HSC 2014 MX2 Marathon (archive) (1 Viewer)

hit patel · Apr 11, 2014

Re: HSC 2014 4U Marathon

Davo_01 said:
you forgot the x

$2\pi\int_0^4 x\sqrt{4-(x-2)^2}dx$

Wait why do you have such a different limits and the constant coefficient is so different/. Wait btw I shifted the curve so that I think makes it such that only y remains and u can simply rearrange to form a simple radical with nothing mutliplied.

Davo_01 · Apr 11, 2014

Re: HSC 2014 4U Marathon

dunjaaa said:
So what I should of done:
Multiple top and bottom by sqrt(x^2+4x+7)-x,
You get lim(n->-infinity) 4x+7/(sqrt(x^2+4x+7)-x),
Divide everything by x,
lim(n->-infinity) [4+7/x]/[-sqrt(1+4/x+7/x^2)-1] (for x negative)
=-2

Its right but the way i would have preferred is dividing by $\vert x \vert (=\sqrt{x^2})$

And using $\lim_{x \to -\infty}\frac{x}{\vert x \vert }=-1$

hit patel · Apr 11, 2014

Re: HSC 2014 4U Marathon

Davo_01 said:
you forgot the x

$2\pi\int_0^4 x\sqrt{4-(x-2)^2}dx$

edit: my bad my answer is $8\pi^2$

Does that mean I am right?
8pi integral _2_0 sqrt (4-x^2)
Letting x= 2sintheta and then integrating after using cos double angle formula
16pi ((sin2theta)/2+theta)_pi/2 _ 0 = 8pi^2.

hit patel · Apr 11, 2014

Re: HSC 2014 4U Marathon

hit patel said:
Does that mean I am right?
8pi integral _2_0 sqrt (4-x^2)
Letting x= 2sintheta
16pi ((sin2theta)/2+theta)_pi/2 _ 0 = 8pi^2. yayyyyy

Should i attach a hand written solution so you can see if correctly done?

Davo_01 · Apr 11, 2014

Re: HSC 2014 4U Marathon

hit patel said:
Does that mean I am right?
8pi integral _2_0 sqrt (4-x^2)
Letting x= 2sintheta
16pi ((sin2theta)/2+theta)_pi/2 _ 0 = 8pi^2. yayyyyy

Im not sure if the working out is correct, as the preferred substitutions were either $x-2=2\sin{\theta}$ or $u=x-2$

dunjaaa · Apr 11, 2014

Re: HSC 2014 4U Marathon

ahh i see now, I thought my first approach was wrong lol..

Davo_01 · Apr 11, 2014

Re: HSC 2014 4U Marathon

hit patel said:
Should i attach a hand written solution so you can see if correctly done?

Sure, while you make one il type up my solution, I know two different pathways that work but one works very nicely in this example.

hit patel · Apr 11, 2014

Re: HSC 2014 4U Marathon

Davo_01 said:
Im not sure if the working out is correct, as the preferred substitutions were either $x-2=2\sin{\theta}$ or $u=x-2$

Ok davo. So what I did was the quicker way:
Shifting the curve and its reflection to the right two units meant that the equation now changed to (x^2)/4 +y^2 =1 and therefore the axis through which it is revolved changes to 2. Then finding the volumes of the shells and adding them up to find total volume i simply subbed in y and then i got sqrt( 4-x^2).
Edit: and the radius remains the same for the volume calculation but waht changes is the equation and axis of revolution so you dont have to integration that..

Davo_01 · Apr 11, 2014

Re: HSC 2014 4U Marathon

$V= 2\pi\int_0^4 x\sqrt{4-(x-2)^2}dx$

Let $u=x-2$ $dx=du$

Limits change to -2 and 2

$V=2\pi\int_{-2}^{2}(u+2)\sqrt{4-u^2}du$

$V=2\pi\int_{-2}^{2}u\sqrt{4-u^2}du +4\pi\int_{-2}^{2}\sqrt{4-u^2}du$

$=0+4\pi(\frac{4\pi}{2})$ (Odd function, Area of semicircle)

$=8\pi^2$ $units^3$

Davo_01 · Apr 11, 2014

Re: HSC 2014 4U Marathon

hit patel said:
Ok davo. So what I did was the quicker way:
Shifting the curve and its reflection to the right two units meant that the equation now changed to (x^2)/4 +y^2 =1 and therefore the axis through which it is revolved changes to 2. Then finding the volumes of the shells and adding them up to find total volume i simply subbed in y and then i got sqrt( 4-x^2).
Edit: and the radius remains the same for the volume calculation but waht changes is the equation and axis of revolution so you dont have to integration that..

Thing is i think you forgot to multiply by the shell radius. Remember $V=2\pi xy$ , It looks like ur missing the x, if you shift the axis of rotation by 2 then instead of $V=2\pi xy$ , $V=2\pi(x--2)y$

Its actually similar to using substitution u=x-2

hit patel · Apr 11, 2014

Re: HSC 2014 4U Marathon

Davo_01 said:
$V= 2\pi\int_0^4 x\sqrt{4-(x-2)^2}dx$

Let $u=x-2$ $dx=du$

Limits change to -2 and 2

$V=2\pi\int_{-2}^{2}(u+2)\sqrt{4-u^2}du$

$V=2\pi\int_{-2}^{2}u\sqrt{4-u^2}du +4\pi\int_{-2}^{2}\sqrt{4-u^2}du$

$=0+4\pi(\frac{4\pi}{2})$ (Odd function, Area of semicircle)

$=8\pi^2$ $units^3$

*facedesk*. What was I thinking. Yes caffeine has gotten to me. I though about using the area of semicircle for integration but then this is me 'hmmm so its a semicircle and i have to find area of a quarter but limits are 2 and 0 for me and the integral says 4-x^2, so i cant do it' hahahhaa. well forgive my stupidity or caffeneity. And sorry how do you attach stuff?

Davo_01 · Apr 11, 2014

Re: HSC 2014 4U Marathon

hit patel said:
*facedesk*. What was I thinking. Yes caffeine has gotten to me. I though about using the area of semicircle for integration but then this is me 'hmmm so its a semicircle and i have to find area of a quarter but limits are 2 and 0 for me and the integral says 4-x^2, so i cant do it' hahahhaa. well forgive my stupidity or caffeneity. And sorry how do you attach stuff?

ahahha no worries, there is a button 3rd from the right

hit patel · Apr 11, 2014

Re: HSC 2014 4U Marathon

hit patel said:
*facedesk*. What was I thinking. Yes caffeine has gotten to me. I though about using the area of semicircle for integration but then this is me 'hmmm so its a semicircle and i have to find area of a quarter but limits are 2 and 0 for me and the integral says 4-x^2, so i cant do it' hahahhaa. well forgive my stupidity or caffeneity. And sorry how do you attach stuff?

oh ok it was just some javascript problem that didnt let me attach.

Sorry had to attach in 4 pics. phone is a touchscreen but camera quality is like that of a bricks.

Davo_01 · Apr 12, 2014

Re: HSC 2014 4U Marathon

hit patel said:
oh ok it was just some javascript problem that didnt let me attach. View attachment 30265 View attachment 30266
View attachment 30267 View attachment 30268

Sorry had to attach in 4 pics. phone is a touchscreen but camera quality is like that of a bricks.

Again remember to multiply by shell radius, your answer was still right because the extra part of the integral happened to be zero but in general ur answer will probably change. Other than that good work

hit patel · Apr 12, 2014

Re: HSC 2014 4U Marathon

Davo_01 said:
Again remember to multiply by shell radius, your answer was still right because the extra part of the integral happened to be zero but in general ur answer will probably change. Other than that good work

But I did. It just happened that when I added the volumes of the two shells. The x part became zero due to symmetry. And this will always happen if you shift the curve I think. Thanks for the question though. It was a well set out and straight to the point question.

Davo_01 · Apr 12, 2014

Re: HSC 2014 4U Marathon

hit patel said:
oh ok it was just some javascript problem that didnt let me attach. View attachment 30265 View attachment 30266
View attachment 30267 View attachment 30268

Sorry had to attach in 4 pics. phone is a touchscreen but camera quality is like that of a bricks.

Ooh why did you take two seperate partial volumes

Davo_01 · Apr 12, 2014

Re: HSC 2014 4U Marathon

Hmm ive never used this symmetry method before but if it works it works,
anyways man i gotta go to sleep, work tomoro so il catch ya later. take care bro

hit patel · Apr 12, 2014

Re: HSC 2014 4U Marathon

Davo_01 said:
Ooh why did you take two seperate partial volumes

Doesnt it help ? Well isnt that ok?

Davo_01 · Apr 12, 2014

Re: HSC 2014 4U Marathon

No wait i see how the symetry method works, both partial volumes are meant to be equal so you should add them and divide by 2 to get v, you didn't divide by 2 so your answer would have been double but you made ur limits 0 to 2 instead of -2 to 2 so the volume halved and overall answer stayed same but the working out isnt fully correct do you see what i mean?

hit patel · Apr 12, 2014

Re: HSC 2014 4U Marathon

Davo_01 said:
No wait i see how the symetry method works, both partial volumes are meant to be equal so you should add them and divide by 2 to get v, you didn't divide by 2 so your answer would have been double but you made ur limits 0 to 2 instead of -2 to 2 so the volume halved and overall answer stayed same but the working out isnt fully correct do you see what i mean?

Yes I guess but yes I should have showed that by dividing by 2. I will do that next time. Hahaha next one when you have time please.

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