Question

Consider f and c below. f(x, y) = x2 i + y2 j c is the arc of the parabola y = 2x^2 from (−1, 2) to (0, 0)

Answer 1

(a) If there is a scalar function f(x, y) such that ∇ f(x, y) = x² i + y² j, then

∂f/∂x = x²

∂f/∂y = y²

Integrating both of these equations with respect to x and y (respectively) gives

f(x, y) = 1/3 x³ + g(y)

f(x, y) = 1/3 y³ + h(x)

Differentiating with respect to the other variable gives

∂f/∂y = g'(y) = y²

∂f/∂x = h'(x) = x²

so it follows that

f(x, y) = 1/3 x³ + 1/3 y³ + C

for some constant C.

(b) By the gradient theorem,

$\displaystyle \int_C \nabla f \cdot d\vec r = f(0,0) - f(-1,2) = \boxed{-\dfrac73}$