Mathematical Immaturity

14.15 Exercises

• A particle moves along a plane curve with constant speed 5. It starts at the origin at time $t = 0$ with initial velocity $5\,\mathbf{j},$ and it never goes to the left of the $y$-axis. At every instant the curvature of the path is $\kappa(t) = 2t.$ Let $\alpha(t)$ denote the angle that the velocity vector makes with the positive $x$-axis at time $t.$ (a) Determine $\alpha(t)$ explicitly as a function of $t.$ (b) Determine the velocity $\mathbf{v}(t)$ in terms of $\,\mathbf{i}$ and $\,\mathbf{j}.$
• (a) Refer to Example 2 of Section 14.14, "Curvature of a plane curve", to review the relation between curvature $\kappa(t)$ and the angle $\alpha(t)$ between the velocity vector and the positive $x$-axis. (b) Recall from Section 14.8 that the unit tangent $T(t)$ can be expressed in terms of the angle of inclination as follows: $$ \begin{align*} \\ T(t) &= \cos\alpha(t)\mathbf{i} + \sin\alpha(t)\mathbf{j} \end{align*} $$ Then, we can multiply the unit tangent by the speed to obtain the velocity in terms of $\mathbf{i}$ and $\mathbf{j}.$
• (a) As discussed in Example 2 of Section 14.14, the angle of inclination $\alpha(t)$ between the tangent vector and the positive $x$-axis is related to curvature as follows: $$ \begin{align*} \\ \kappa(t) &= \left|\frac{d\alpha}{ds}\right| \end{align*} $$ Using the chain rule, noting that $v(t) = ds/dt,$ we can deduce that $$ \begin{align*} \\ |\alpha'(t)| &= \left|\frac{d\alpha}{dt}\right| \\ &= \left|\frac{d\alpha}{ds}\right|\frac{ds}{dt} \\ &= \left|\frac{d\alpha}{ds}\right|v(t) \\ &= \kappa(t)v(t) \end{align*} $$ Given that the velocity has a starting angle of $\pi/2$ and that the particle never crosses the $y$-axis, this implies that the angle of inclination always moves towards $0.$ In other words, $\alpha'(t) = -\kappa(t)v(t).$ Plugging in our values for speed and curvature, and noting that the starting angle is $\pi/2,$ we can express the angle of inclination explicitly as a function of $t$ by integrating its derivative from $0$ to $t$ $$ \begin{align*} \\ \alpha(t) &= \int_0^t \alpha'(u)\,du + \frac{\pi}{2} \\ \\ &=\frac{\pi}{2} - \int_0^t 10u\,du \\ \\ &= \frac{\pi}{2} + 5u^2\Biggr|_t^0 \\ &= \frac{\pi}{2} -5t^2 \quad \blacksquare \end{align*} $$ (b) Recall from Section 14.8 that the unit tangent $T(t)$ can be expressed in terms of the angle of inclination as follows: $$ \begin{align*} \\ T(t) &= \cos\alpha(t)\mathbf{i} + \sin\alpha(t)\mathbf{j} \end{align*} $$ In other words: $$ \begin{align*} \\ T(t) &= \cos \left(\frac{\pi}{2} -5t^2\right)\mathbf{i} + \sin\left(\frac{\pi}{2} -5t^2\right)\mathbf{j} \end{align*} $$ Applying the trigonometric identities $$ \begin{align*} \\ \cos\left(x + \frac{\pi}{2}\right) &= -\sin x \\ \sin\left(x + \frac{\pi}{2}\right) &= \cos x \\ \cos(-x) &= \cos x \\ \sin(-x) &= -\sin x \end{align*} $$ The unit tangent becomes $$ \begin{align*} \\ T(t) &= \cos \left(\frac{\pi}{2} -5t^2\right)\mathbf{i} + \sin\left(\frac{\pi}{2} -5t^2\right)\mathbf{j} \\ &= -\sin \left(-5t^2\right)\mathbf{i} + \cos\left(-5t^2\right)\mathbf{j} \\ &= \sin \left(5t^2\right)\mathbf{i} + \cos\left(5t^2\right)\mathbf{j} \end{align*} $$ Multiplying by speed gives us the velocity in terms of $\mathbf{i}$ and $\mathbf{j}$ $$ \begin{align*} \\ \mathbf{v}(t) &= 5\sin \left(5t^2\right)\mathbf{i} + 5\cos\left(5t^2\right)\mathbf{j} \quad \blacksquare \end{align*} $$