PR Exercise 24 pag 164

Forum for the GRE subject test in mathematics.
ardroc
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PR Exercise 24 pag 164

Postby ardroc » Mon Sep 22, 2014 10:04 am

A vertical fence is constructed whose base is the curve y= x* sqrt(x) from (0,0) to (1,1) and whose height above each point (x,y) along the curve is x^3 - y^2 + 27. Find the area of this fence.

The correct answer given by the book is D. 13*sqrt(13) - 8, which is obtained by solving a path integral.
However I was a bit confused by this and I thought that a path integral in this case wouldn't give you the "area" of the fence but something else.

I thought the height above each point on the curve is simply 27 by plugging in y=x*sqrt(x) and so the area of the fence should simply be obtained by integrating 27 + x*sqrt(x) - x*sqrt(x) = 27 from 0 to 1.
So my answer would be E. 27

Any thoughts on this?
Thanks!

Austin
Posts: 26
Joined: Fri Jun 01, 2012 9:32 am

Re: PR Exercise 24 pag 164

Postby Austin » Tue Sep 23, 2014 12:05 am

You're right that the height is constantly 27 on this fence, but the base is not of unit length. You need to find the arc length of the base of the fence.

You'll want to integrate \sqrt[ 1+(y'(t))^2 ] from 0 to 1.

I hope this helps.

DDswife
Posts: 58
Joined: Thu Aug 14, 2014 5:29 pm

Re: PR Exercise 24 pag 164

Postby DDswife » Tue Sep 23, 2014 5:34 am

The line integral oes give you the area of fence for a scalar funtion, yes.

ardroc
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Joined: Tue Sep 24, 2013 8:55 am

Re: PR Exercise 24 pag 164

Postby ardroc » Tue Sep 23, 2014 8:49 am

But isn't the area of the fence simply the area between curves y=x*sqrt(x) + 27 and y=x*sqrt(x), between 0 and 1.
(And so simply 27 from 0 to 1)?

DDswife
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Joined: Thu Aug 14, 2014 5:29 pm

Re: PR Exercise 24 pag 164

Postby DDswife » Tue Sep 23, 2014 10:15 am

Find the length of the base curve. Is it 1?

ardroc
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Re: PR Exercise 24 pag 164

Postby ardroc » Tue Sep 23, 2014 11:08 am

The base curve length is not 1 but shouldn't that be irrelevant to the area of the fence since the fence is bounded above by the same curve (just shifted upwards by 27)?

DDswife
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Joined: Thu Aug 14, 2014 5:29 pm

Re: PR Exercise 24 pag 164

Postby DDswife » Tue Sep 23, 2014 11:11 am

I don't think so

I did the arclength, multiplied by 27 (it was smart of you to realise about that) and got exactly the answer. Just give it a try.

Where did you got this problem from?

ardroc
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Joined: Tue Sep 24, 2013 8:55 am

Re: PR Exercise 24 pag 164

Postby ardroc » Tue Sep 23, 2014 11:44 am

It's from the Princeton Review book. They provide a solution as well by doing the line integral and getting answer D. But somehow I think they made a mistake and answer should be 27 which is also amongst the possible answers.

Well, if you agree that the fence is bounded above and below by the two curves with the same shape (shifted by 27) then I think the area of this thing should be 27! It's the same as the area of a rectangle with height 27 because whatever you add below you take away above.

DDswife
Posts: 58
Joined: Thu Aug 14, 2014 5:29 pm

Re: PR Exercise 24 pag 164

Postby DDswife » Tue Sep 23, 2014 11:55 am

We can consider that a rectangle that is a bit bended. And the hieght is 27, but the base is not sqrt 2 and you need to calculate it.

But, to clarify this to you, I will change the example:

What if you have a fence of height 27 over the unit circle? What would be its area? Do you think that the area is 0?

What if you have the same but only from 0 to pi? Do you think that the area is now
(1-(-1)) x 27?

This instgeral is NOT path independent. You actually need to calculate it.

If you still don't see it, do a drawing and think of you fencing your yard. Forget about integrals, pretend you only attended the primary school. How would you alculate the amount of paint in gallons for this fence?

ardroc
Posts: 10
Joined: Tue Sep 24, 2013 8:55 am

Re: PR Exercise 24 pag 164

Postby ardroc » Tue Sep 23, 2014 1:33 pm

OMG All this time I was confusing the "area of the fence" with the "area ENCLOSED by the fence"!
I thought there was a hypothetical garden enclosed by fences on x=0, x=1, y=x*sqrt(x), y=x*sqrt(x) +27.

Sorry I must have looked like a fool! Now it's completely obvious!
Thanks!

ardroc
Posts: 10
Joined: Tue Sep 24, 2013 8:55 am

Re: PR Exercise 24 pag 164

Postby ardroc » Tue Sep 23, 2014 1:41 pm

Oh wait! That's not even what I thought.
I actually thought the fence was lying flat on the x-y plane bounded by those four curves I mentioned! which brought me to calculate the area described above.

Thanks and sorry!

DDswife
Posts: 58
Joined: Thu Aug 14, 2014 5:29 pm

Re: PR Exercise 24 pag 164

Postby DDswife » Tue Sep 23, 2014 1:57 pm

Don't be. This happens to me too at times.

Not sure how the fence is. It says a vertical fence and you didn't post equals to what. So, I think they mean z
Last edited by DDswife on Tue Sep 23, 2014 3:48 pm, edited 1 time in total.

ardroc
Posts: 10
Joined: Tue Sep 24, 2013 8:55 am

Re: PR Exercise 24 pag 164

Postby ardroc » Tue Sep 23, 2014 3:28 pm

Yeah it doesn't say equal z but you made me realise that the height should definitely be in the z direction.

However if the fence is flat on the x-y plane (i.e flat on the floor) then the area should definitely be 27 if the fence is bounded by the curves I wrote just now! :wink:

DDswife
Posts: 58
Joined: Thu Aug 14, 2014 5:29 pm

Re: PR Exercise 24 pag 164

Postby DDswife » Tue Sep 23, 2014 3:46 pm

I see now what you understood. And in this case you should be right.




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