Consider f and c below. f(x, y) = (3 + 4xy2)i + 4x2yj, c is the arc of the hyperbola y = 1/x from (1, 1) to 3, 1 3 (a) find a function f such that f = ∇f. f(x, y) = (b) use part (a) to evaluate c f · dr along the given curvec.
Accepted Solution
A:
We want to find a scalar function [tex]f(x,y)[/tex] such that [tex]\nabla f(x,y)=\mathbf f(x,y)=\dfrac{\partial f}{\partial x}\,\mathbf i+\dfrac{\partial f}{\partial y}\,\mathbf j[/tex]. So we need to have [tex]\dfrac{\partial f}{\partial x}=3+4xy^2[/tex] Integrating both sides with respect to [tex]x[/tex] gives [tex]f(x,y)=3x+2x^2y^2+g(y)[/tex] Differentiating with respect to [tex]y[/tex] gives [tex]\dfrac{\partial f}{\partial y}=4x^2y+\dfrac{\mathrm dg}{\mathrm dy}=4x^2y[/tex][tex]\implies\dfrac{\mathrm dg}{\mathrm dy}=0[/tex][tex]\implies g(y)=C[/tex] So we find that [tex]f(x,y)=3x+2x^2y^2+C[/tex] By the fundamental theorem of calculus, we then know the line integral depends only on the values of [tex]f(x,y)[/tex] at the endpoints of the path. Therefore