# Work and Line Integral

A two-dimensional force field is given by the equation $$f(x,y)=cxy\textbf{i}+x^6y^2\textbf{j}$$, where $c$ is a positive constant. This force acts on a particle which must move from $(0,0)$ to the line $x=1$ along a curve of the form $$y=ax^b$$ where $a>0$ and $b>0$

Find a value of a (in terms of c) such that the work done by this force is independent of b.

I started by expressing the parametric equations of the path namely: $x(t)=t$ and $y(t)=at^b$

Then we have: $\int\limits_{0}^1cxydx+x^6y^2dy=\int\limits_{0}^1(a^3bt^{3b+5}+act^{b+1})dt=\frac{a^3b}{3b+6}+\frac{ac}{b+2}$

From here I am not sure how to continue.

Let $g(b)=\frac{a^3b}{3b+6}+\frac{ac}{b+2}$. If $g$ is independent of $b$, then $g(b)$ must be constant. Thus, $$0=g'(b)=\frac{2a^3-3ac}{3(b+2)^2}$$ so that $a=\sqrt{\frac{3c}{2}}$.