Volume 2011 JRAHS Trial (1 Viewer)

[asianese]

Done part i...I'm unsure about part ii and iii). Just started volumes...

Can anyone help?

Thanks

 

[lolcakes52]

Okay, it does take a while to get your head around it but its easy enough once you do some doodling. I found the y value of the elipse for any value of x, ((4-x^2)^(1/2))/2 and the value of x for the corresponding point on the parabola mentioned in part i. Simply put, I found the corresponding focal lengths for each parabola and let them be equal so I could find the value of x in terms of a.
This yielded 2a=((4-x^2)^(1/2))/2
which became a=((4-x^2)^(1/2))/4.
I then subbed this into the equation in part i and gained the correct expression for A(x).
Part ii is quite easy after this.

∂V=∂x*A(x)
V is the integral of A(x)dx from 2 to -2

 

[AbsoluteValue]

It says the focal chord is perpendicular to the axis, therefore you are looking at the area between y=0 and y=a. So just integrate x=(4ay)^0.5 in that limit, I think you should get the area.

 

[AbsoluteValue]

I get 4a^2/3 which is half the required, something went wrong.

 

[asianese]

THANKS!! I knew I had to use the latus rectum length somewhere (4a)...Thanks

From the same paper...can't get my head around this one!

 

[AbsoluteValue]

You are integrating this: 2\pi \int_{0}^{\frac{\pi }{2}} x(1-sinx)dx
Type this in http://www.codecogs.com/latex/eqneditor.php to make it easier to read. For some reason I can't post latex.

 

[asianese]

Well I did it via the rectangle from -2a to 2a minus the the integral

 

[AbsoluteValue]

How can you copy/past latex? It doesn't work for me.

 

[asianese]

You have to type [ tex ] [ /tex ] around what you want latexed.

And yes, I have that expression, but how do I get it to ?

 

[AbsoluteValue]

[ tex ] 2\pi \int_{0}^{\frac{\pi }{2}} x(1-sinx)dx [ /tex ]
Doesn't work

 

[rolpsy]

THANKS!! I knew I had to use the latus rectum length somewhere (4a)...Thanks

From the same paper...can't get my head around this one!

View attachment 25801

well you know that



so you can say



use the slicing method and you should get it

edit:

[ tex ] 2\pi \int_{0}^{\frac{\pi }{2}} x(1-sinx)dx [ /tex ]
Doesn't work

you have to put
[noparse][/noparse] (no spaces)

 

[Siddy123]

iv seen that question before in a really old sydney tech paper.
i think 80's

 

[asianese]

Absolute, without the spaces! (in the tex tags)

 

[AbsoluteValue]

WORKS !!!

 

[AbsoluteValue]

WORKS !!!

How do you change this to

 

[asianese]

well you know that



so you can say



use the slicing method and you should get it

edit:

you have to put
[noparse][/noparse] (no spaces)

HO. LY. WHAT THE HELL MIND BLOWN O__________________________________O. Did not know you could do that. It's so simple. Yet amazing. Whoa. Thankyou. I will put that in the books.

 

[AbsoluteValue]

HO. LY. WHAT THE HELL MIND BLOWN O__________________________________O. Did not know you could do that. It's so simple. Yet amazing. Whoa. Thankyou. I will put that in the books.

whaaat, they never teach this technique to us

 

[asianese]

Yeah IKR...Wuttt

So slices, just normal discs, holy...

 

[AbsoluteValue]

Yeah IKR...Wuttt

So slices, just normal discs, holy...

No need to ever use cylindrical shells like this

 

[rolpsy]

HO. LY. WHAT THE HELL MIND BLOWN O__________________________________O. Did not know you could do that. It's so simple. Yet amazing. Whoa. Thankyou. I will put that in the books.

Oh lol

Terry Lee's book makes quite a fuss about it

anyway the solutions are here