Suppose \nabla f (x, y) = 3 y \sin(xy) \vec{i} + 3 x \sin(xy)\vec{j}, \vec{f} = \nabla f(x, y), and c is the segment of the parabola y = 3 x^2 from the point (1,3) to (4,48). then

I'll assume you're supposed to compute the line integral of over the given path . By the fundamental theorem of calculus,

so evaluating the integral is as simple as evaluting at the endpoints of . But first we need to determine given its gradient.

We have

Differentiating with respect to gives

and we end up with

for some constant . Then the value of the line integral is .

the thermometer is precise but not accurate

step-by-step explanation: