Mathematical Immaturity

14.13 Exercises

• A particle moves with position vector $$\mathbf{r}(t) = t\mathbf{A} + t^2\mathbf{B} + 2\left(\frac{2}{3}t\right)^{3/2}\mathbf{A} \times \mathbf{B}$$ where $\mathbf{A}$ and $\mathbf{B}$ are two fixed unit vectors making an angle of $\pi/3$ radians with each other. Compute the speed of the particle at time $t$ and find how long it takes for it to move a distance of 12 units of arc length from the initial position $\mathbf{r}(0).$
• In the calculation of $v(t),$ remember that $A \cdot A = 1,$ $B \cdot B = 1,$ $A \cdot (A \times B) = O,$ and $B \cdot (A \times B) = O.$ You can also use Lagrange's identity for the norm of a cross product: $$ \|A \times B \|^2 = \|A\|^2\|B\|^2 - (A \cdot B)^2 $$
• Taking the derivative of $\mathbf{r}$ with respect to $t,$ we get: $$ \begin{align*} \mathbf{v}(t) &= A + 2tB + 3\left(\frac{2}{3}\right)^{3/2}\sqrt{t}\,(A \times B) \end{align*} $$ Speed $v(t)$ is then: $$ \begin{align*} v(t) &= \left\|A + 2tB + 3\left(\frac{2}{3}\right)^{3/2}\sqrt{t}\,(A \times B)\right\| \\ &= \left[\left(A \cdot A + 2t A\cdot B\right) + \left(2t B \cdot A + 4t^2 B \cdot B\right) + \frac{8}{3}t\,\|A \times B\|^2 \right]^{1/2} \\ &= \left[1 + 4t(A\cdot B) + 4t^2 + \frac{8}{3}t\left[1 - \left(A \cdot B\right)^2\right]\right]^{1/2} \\ \end{align*} $$ Noting that the angle between $A$ and $B$ is $\pi/3,$ and that the lengths of $A$ and $B$ are both $1:$ $$ \begin{align*} v(t) &= \left[1 + 4t\cos\frac{\pi}{3} + 4t^2 + \frac{8}{3}t\left(1 - \cos^2\frac{\pi}{3}\right)\right]^{1/2} \\ &= \left[1 + 2t + 4t^2 + \frac{8}{3}t\left(1 - \frac{1}{4}\right)\right]^{1/2} \\ &= \sqrt{1 + 4t + 4t^2} \\ &= 1 + 2t \end{align*} $$ To find how long it takes for the particle to move 12 units of distance from the initial position $\mathbf{r}(0),$ we find $t_0$ such that the arc length traced out by $t_0$ has length $12.$ In other words, we wish to find $t_0$ such that: $$ \begin{align*} \int_0^{t_0} 1 + 2t\,dt= 12 \end{align*} $$ Evaluating the integral, we see that $t_0 + t_0^2 = 12,$ which is satisfied by $t_0 = 3$ and $t_0 = -4.$ Assuming we do not travel backwards in time, this means that it takes $3$ units of time for the particle to move $12$ units of length from its initial position. $\quad \blacksquare$