Find the work done by the force field $F(x,y) = -xi + 6yj$ along the path $C:y = x^3$ from $(0,0)$ to $(6,216)$
I tried parameterizing C which gave me $x(t) = t$ and $y(t) = t^3$ but do I use those two to find $ds$ or do I use the original function $F(x,y)$?
Any help is appreciated!
$\endgroup$ 12 Answers
$\begingroup$Use the definition of a Line Integral for Vector Fields ($\mathrm ds$ is for integrals with respect to arc length).
HINT: In plain words, denote and evaluate $F$ at $\alpha$ (substitute, take the dot product and integrate!)
SPOILER!
$\endgroup$ 2 $\begingroup$$W=\displaystyle\int_0^6[-t,6t^3]\cdot[1,3t^2]\,\mathrm dt$
$\alpha:(0,6)\rightarrow C\in\mathbb{R}^2,\,t\mapsto[t,t^3]$
$dr=x(t) \hat{\imath}+y(t) \hat{\jmath}$
$$\text{ Work }=\int_C F \cdot dr=\int_0^6 F(t) \cdot \frac{dr}{dt}dt= \int_0^6 (-t \hat{\imath}+6t^3 \hat{\jmath}) \cdot (\hat{\imath}+3t^2 \hat{\jmath})dt=\int_0^6 (-t+18t^5)dt=\left [-\frac{t^2}{2}+3t^6 \right ]_0^6$$
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