work done by a force field

[wakingrufus]

Find the work done by the force field F in moving an object from P to Q.

F(x,y) = (x^2)(y^3)i + (x^3)(y^2)j
P(0,0) Q(2,1)

so i need to integrate the gradient dot dr right?
how do i do that?

[cookiemonster]

You can't do the gradient dot dr. dr is a vector, whereas the gradient is a scalar.

You're looking for a line integral.

http://mathworld.wolfram.com/LineIntegral.html

cookiemonster

[HallsofIvy]

WHAT?? Of course, the gradient is a vector! The "gradient" of the scalar function f(x,y) is defined as the vector fxi+ fyj. Wakingrufus was referring to vector quantity F(x,y) = (x^2)(y^3)i + (x^3)(y^2)j AS a gradient, not taking the gradient of it. (At least I hope not!)

Yes, you could do this by picking some arbitrary path from P to Q (straight line might be easy), since the problem did not give you one. If this is not a "conservative" force field then the work depends on the path and there is not enough information.

Assuming that this a good problem and the force field is conservative, then a simpler way to do this is to find the potential (the function f so that this vector is the gradient of f).

Such a function would have to have fx= x2y[/sup]3[/sup] and fy= x3y2.
From fx= x2y[/sup]3[/sup], f(x,y)= (1/3)x3y3+ g(y) (Since differentiating wrt x treats y as a constant). Differentiating that, fy= x3y2+ g' and that must be equal to x3y2. Okay, that just tells us g is a constant so the "potential function" (really and anti-derivative) is (1/3)x3y3+ C
Evaluate that at P and Q and subtract.

[wakingrufus]

thank you. yes it is conservative. i forgot to mention that i guess.

[HallsofIvy]

Actually, I checked that it was "conservative" (REAL mathematicians would say "exact differential"!) before I did the problem:
The derivative of x2y3 with respect to y is
3x2y2 and the derivative of x3y2 with respect to y is 3x2y[/sup]. Those are the same so the differential is "exact".

[cookiemonster]

Heh, heh, heh... cookiemonster's confusing himself again. Oops?

Mixed up gradient and divergence. Was thinking \nabla applied to a vector (which, little to my credit, is a scalar...).

Moral of the day: read slower and think more. Sorry about that. Please excuse me while I jump off a bridge.

cookiemonster