00:01
We're given a vector field big f and a curve c.
00:06
In part a, we're asked to find a function little f, such that big f is equal to the gradient of little f.
00:17
Well, this means that the partial derivative of f with respect to y, well, this should be equal to the component, the y component of big f, which is x squared e to the xy.
00:34
I guess i should say first big f is the vector field 1 plus xy times e to the xy i plus x squared times e to the xy j, and c is the curve vector equation r of t equals cosine t plus sine t, 2 sine t j, or t ranges from 0 to pi over 2.
01:01
So back to part a.
01:04
Well, this implies that our function f, taking the antiderivative with respect to y should be equal to x times e to the x y plus some function of g i'm sorry plus g some function of x in this case and therefore the partial derivative of f with respect to x is equal to by the product rule.
01:44
This is x times y times e to the x y plus e to the xy plus g prime of x but this is and this can of course also be written as one plus x y times e to the xy plus g prime of x but we also know that this is equal to the y component of big f which was sorry the x component of big f which is 1 plus xy, e to the xy...