MATH SOLVE

4 months ago

Q:
# Consider f and c below. f(x, y) = x2 i + y2 j c is the arc of the parabola y = 4x2 from (−1, 4) to (0, 0) (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're looking for a scalar function [tex]f(x,y)[/tex] such that

[tex]\nabla f(x,y)=\dfrac{\partial f}{\partial x}\,\mathbf i+\dfrac{\partial f}{\partial y}\,\mathbf j=x^2\,\mathbf i+y^2\,\mathbf j=\mathbf f(x,y)[/tex]

We have

[tex]\dfrac{\partial f}{\partial x}=x^2\implies f(x,y)=\dfrac{x^3}3+g(y)[/tex]

[tex]\implies\dfrac{\partial f}{\partial y}=\dfrac\partial{\partial y}\left[\dfrac{x^3}3+g(y)\right][/tex]

[tex]\implies y^2=\dfrac{\mathrm dg}{\mathrm dy}[/tex]

[tex]\implies g(y)=\dfrac{y^3}+C[/tex]

So

[tex]f(x,y)=\dfrac{x^3+y^3}3+C[/tex]

By the fundamental theorem of calculus (for line integrals), we then have

[tex]\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{\mathcal C}\nabla f\cdot\mathrm d\mathbf r=f(0,0)-f(-1,4)=0-21=-21[/tex]

[tex]\nabla f(x,y)=\dfrac{\partial f}{\partial x}\,\mathbf i+\dfrac{\partial f}{\partial y}\,\mathbf j=x^2\,\mathbf i+y^2\,\mathbf j=\mathbf f(x,y)[/tex]

We have

[tex]\dfrac{\partial f}{\partial x}=x^2\implies f(x,y)=\dfrac{x^3}3+g(y)[/tex]

[tex]\implies\dfrac{\partial f}{\partial y}=\dfrac\partial{\partial y}\left[\dfrac{x^3}3+g(y)\right][/tex]

[tex]\implies y^2=\dfrac{\mathrm dg}{\mathrm dy}[/tex]

[tex]\implies g(y)=\dfrac{y^3}+C[/tex]

So

[tex]f(x,y)=\dfrac{x^3+y^3}3+C[/tex]

By the fundamental theorem of calculus (for line integrals), we then have

[tex]\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{\mathcal C}\nabla f\cdot\mathrm d\mathbf r=f(0,0)-f(-1,4)=0-21=-21[/tex]