Consider F and C below. F(x, y) = x4y5i + x5y4j, C: r(t) = t3 − 2t, t3 + 2t , 0 ≤ t ≤ 1 (a) Find a function f such that F = ∇f. f(x, y) = (b) Use part (a) to evaluate C F · dr along the given curve C.

[tex]\mathbf f(x,y)=x^4y^5\,\mathbf i+x^5y^4\,\mathbf j[/tex]

We want a scalar function [tex]f(x,y)[/tex] whose gradient is equivalent to the vector field. That means

[tex]\dfrac{\partial f}{\partial x}=x^4y^5\implies f(x,y)=\dfrac{x^5y^5}5+g(y)[/tex]
[tex]\dfrac{\partial f}{\partial y}=x^5y^4\implies x^5y^4=x^5y^4+\dfrac{\mathrm dg}{\mathrm dy}[/tex]
[tex]\implies\dfrac{\mathrm dg}{\mathrm dy}=0\implies g(y)=C[/tex]

So (a)

[tex]f(x,y)=\dfrac{x^5y^5}5+C[/tex]

which means (b)

[tex]\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=f(\mathbf r(1))-f(\mathbf r(0))=f(-1,3)-f(0,0)=-\dfrac{243}5-0=-\dfrac{243}5[/tex]