Mathematical Immaturity

14.9 Exercises

• Given two fixed nonzero vectors $A$ and $B$ making an angle $\theta$ with each other, where $0 < \theta < \pi.$ A motion with position vector $\mathbf{r}(t)$ at time $t$ satisfies the differential equation $\mathbf{r}'(t) = A \times \mathbf{r}(t)$ and the initial condition $\mathbf{r}(0) = B.$ (a) Prove that the acceleration $\mathbf{a}(t)$ is orthogonal to $A.$ (b) Prove that the speed is constant and compute this speed in terms of $A,$ $B,$ and $\theta.$ (c) Make a sketch of the curve, showing its relation to the vectors $A$ and $B.$
• (a) What is the derivative of $\mathbf{r}'(t)?$ What do we know about the dot product of a cross product with its components $(\text{ie},$ what is $A \cdot (A \times B))?$ (b) Use the result of part (a) using the velocity vector $\mathbf{r}'(t)$ with the component definition of acceleration: $$ \mathbf{a}(t) = v'(t)T(t) + v(t)T'(t) $$ recalling that $T(t) = \mathbf{v}(t)/v(t).$
• (a) We know that the derivative of the cross product of two vectors $F,$ $G$ is: $$ (F \times G)' = F' \times G + F \times G' $$ Applying this to $\mathbf{r}'(t) = A \times \mathbf{r}(t),$ noting that $A$ is a constant vector, we get $$ \begin{align*} \mathbf{a}(t) &= A' \times \mathbf{r}(t) + A \times \mathbf{r}'(t) \\ &= A \times \mathbf{r}'(t) \end{align*} $$ But we know from Theorem 13.12 that the cross product of two vectors is orthogonal to those same two vectors, thus $A$ and $\mathbf{a}(t)$ are orthogonal for all $t.$ (b) From part (a), we know that $\mathbf{a}(t) = A \times \mathbf{r}'(t).$ Now, we write $\mathbf{a}(t)$ as the linear combination of its components: $$ \mathbf{a}(t) = r''(t)T(t) + r'(t)T'(t) $$ where $r'(t) = \|\mathbf{r}'(t)\|$ is the speed and $r''(t) = (r'(t))'$ is the derivative of the speed. Using the result of part (a), we know that $\mathbf{a}(t) \cdot \mathbf{r}'(t) = 0.$ Applying this to the above equation for acceleration, we get: $$ \begin{align*} \mathbf{a}(t) \cdot \mathbf{r}'(t) &= r''(t)T(t)\cdot \mathbf{r}'(t) + r'(t)T'(t) \cdot \mathbf{r}'(t) \\ &= 0 \end{align*} $$ But we know that the unit tangent $T(t) = \mathbf{r}'(t)/\|\mathbf{r}'(t)\|,$ and that $T'(t) \cdot \mathbf{r}'(t) = 0$ since the principal normal is orthogonal to any scalar multiple of the unit tangent vector. As such, the above dot product simplifies to: $$ r''(t)r'(t) = 0 $$ for all $t.$ Then, two possible cases arise. (1) $r'(t) = 0$ for all $t,$ in which case position is constant, with $A$ and $B$ being linearly dependent. (2) $r'(t) \neq 0 $ for all $t,$ in which case $r''(t) = 0,$ implying that speed is constant for all $t,$ with $$ \begin{align*} r'(t) &= \|\mathbf{r}'(t)\| \\ &= \|A \times \mathbf{r}(t)\| \\ &= \|A\|\|\mathbf{r}(t)\|\sin \theta_t \\ &= k \end{align*} $$ where $\theta_t$ is the angle between $A$ and $\mathbf{r}(t)$ at time $t,$ and where $k$ is some real constant. But if $r'(t)$ is constant for all $t,$ then for any given $t,$ $r'(t) = r'(0),$ or $$ r'(t) = \|A\|\|B\|\sin \theta $$ for all $t.$