- Calculus, Volume 1: One Variable Calculus, with an Introduction to Linear Algebra
- Tom M. Apostol
- Second Edition
- 1967
- 978-1-119-49673-1
14.9 Exercises
-
For each of the following statements about a curve traced out by a particle moving in 3-space, either give a proof or exhibit a counter example.
(a) If the velocity is constant, the curve lies in a plane.
(b) If the speed is constant, the curve lies in a plane.
(c) If the acceleration is constant, the curve lies in a plane.
(d) If the velocity is perpendicular to the acceleration, the curve lies in a plane.
-
(a) Proof.$\quad$ If velocity is constant, then position $\mathbf{r}(t)$ can be described as the integral of $\mathbf{v}(x)$ from $x = 0$ to $x = t.$ Let $\mathbf{v}(x) = A,$ where $A$ is a constant vector, and let $\mathbf{r}(0) = P.$ Then $\mathbf{r}(t)$ is defined as:
$$
\begin{align*}
\mathbf{r}(t) &= \int_0^t A\,dx + \mathbf{r}(0)\\
&= P + At
\end{align*}
$$
for $t > 0.$ Thus, we have shown that if velocity is constant, not only does the curve lie in a plane, but the curve is in fact a line.
\begin{align*}
\end{align*}
(b) Counterexample.$\quad$ Take, for example, helical motion, whose velocity is
$$
-\omega a \sin\, \omega t\, \mathbf{i} + \omega a \cos\, \omega t\, \mathbf{j} + \omega b\, \mathbf{k}
$$
and whose speed is a constant $|\omega|\sqrt{a^2 + b^2},$ but whose curve does not lie in a plane.
\begin{align*}
\end{align*}
(c) Proof.$\quad$ Let $\mathbf{a}(t) = A$ for all $t,$ then $\mathbf{v}(t) = At + \mathbf{v}(0)$ and $\mathbf{r}(t) = \frac{t^2}{2}A + tB + C$ where $B = \mathbf{v}(0)$ and $C = \mathbf{r}(0).$ But every point on $\mathbf{r}(t)$ is a member of some set:
$$
\{C + uB + vA \mid u, v\ \text{real}\}
$$
which is, by definition, a plane.
\begin{align*}
\end{align*}
(d) Counterexample.$\quad$ Consider again helical motion, whose velocity is
$$
\mathbf{v}(t) = -\omega a \sin\, \omega t\, \mathbf{i} + \omega a \cos\, \omega t\, \mathbf{j} + \omega b\, \mathbf{k}
$$
and whose acceleration is
$$
\mathbf{a}(t) = -\omega^2 a \cos\, \omega t\, \mathbf{i} - \omega^2 a \sin\, \omega t\, \mathbf{j}
$$
The dot product $\mathbf{v} \cdot \mathbf{a}$ is zero, but the motion does not lie within a plane.
