Mathematical Immaturity

14.19 Exercises

• Let $\phi$ denote the angle, $0 \leq \phi \leq \pi,$ between the position vector and the velocity vector of a curve. If the curve is expressed in polar coordinates, prove that $v \sin \phi = r$ and $v \cos \phi = dr/d\theta,$ where $v$ is the speed.
• If $\phi$ denotes the angle between the position vector and velocity vector, then $\sin \phi$ and $\cos\phi$ can be expressed as: \begin{align*} \\ \sin \phi &= \frac{\|\mathbf{r}\times\mathbf{v}\|}{\|\mathbf{r}\|\|\mathbf{v}\|} \\ \\ \cos \phi &= \frac{\mathbf{r}\cdot\mathbf{v}}{\|\mathbf{r}\|\|\mathbf{v}\|} \end{align*} If the curve is expressed in polar coordinates, this means that $r = f(\theta)$ where $\theta = t.$ What does this imply about $d\theta/dt$ and $dr/dt?$
• If $\phi$ denotes the angle between the position vector and velocity vector, then $\sin \phi$ and $\cos\phi$ can be expressed as: \begin{align*} \\ \sin \phi &= \frac{\|\mathbf{r}\times\mathbf{v}\|}{\|\mathbf{r}\|\|\mathbf{v}\|} \\ \\ \cos \phi &= \frac{\mathbf{r}\cdot\mathbf{v}}{\|\mathbf{r}\|\|\mathbf{v}\|} \end{align*} Expressing position and velocity in terms of their radial and transverse components, noting that $\|\mathbf{r}\| = r$ and $\|\mathbf{v}\| = v,$ we get: \begin{align*} \\ \mathbf{r} &= r\mathbf{u}_r \\ \\ \mathbf{v} &= \frac{dr}{dt}\mathbf{u}_r + r\frac{d\theta}{dt}\mathbf{u}_{\theta}. \end{align*} Then, we can use Lagrange's identity to express $\|\mathbf{r}\times\mathbf{v}\|$ in terms of these components to give us: \begin{align*} \\ \sin \phi &= \frac{\|\mathbf{r}\times\mathbf{v}\|}{\|\mathbf{r}\|\|\mathbf{v}\|} \\ \\ &= \frac{\sqrt{r^2\left[\left(\frac{dr}{dt}\right)^2 + r^2\left(\frac{d\theta}{dt}\right)^2\right] - r^2\left(\frac{dr}{dt}\right)^2}}{rv} \\ \\ &= \frac{r}{v}\frac{d\theta}{dt}. \\ \\ \cos \phi &= \frac{\mathbf{r}\cdot\mathbf{v}}{\|\mathbf{r}\|\|\mathbf{v}\|} \\ \\ &= \frac{r\frac{dr}{dt}}{rv} \\ \\ &= \frac{dr}{dt}\frac{1}{v}. \end{align*} Multiplying both sides by v gives us \begin{align*} \\ v\sin\phi &= r\frac{d\theta}{dt} \\ \\ v\cos\phi &= \frac{dr}{dt}. \end{align*} But, the curve is expressed in polar coordinates, which means $r = f(\theta),$ where $\theta(t) = t.$ By application of the chain rule, with $d\theta/dt = 1,$ the above equations become \begin{align*} \\ v\sin\phi &= r\frac{d\theta}{dt} = r \\ \\ v\cos\phi &= \frac{dr}{dt} = \frac{dr}{d\theta}\frac{d\theta}{dt}= \frac{dr}{d\theta}. \quad \blacksquare \end{align*}