- Calculus, Volume 1: One Variable Calculus, with an Introduction to Linear Algebra
- Tom M. Apostol
- Second Edition
- 1967
- 978-1-119-49673-1
14.7 Exercises
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Let $\mathbf{c}$ be a fixed unit vector. A particle moves in space in such a way that its position vector $\textbf{r}(t)$ satisfies the equation $\textbf{r}(t) \cdot \mathbf{c} = e^{2t}$ for all $t,$ and its velocity vector $\textbf{v}(t)$ makes a constant angle $\theta$ with $\mathbf{c},$ where $0 \lt \theta \lt \frac{\pi}{2}.$
\begin{align*}
\end{align*}
(a) Prove that the speed at time $t$ is $2e^{2t}/\cos \theta.$
(b) Compute the dot product $\textbf{a}(t) \cdot \textbf{v}(t)$ in terms of $t$ and $\theta.$
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(a) Note that speed is the norm of the velocity vector, and that
$$\cos \theta = \frac{\mathbf{v}(t) \cdot \mathbf{c}}{\|\mathbf{v}(t)\|\|\mathbf{c}\|}$$
(b) Note that if we differentiate the dot product $\mathbf{v} \cdot \mathbf{v}$ we get: $$\displaylines{\left(\mathbf{v} \cdot \mathbf{v}\right)' = \mathbf{v}' \cdot \mathbf{v} + \mathbf{v} \cdot \mathbf{v}'\\ = 2\ \mathbf{a} \cdot \mathbf{v}}$$
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(a) Differentiating both sides of the equation $\mathbf{r}(t) \cdot \mathbf{c} = e^{2t},$ we get:
$$\mathbf{v}(t) \cdot \mathbf{c} = 2\ e^{2t}$$
Written another way,
$$\|\mathbf{v}(t)\|\|\mathbf{c}\|\cos \theta = 2\ e^{2t}$$
And since we know $\mathbf{c}$ is of unit length, speed $\|\mathbf{v}(t)\|$ is
$$\|\mathbf{v}(t)\| = \frac{2e^{2t}}{\cos \theta}\quad\blacksquare$$
(b) First, we note that $$\|\mathbf{v}(t)\|^2 = \mathbf{v}(t) \cdot \mathbf{v}(t) = \frac{4e^{4t}}{\cos^2 t}.$$Then, differentiating with respect to $t$ gives us:
$$\displaylines{2\,\mathbf{a}(t) \cdot \mathbf{v}(t) = \frac{16\,e^{4t}}{\cos^2 \theta}}$$
Or in other words:
$$\mathbf{a}(t) \cdot \mathbf{v}(t) = \frac{8\,e^{4t}}{\cos^2 \theta}\quad\blacksquare$$
