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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!

Coop
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  • Possibly by definition $\displaystyle \int _CF\mid \mathrm d \vec r=\int \limits _a^b\left \langle F(\gamma (t)),\gamma'(t)\right\rangle\mathrm dt$, where $\gamma \colon [a,b]\to \mathbb R^2$ is a parametrization of $C$. – Git Gud Jul 20 '14 at 22:41

2 Answers2

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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!

$W=\displaystyle\int_0^6[-t,6t^3]\cdot[1,3t^2]\,\mathrm dt$
$\alpha:(0,6)\rightarrow C\in\mathbb{R}^2,\,t\mapsto[t,t^3]$

Daniel Charry
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  • Thank you! Following your hint, I ended up with $int_0^6[-1/2t+18t^5 dt]$ which gave me 13,9950 when evaluated. That seems like a very large value for the work done. Did I make a mistake somewhere? – Coop Jul 20 '14 at 22:54
  • Do not worry, it is right, for the vector field given and the curve you are taking. – Daniel Charry Jul 20 '14 at 22:57
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$dr=x(t) \hat{\imath}+y(t) \hat{\jmath}$

$$\text{ Work }=\int_C F \cdot dr=\int_0^6 F(t) \cdot \frac{dr}{dt}dt= \int_0^6 (-t \hat{\imath}+6t^3 \hat{\jmath}) \cdot (\hat{\imath}+3t^2 \hat{\jmath})dt=\int_0^6 (-t+18t^5)dt=\left [-\frac{t^2}{2}+3t^6 \right ]_0^6$$

Mary Star
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