Question 7 of 8: Work Done by a Conservative Field Along a Parametrized Path

Question 7: Work Done by a Conservative Field Along a Parametrized Path (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $\mathbf F=(x^2,\,y,\,-z)$; path $(6t,\,2\cos t,\,2\sin t)$ from $(0,2,0)$ to $(3\pi,0,2)$.

Find. The work $W=\int_C\mathbf F\cdot d\mathbf r$.

Approach. Each component of $\mathbf F$ depends only on its own variable, so $\mathbf F$ is conservative with an easily found potential $\varphi$; then $W=\varphi(\text{end})-\varphi(\text{start})$, avoiding the direct trigonometric line integral entirely.

1. Recognize $\mathbf F$ as conservative. Since $F_1=x^2$ depends only on $x$, $F_2=y$ only on $y$, $F_3=-z$ only on $z$, a potential is $$\varphi(x,y,z)=\frac{x^3}3+\frac{y^2}2-\frac{z^2}2,\qquad\text{since }\nabla\varphi=(x^2,\,y,\,-z)=\mathbf F.$$
2. Locate the parameter values. At the start $(0,2,0)$: $6t=0\Rightarrow t=0$ (and $2\cos0=2$, $2\sin0=0$ ✓). At the end $(3\pi,0,2)$: $6t=3\pi\Rightarrow t=\tfrac\pi2$ (and $2\cos\tfrac\pi2=0$, $2\sin\tfrac\pi2=2$ ✓).
3. Evaluate the potential at both endpoints. $$\varphi(0,2,0)=0+\frac{4}2-0=2,\qquad\varphi(3\pi,0,2)=\frac{(3\pi)^3}3+0-\frac{4}2=9\pi^3-2.$$
4. Work as the potential difference. $$W=\varphi(\text{end})-\varphi(\text{start})=(9\pi^3-2)-2=9\pi^3-4.$$

$$W=\boxed{9\pi^3-4}\approx275.06$$