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I am trying to find the work done by a field $$F(x,y)=(e^ysin(x))i-(e^ycos(x)-\sqrt{1+y})j$$ in a moving particle from $(-\pi,\pi^2)$ to $(\pi,\pi^2)$ on the parabola $y=x^2$. I seem to be having difficulty determining if there is a potential function for this and what is it.

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  • $\begingroup$ I suggest looking up potentials. I will attempt an answer. I am uncertain of why you were down voted, I have corrected it. Perhaps you were down voted because you didn't show your working. Please do, and show it in future. This site is not wolfram. $\endgroup$ – Almentoe Dec 1 '15 at 11:43
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"I am trying to find the work done by a field $$F(x,y)=(e^ysin(x))i-(e^ycos(x)-\sqrt{1+y})j$$ in a moving particle from $(-\pi,\pi^2)$ to $(\pi,\pi^2)$ on the parabola $y=x^2$. I seem to be having difficulty determining if there is a potential function for this and what is it."

Methodologically (physics):

This is the field equation: $$ F(x,y)=(e^ysin(x)) \mathbf{i}-(e^ycos(x)-\sqrt{1+y})\mathbf{j}$$ We'd like to find a Potential, $P$. In physics (not applied maths, look up the appropriate definitions.) $P$ must satisfy $-\nabla P = F $

We know that if there is a potential, then the force must be conservative. This is the same as $\text{curl} (F) = 0$. Clearly, this is not the case (exercise).

So we must do it the long way around!

Let $c$ denote the path along the parabola $y=x^2$ which starts at $(-\pi,\pi^2)$ and ends at $(\pi,\pi^2)$. We parametrise it thus:

$c(t) = t\mathbf{i} + t^2 \mathbf{j} $.

And the rest is up to you. I do hope you can do it.

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