Mathematical Immaturity

14.19 Exercises

• A particle moves in a plane so that its position at time $t$ has polar coordinates $r = t,$ $\theta = t.$ Find formulas for the velocity $\mathbf{v},$ the acceleration $\mathbf{a},$ and the curvature $\kappa$ at any time $t.$
• Recall from Section 14.16 that position $\mathbf{r}$ can be parameterized by $\theta$ using the equation $$ \begin{align*} \\ \mathbf{r} &= r\mathbf{u}_r, \quad \text{where} \quad \mathbf{u}_r = \cos\theta\,\mathbf{i} + \sin\theta\,\mathbf{j} \end{align*} $$ We also define the unit vector $\mathbf{u}_\theta\, ,$ perpendicular to $\mathbf{u}_r\, ,$ as $$ \begin{align*} \\ \mathbf{u}_\theta &= \frac{d\mathbf{u}_r}{d\theta} = -\sin\theta\,\mathbf{i} + \cos\theta\,\mathbf{j} \end{align*} $$ Then, velocity with respect to $t$ is given by: $$ \begin{align*} \\ \mathbf{v} &= \frac{d\mathbf{r}}{dt} = \frac{d(r\mathbf{u}_r)}{dt} = \frac{dr}{dt}\mathbf{u}_r + r\frac{d\mathbf{u}_r}{dt}, \end{align*} $$ Using the chain rule, we can write $\frac{d\mathbf{u}_r}{dt}$ as: $$ \begin{align*} \\ \frac{d\mathbf{u}_r}{dt} &= \frac{d\mathbf{u}_r}{d\theta} \frac{d\theta}{dt} = \mathbf{u}_\theta \frac{d\theta}{dt} \end{align*} $$ Thus, we can express $\mathbf{v}$ in terms of the perpendicular unit vectors $\mathbf{u}_r$ and $\mathbf{u}_\theta$ $$ \begin{align*} \\ \mathbf{v} &= \frac{dr}{dt}\mathbf{u}_r + r\frac{d\theta}{dt}\mathbf{u}_\theta \end{align*} $$ Where the factors multiplying $\mathbf{u}_r$ and $\mathbf{u}_\theta$ are called the radial and transverse components of velocity, respectively. Acceleration can also be expressed in terms of its radial and transverse components: $$ \begin{align*} \\ \mathbf{a} &= \left(\frac{d^{2}r}{dt^2} - r\left(\frac{d\theta}{dt}\right)^2\right)\mathbf{u}_r + \left(r\frac{d^{2}\theta}{dt^2} + 2\frac{dr}{dt}\frac{d\theta}{dt}\right)\mathbf{u}_\theta \end{align*} $$ What happens to the second derivatives of $r$ and $\theta$ with respect to $t$ when $r = \theta = t?$
• Let $\mathbf{u}_r$ and $\mathbf{u}_\theta$ be the perpendicular unit vectors corresponding to the radial and transverse components of motion: $$ \begin{align*} \\ \mathbf{u}_r &= \cos\theta\,\mathbf{i} + \sin\theta\,\mathbf{j} \\ \\ \mathbf{u}_\theta &= \frac{d\mathbf{u}_r}{d\theta} = -\sin\theta\,\mathbf{i} + \cos\theta\,\mathbf{j} \end{align*} $$ Then, radial position is defined by $\mathbf{r} = r\mathbf{u}_r.$ Using equations (14.25) and (14.26), we can express velocity and acceleration in terms of their radial and transverse components as follows: $$ \begin{align*} \\ \mathbf{v} &= \frac{dr}{dt}\mathbf{u}_r + r\frac{d\theta}{dt}\mathbf{u}_\theta \\ \\ &= \mathbf{u}_r + r\,\mathbf{u}_\theta \\ \\ \mathbf{a} &= \left(\frac{d^{2}r}{dt^2} - r\left(\frac{d\theta}{dt}\right)^2\right)\mathbf{u}_r + \left(r\frac{d^{2}\theta}{dt^2} + 2\frac{dr}{dt}\frac{d\theta}{dt}\right)\mathbf{u}_\theta \\ \\ &= -r\,\mathbf{u}_r + 2\,\mathbf{u}_\theta \end{align*} $$ The curvature of the curve described by $\mathbf{r}$ can be obtained by using equation (14.22): $$ \begin{align*} \\ \kappa(t) &= \frac{\|\mathbf{a}\times \mathbf{v}\|}{v^3(t)} \end{align*} $$ Using Lagrange's identity, noting that $\mathbf{u}_r$ and $\mathbf{u}_\theta$ are perpendicular unit vectors, we obtain $$ \begin{align*} \\ \|\mathbf{a}\times \mathbf{v}\| &= \sqrt{\|\mathbf{a}\|^2\|\mathbf{v}\|^2 - \left(\mathbf{a} \cdot \mathbf{v}\right)^2} \\ &= \sqrt{(r^2 + 4)(r^2 + 1) - r^2} \\ &= \sqrt{r^4 + 4r^2 + 4} \\ &= r^2 + 2 \end{align*} $$ In the above derivation, we also see that $\|\mathbf{v}\|^2 = r^2 + 1.$ Combining these results, we find that curvature is: $$ \begin{align*} \\ \kappa &= \frac{r^2 + 2}{\left(r^2 + 1\right)^{3/2}} \end{align*} $$ Noting that $r(t) = t$ and $\theta(t) = t$ for all $t,$ we can express velocity, acceleration, and curvature in terms of $t$ to give us: $$ \begin{align*} \mathbf{v}(t) &= \mathbf{u}_r + t\,\mathbf{u}_\theta \\ \\ \mathbf{a}(t) &= -t\,\mathbf{u}_r + 2\,\mathbf{u}_\theta \\ \\ \kappa(t) &= \frac{t^2 + 2}{\left(t^2 + 1\right)^{3/2}} \quad \blacksquare \end{align*} $$