Find the work done by the force field F on particle moving along path

opaquin
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Homework Statement


Find the work done by the force field F on particle moving along path

F(x,y) = -xi + 6yj, <I>C</I>: y = x^3 from (0,0) to (6,216)

The Attempt at a Solution


Drew the graph in the xy plane (y = x^3 is upper limit, x = 0 is lower limit)
W = F ° r'(t) dt
Vector field is not conservative
I want to find r'(t) but am confused as to how, just not sure where to go from here. I have a habit of over thinking problems and am probably doing so. Any assistance is greatly appreciated.
 
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If you're given a path of the form y = f(x), then you can always parametrize it as (x(t),y(t)) = (t, f(t)).
 
in that case, would I just evaluate the line integral?
∫ f(x(t),y(t))||r'||
 
r(t) = _ i + _ j, _ <= t <= _

r(t) = (t)i + (t^3)j, 0 <= t <= 6
r'(t) = <1, 3t^2)

am I on the right track?
 
opaquin said:
in that case, would I just evaluate the line integral?
∫ f(x(t),y(t))||r'||

You need to calculate
\int_0^6 \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}&#039;(t)\,\mathrm{d}t

opaquin said:
r(t) = _ i + _ j, _ <= t <= _

r(t) = (t)i + (t^3)j, 0 <= t <= 6
r'(t) = <1, 3t^2)

am I on the right track?

Yes.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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