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F(x, y) = x2 i + y2 j c is the arc of the parabola y = 2x2 from (1, 2) to (2, 8) (a) find a function f such that f = ∇f.

1 Answer

6 votes

\mathbf F(x,y)=x^2\,\mathbf i+y^2\,\mathbf j

If a scalar function
f(x,y) exists for which
\\abla f(x,y)=\mathbf F(x,y), then we'd have


(\partial f)/(\partial x)=x^2\implies f(x,y)=\frac{x^3}3+g(y)

\implies(\partial f)/(\partial y)=y^2=g'(y)\implies g(y)=\frac{y^3}3+C

so the scalar function would be given by


f(x,y)=\frac{x^3+y^3}3+C

where
C is an arbitrary constant.

Presumably, this question is being asked in the context of the line integral of
\mathbf F(x,y) along the given contour
C. Since
\mathbf F(x,y) is conservative - and consequently a scalar potential function
f(x,y) exists - we can simply use the gradient theorem to evaluate the line integral. We would get


\displaystyle\int_C\mathbf F\cdot\mathrm d\mathbf r=f(2,8)-f(1,2)=\frac{511}3

(and we get the same answer by parameterizing
C and computing the integral as we usually would)
User Yochi
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