7 10 points let vecfx y 1 xyexy x2exy be a given force field...

7. (10 points) Let $\vec{F}(x, y) = ((1 + xy)e^{xy}, x^2e^{xy})$ be a given force field and let C: $\vec{r}(t) = (\cos t, 2\sin t)$, $0 \le t \le 1$ be a given path. (a) (5 points) Find a potential function for $\vec{F}$. (b) (5 points) Find $\int_C \vec{F} \cdot d\vec{r}$, the work done by $\vec{F}$ in moving an object along C.

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00:01 Okay, so let's get started by finding the potential function lowercase f of x.

00:06 Well, such a function must be a function such that the gradient of f must be equal to our vector field.

00:15 In particular, this shows that the partial derivative of f with respect to y must be equal to x squared e to the x y.

00:26 Perfect.

00:27 So, this implies what? well, this implies that f of x comma y must be a function of the form x squared y, x squared over x.

00:43 Okay.

00:45 X squared over x multiplied by e to the x y plus a function depending on x only.

00:56 Let me call it g of x, g of x.

01:00 Okay.

01:01 Perfect.

01:02 Now, let's determine g of x.

01:06 Well, to determine g of x, we just need to observe that the partial derivative of f with respect to x must be equal to the x coordinate of our vector field.

01:18 So, 1 plus xy multiplied by e to the xy.

01:25 Okay, let's differentiate this one with respect to x.

01:29 Before we do this, let me simplify.

01:33 This one is x.

01:35 Okay, now differentiating what do we get? we get the derivative of this one with respect to x is e to the x y plus x y e to the x y plus the derivative of g with respect to x, g prime of x...

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