Consider f and c below. f(x, y) = x2 i + y2 j c is the arc of the parabola y = 2x2 from (−1, 2) to (1, 2) (a) find a function f such that f = ∇f. f(x, y) = x3 3+ y3 3 correct: your answer is correct. (b) use part (a) to evaluate c ∇f · dr along the given curvec.
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A:
In this exercise we have to perform the parameterization of the given function and we will have:[tex]\int\limits_C {f} \, dr = f(1, 2)-f(-1, 2)= 2/3[/tex]In there is some scalar function [tex]f(x, y)[/tex] such that:[tex]\nabla f(x,y) = f(x, y) = x^2i+y^2j[/tex]Then we want to find [tex]f[/tex] such that:[tex]f'(x)= x^2= f(x,y) = \frac{x^3}{3} + g(y)\\f'(y)= y^2= f(x,y)= \frac{y^3}{3} + C\\f(x, y)= \frac{x^3}{3} +\frac{y^3}{3} + C[/tex]So the vector filed [tex]f(x, y)[/tex] is conservative, which means the fundamental theorem applies; the line integral of [tex]f[/tex] along any path [tex]C[/tex] parameterized by some vector-valued function [tex]r(t)[/tex] over [tex]a\leq t\leq b[/tex] is given by: [tex]\int\limits_C {f} \, dr\\\int\limits^t=a_t=b {f(r(t))} \, dt= f(r(b))-f(r(a))[/tex]In the case:[tex]\int\limits_C {f} \, dr = f(1, 2)-f(-1, 2)= 2/3[/tex]Learn more: brainly.com/question/14770282