Question 9.8.6: Find the work done by (a) F =xi+ yj and (b) F = 3/4 i + 1/2 ......
Find the work done by (a) \mathrm{F}=x \mathrm{i}+y \mathrm{j} and (b) \mathrm{F}=\frac{3}{4} \mathrm{i}+\frac{1}{2} \mathrm{j} along the curve C traced by \mathrm{r}(t)=\cos t \mathrm{i}+\sin t \mathrm{j} from t=0 to t=\pi.
(a) The vector function \mathrm{r}(t) gives the parametric equations x=\cos t, y=\sin t, 0 \leq t \leq \pi, which we recognize as a half-circle. As seen in FIGURE 9.8.9, the force field \mathrm{F} is perpendicular to C at every point. Since the tangential components of \mathrm{F} are zero, the work done along C is zero. To see this we use (12):
W=\int_C P(x, y) d x+Q(x, y) d y \quad \text { or } \quad W=\int_C F \cdot d r . (12)
\begin{aligned} W=\int_{C} \mathrm{F} \cdot d \mathrm{r} & =\int_{C}(x \mathrm{i}+y \mathrm{j}) \cdot d \mathrm{r} \\ & =\int_{0}^{\pi}(\cos t \mathrm{i}+\sin t \mathrm{j}) \cdot(-\sin t \mathrm{i}+\cos t \mathrm{j}) d t \\ & =\int_{0}^{\pi}(-\cos t \sin t+\sin t \cos t) d t=0 .\end{aligned}
(b) In FIGURE 9.8.10 the vectors in red are the projections of \mathrm{F} on the unit tangent vectors. The work done by \mathrm{F} is
\begin{aligned} W=\int_{C} \mathrm{F} \cdot d \mathrm{r} & =\int_{C}\left(\frac{3}{4} \mathrm{i}+\frac{1}{2} \mathrm{j}\right) \cdot d \mathrm{r} \\ & =\int_{0}^{\pi}\left(\frac{3}{4} \mathrm{i}+\frac{1}{2} \mathrm{j}\right) \cdot(-\sin t \mathrm{i}+\cos t \mathrm{j}) d t \\ & =\int_{0}^{\pi}\left(-\frac{3}{4} \sin t+\frac{1}{2} \cos t\right) d t \\ & \left.=\left(\frac{3}{4} \cos t+\frac{1}{2} \sin t\right)\right]_{0}^{\pi}=-\frac{3}{2} . \end{aligned}
The units of work depend on the units of \|\mathrm{F}\| and on the units of distance.
