \text{ d}h$ (using the formula for the volume of a cylinder).

Then the total volume is found by integrating from h = 0 to h = 4.

Drsoccerball · Apr 21, 2015

Re: HSC 2015 4U Marathon

so y=x^2, y=1, x=3 about x axis would be
$pi \int_1^3(3-x)^2dx$ ?

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