Mathematical Immaturity

14.19 Exercises

• If a curve has the polar equation $r = f(\theta),$ show that its radius of curvature $\rho$ is given by the formula \begin{align*} \\ \rho = \frac{(r^2 + r'^2)^{3/2}}{|r^2 + 2r'^2 - rr''|}, \end{align*} where $r' = f'(\theta)$ and $r'' = f''(\theta).$
• Refer to Section 14.16 where velocity, speed, and acceleration are simplified in the case where $t = \theta$ and $r = f(\theta).$ When calculating the cross product $\mathbf{a} \times \mathbf{v},$ you can handle the unit vectors $\mathbf{u}_r$ and $\mathbf{u}_{\theta}$ in the same way as $\mathbf{i}$ and $\mathbf{j}$ since they are perpendicular to eachother and to the unit vector $\mathbf{k}.$
• When $t = \theta,$ we can express velocity and acceleration in terms of their radial and transverse components as follows: \begin{align*} \\ \mathbf{v} &= \frac{dr}{d\theta}\mathbf{u}_r + r\mathbf{u}_{\theta} \\ \\ &= r'\mathbf{u}_r + r\mathbf{u}_{\theta} \\ \\ \mathbf{a} &= \left(\frac{d^2r}{d\theta^2} - r\right)\mathbf{u}_r + 2\frac{dr}{d\theta}\mathbf{u}_{\theta} \\ \\ &= \left(r'' - r\right)\mathbf{u}_r + 2r'\mathbf{u}_{\theta}. \end{align*} And as we saw in Exercise 4, we can express speed $v$ in terms of $f(\theta)$ as: \begin{align*} \\ v &= \sqrt{\left(\frac{dr}{d\theta}\right)^2 + r^2} \\ \\ &= \sqrt{\left(r'\right)^2 + r^2}. \end{align*} We can use Equation 14.22 to express curvature in terms of $\|\mathbf{a} \times \mathbf{v}\|$ and $v:$ \begin{align*} \\ \kappa &= \frac{\|\mathbf{a} \times \mathbf{v}\|}{v^3} \\ \\ &= \frac{\|\left(r''r - r^2 - 2r'^2\right)\mathbf{k}\|}{\left(r^2 + r'^{2}\right)^{3/2}} \\ \\ &= \frac{\left|r''r - r^2 - 2r'^2\right|}{\left(r^2 + r'^{2}\right)^{3/2}} \\ \\ &= \frac{\left|-\left(r^2 + 2r'^2 - rr''\right)\right|}{\left(r^2 + r'^{2}\right)^{3/2}} \\ \\ &= \frac{\left|r^2 + 2r'^2 - rr''\right|}{\left(r^2 + r'^{2}\right)^{3/2}}. \end{align*} By definition, for $\|\mathbf{a} \times \mathbf{v}\| \neq 0,$ the radius of curvature $\rho$ is the reciprocal of curvature $\kappa,$ thus: \begin{align*} \\ \rho &= \frac{\left(r^2 + r'^{2}\right)^{3/2}}{\left|r^2 + 2r'^2 - rr''\right|}. \quad \blacksquare \end{align*}