Work Done by a Vector Field Along a Helical Path — Conservative-Field Shortcut: 04-BS-1, May 2015, Q4

Question 4: Work Done by a Vector Field Along a Helical Path — Conservative-Field Shortcut (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. Field $\mathbf F(x,y,z)=(x^2,\,y,\,-z)$; helical path $x=6t,\ y=2\cos t,\ z=2\sin t$; endpoints $(0,2,0)$ and $(3\pi,0,2)$.

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

Approach. Each component of $\mathbf F$ depends on only one variable — $x^2$ on $x$ alone, $y$ on $y$ alone, $-z$ on $z$ alone — so $\mathbf F$ is a gradient field, $\mathbf F=\nabla\varphi$, and the work is path-independent: $W=\varphi(\text{end})-\varphi(\text{start})$. This is confirmed by directly integrating along the given parametrization as a cross-check.

Locate the endpoints on the path. At $t=0$: $(x,y,z)=(0,2,0)$ — the start point. At $t=\pi/2$: $(x,y,z)=(6\cdot\tfrac\pi2,\,2\cos\tfrac\pi2,\,2\sin\tfrac\pi2)=(3\pi,0,2)$ — the end point. So $t$ runs $0\to\pi/2$.
Recognize $\mathbf F$ as conservative and find a potential. Since $\partial(x^2)/\partial y=\partial(x^2)/\partial z=0$ (and similarly for the other components), a potential $\varphi$ with $\nabla\varphi=\mathbf F$ exists: $$\varphi(x,y,z)=\int x^2\,dx+\int y\,dy-\int z\,dz=\frac{x^3}{3}+\frac{y^2}{2}-\frac{z^2}{2}$$ (the three integrations are independent since each $\mathbf F$-component depends on a single variable, so no cross terms or "unknown function of the others" arise).
Evaluate $\varphi$ at the endpoints. $$\varphi(3\pi,0,2)=\frac{(3\pi)^3}{3}+0-\frac{4}{2}=9\pi^3-2,\qquad \varphi(0,2,0)=0+\frac{4}{2}-0=2.$$
Work as a potential difference. $$W=\varphi(\text{end})-\varphi(\text{start})=(9\pi^3-2)-2=\boxed{9\pi^3-4}.$$
Cross-check by direct parametrization. With $\mathbf r'(t)=(6,\,-2\sin t,\,2\cos t)$ and $\mathbf F(t)=(36t^2,\,2\cos t,\,-2\sin t)$, $$\mathbf F\cdot\mathbf r'=216t^2-4\sin t\cos t-4\sin t\cos t=216t^2-4\sin(2t).$$ $$W=\int_0^{\pi/2}\!\big(216t^2-4\sin2t\big)dt=\Big[72t^3+2\cos2t\Big]_0^{\pi/2}=\big(9\pi^3-2\big)-\big(0+2\big)=9\pi^3-4$$ — matches the potential-function result exactly.

Quantity	Result
Potential $\varphi(x,y,z)$	$\tfrac{x^3}{3}+\tfrac{y^2}{2}-\tfrac{z^2}{2}$
$\varphi(\text{end})-\varphi(\text{start})$	$(9\pi^3-2)-2$
Work $W$	$9\pi^3-4\approx275.06$

Concept & Theory

Topic. Work as a line integral, and the shortcut for conservative (gradient) vector fields.

Key relations. If $\mathbf F=\nabla\varphi$ on a domain containing $C$, then $\int_C\mathbf F\cdot d\mathbf r=\varphi(\text{end})-\varphi(\text{start})$, independent of the path's shape — only the endpoints matter. A field with each Cartesian component depending on only its own variable is always of this form.

Why this works. Path-independence follows from the fundamental theorem for line integrals: $\int_C\nabla\varphi\cdot d\mathbf r=\varphi(\mathbf r(b))-\varphi(\mathbf r(a))$ by the chain rule, exactly mirroring the single-variable fundamental theorem of calculus. Recognizing $\mathbf F$'s separable structure before touching the parametrization turns a two-line trigonometric integral into two function evaluations.

Common pitfall. Committing to the full parametrized integral (correct, but slower and more error-prone through the $\sin(2t)$ algebra) without first checking whether $\mathbf F$ is conservative — always test $\partial F_i/\partial x_j=\partial F_j/\partial x_i$ (or, as here, note each component depends on a single variable) before integrating along a curve.

Source. Stewart, Calculus: Early Transcendentals, Ch. 16 (§16.3 The Fundamental Theorem for Line Integrals).

Question 4 of 8: Work Done by a Vector Field Along a Helical Path — Conservative-Field Shortcut

Question 4: Work Done by a Vector Field Along a Helical Path — Conservative-Field Shortcut (20 marks)