Question 9.2.2: Suppose the vector function in Example 2 of Section 9 .1 rep......
Suppose the vector function in Example 2 of Section 9.1 represents the position of a particle moving in a circular orbit. Graph the velocity and acceleration vector at t=\pi / 4.
Recall that \mathrm{r}(t)=2 \cos t \mathrm{i}+2 \sin t \mathrm{j}+3 \mathrm{k} is the position vector of a particle moving in a circular orbit of radius 2 in the plane z=3. When t=\pi / 4 the particle is at the point P(\sqrt{2}, \sqrt{2}, 3). Now,
\begin{aligned} \mathrm{v}(t) & =\mathrm{r}^{\prime}(t)=-2 \sin t \mathrm{i}+2 \cos t \mathrm{j} \\ \mathrm{a}(t) & =\mathrm{r}^{\prime \prime}(t)=-2 \cos t \mathrm{i}-2 \sin t \mathrm{j} .\end{aligned}
Since the speed is \|\mathrm{v}(t)\|=2 for all time t, it follows from the discussion preceding this example that \mathrm{a}(t) is perpendicular to \mathrm{v}(t). (Verify this.) As shown in FIGURE 9.2.2, the vectors
\begin{aligned} & \mathrm{v}\left(\frac{\pi}{4} \right)=-2 \sin \frac{\pi}{4} \mathrm{i}+2 \cos \frac{\pi}{4} \mathrm{j}=-\sqrt{2} \mathrm{i}+\sqrt{2} \mathrm{j} \\ & \mathrm{a}\left(\frac{\pi}{4}\right)=-2 \cos \frac{\pi}{4} \mathrm{i}-2 \sin \frac{\pi}{4} \mathrm{j}=-\sqrt{2} \mathrm{i}-\sqrt{2} \mathrm{j} \end{aligned}
are drawn at the point P. The vector \mathrm{v}(\pi / 4) is tangent to the circular path and \mathrm{a}(\pi / 4) points along a radius toward the center of the circle.
