Find 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 f_xi+ f_yj. 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 f_x= x²y[/sup]3[/sup] and f_y= x³y².
From f_x= x²y[/sup]3[/sup], f(x,y)= (1/3)x³y³+ g(y) (Since differentiating wrt x treats y as a constant). Differentiating that, f_y= 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