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Work done by a force field 
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#1
Mar304, 01:43 AM

P: 14

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? 


#2
Mar304, 01:49 AM

P: 991

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 


#3
Mar304, 11:39 AM

Math
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PF Gold
P: 39,311

WHAT?? Of course, the gradient is a vector! The "gradient" of the scalar function f(x,y) is defined as the vector f_{x}i+ f_{y}j. 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}= x^{2}y[/sup]3[/sup] and f_{y}= x^{3}y^{2}. From f_{x}= x^{2}y[/sup]3[/sup], f(x,y)= (1/3)x^{3}y^{3}+ g(y) (Since differentiating wrt x treats y as a constant). Differentiating that, f_{y}= x^{3}y^{2}+ g' and that must be equal to x^{3}y^{2}. Okay, that just tells us g is a constant so the "potential function" (really and antiderivative) is (1/3)x^{3}y^{3}+ C Evaluate that at P and Q and subtract. 


#4
Mar304, 11:59 AM

P: 14

Work done by a force field
thank you. yes it is conservative. i forgot to mention that i guess.



#5
Mar304, 05:54 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,311

Actually, I checked that it was "conservative" (REAL mathematicians would say "exact differential"!) before I did the problem:
The derivative of x^{2}y^{3} with respect to y is 3x^{2}y^{2} and the derivative of x^{3}y^{2} with respect to y is 3x^{2}y[/sup]. Those are the same so the differential is "exact". 


#6
Mar404, 01:32 AM

P: 991

Heh, heh, heh... cookiemonster's confusing himself again. Oops?
Mixed up gradient and divergence. Was thinking [itex]\nabla[/itex] 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 


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