MATH SOLVE

5 months ago

Q:
# 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

[tex]\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=f\left(3,\frac13\right)-f(1,1)=11-5=6[/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

[tex]\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=f\left(3,\frac13\right)-f(1,1)=11-5=6[/tex]