Mathematical Immaturity

14.15 Exercises

• A particle moves along a plane curve with constant speed 2. The motion starts at the origin when $t = 0$ and the initial velocity $\mathbf{v}(0)$ is $2\,\mathbf{i}.$ At every instant it is known that the curvature $\kappa(t) = 4t.$ Find the velocity $\mathbf{v}(t)$ when $t = \frac{1}{4}\sqrt{\pi}$ if the curve never goes below the $x$-axis.
• Use the same setup as Exercise 11, but note this time that the starting angle is $0$ and that the particle moves away from the $x$-axis. What does this imply about the angle of inclination $\alpha(t)$ in relation to the curvature?
• 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 $0$ and that the particle never crosses the $x$-axis, this implies that the angle of inclination always moves towards $\pi/2.$ In other words, $\alpha'(t) = \kappa(t)v(t).$ Plugging in our values for speed and curvature, and noting that the starting angle is $0,$ we can express the angle of inclination explicitly as a function of $t$ by integrating its derivative from $0$ to $t$ $$ \begin{align*} \\ \alpha(t) &= \int_0^t \alpha'(u)\,du \\ \\ &=\int_0^t 8u\,du \\ \\ &= 4u^2\Biggr|_0^t \\ &= 4t^2 \end{align*} $$ Then, we 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: $$ \begin{align*} \\ T(t) &= \cos \left(4t^2\right)\mathbf{i} + \sin\left(4t^2\right)\mathbf{j} \end{align*} $$ Multiplying by speed gives us the velocity in terms of $\mathbf{i}$ and $\mathbf{j}$ $$ \begin{align*} \\ \mathbf{v}(t) &= 2\cos \left(4t^2\right)\mathbf{i} + 2\sin\left(4t^2\right)\mathbf{j} \end{align*} $$ Evaluating at time $t = \frac{1}{4}\sqrt{\pi}$ $$ \begin{align*} \\ \mathbf{v}\left(\frac{1}{4}\sqrt{\pi}\right) &= 2\cos \left(\frac{\pi}{4}\right)\mathbf{i} + 2\sin\left(\frac{\pi}{4}\right)\mathbf{j} \\ &= \sqrt{2}\,\mathbf{i} + \sqrt{2}\,\mathbf{j} \quad \blacksquare \end{align*} $$