## Question47 GR0568

Forum for the GRE subject test in mathematics.
moo5003
Posts: 36
Joined: Mon Oct 06, 2008 7:33 pm

### Question47 GR0568

Let F be a constant unit force that is parallel to the vector (-1,0,1) in xyz-space. What is the work done by F on a particle that moves along the path given by (t,t^2,t^3) between time t=0 and time t=1.

a) -1/4
b) -1/(4root(2))
c) 0
d) root(2)
e) 3root(2)

I did it wrong but my method was to take the projection of (t,t^2,t^3) on (-1,0,1)

Then integrating from 0 to 1. My answer was -1/4 (which is wrong its supposed to be 0) how did you guys do it?

CoCoA
Posts: 42
Joined: Wed Sep 03, 2008 5:39 pm
Work = \int_C{Fdr}; put everything in terms of t:

\int_0^1{(-1,0,1)(1,2t,3t^2)dt}

= \int_0^1{-1+3t^2)dt}

= [-t+t^3]_0^1

= -1+1 = 0

moo5003
Posts: 36
Joined: Mon Oct 06, 2008 7:33 pm
Cocoa I'm a little unsure of what type of notation you are using, if possible could you please explain it.

After looking at it again you basically took the integral from 0 to 1 of the dot product of (-1,0,1) and the rate of change of (t,t^2,t^3).

Where the force on the particle is (1,2t,3t^2) since they want the force only parallel to (-1,0,1) its magnitude is 3t^2-1 at which point you simply integrate from 0 to 1.

Thanks for the help, I made two mistakes in my write up of this problem. Firstly I did not take the rate of change as the force I took the position vector itself and I did not take the magnitude of the projection vector.

CoCoA
Posts: 42
Joined: Wed Sep 03, 2008 5:39 pm
Sorry for the notation problem - Latex symbols (done haphazardly).

Work is the dot product of force and displacement. Force is constant in this problem, but displacement is not.

So if r(t) is position, then total work is the limit of the sum of dot product of force and infitely small change in position i.e., integral over the curve of Fdr (lazy of me to write dot product like this!). Then just push around the symbols to get the answer.

goodtrain
Posts: 11
Joined: Sat Sep 19, 2009 2:57 pm

### Re: Question47 GR0568

why are these physics questions like these

diogenes
Posts: 73
Joined: Sun Aug 31, 2008 9:31 pm

### Re:

CoCoA wrote:$Work = \int_C{Fdr}$

put everything in terms of t:

$\int_0^1{(-1,0,1)(1,2t,3t^2)dt}

= \int_0^1{(-1+3t^2)dt}

= [-t+t^3]_0^1

= -1+1 = 0$

Fixed the tex code.

origin415
Posts: 61
Joined: Fri Oct 23, 2009 11:42 pm

### Re: Question47 GR0568

I find this question weird, as it requires some knowledge of physics. Luckily I am a physics minor...

You can avoid all those nasty integrals (okay, not so bad, but still unnecessary) by knowing that the work done will be the difference in potential, and force is the gradient of potential, so taking U(x, y, z) = z - x to be the potential, U(1, 1, 1) - U(0, 0, 0) = 0.
The actual curve <t, t^2, t^3> is meaningless.