Problem: Find the work done by the force field F(x,y)=<xcos(y), y^2> on a particle that moves along y=x^2 from (-1,1) to (3,9).
So I would be able to do this no problem if it weren't for the y=x^2. That's the part that's messing me up. Normally, you find the equation of the line between the two points, so in this case it would be r(t)=<-1+4t, 1+8t> and then you would find the derivative of that, r'(t)=<4,8> and then find F(r(t))=<(-1+4t)cos(1+8t), (1+8t)^2> then dot the F(r(t)) with r'(t) but what do you do with the y=x^2? I have a feeling it's going to have an effect on r but I can't determine what effect it will have. Cause instead of it having a straight line it's curved so technically finding the equation of the line wouldn't work, correct? So what do I do?! Any help is greatly appreciated!
Eep, okay so I thought I got it but the example doesn't deal with work. It deals with a basic integral where you are already given the formula to integrate. I know the formula comes out to F(r(t)) dotted with r'(t) but in this case, we have the curve to integrate on so we need to make it in terms of the curve. So it should be in terms of x, right? Gah, I'm confusing myself. Can someone help clear this up a bit?
Well the method should be the following:
We want a unit vector in the dx direction, so this will be along the line tangent to y = x^2. So get that unit vector. This will be a function of x. Multiply this unit vector by dx to get your vector dx. Now sub in y = x^2 into your field, take your dot product, and integrate. (Trouble is that I'm getting a seriously bad integral from this, so I'm assuming I did something wrong to the calculation.)
-Dan
This problem is starting to majorly confuse me. My professor said he did it in about an hour, the entire list of problems. So far I'm at over 2 hours working on it and 8 out of the 10 completed.
I know when we worked on these problems in class, we just did the basic work forms, nothing dealing with a curve in the mix and there are no examples. I'm wondering if instead of taking the line with respect to t, would we take it instead with respect to x so the line comes out with -1+4x and 1+8x? Because for regular integral problems, we were told to take dy, in this case it would be 2xdx and that would be the ending part to the integral if we leave it in terms of x and y.