Mathematical Immaturity

14.21 Miscellaneous Review Exercises

• Let $\mathbf{r}$ denote the vector from the origin to an arbitrary point on the parabola $y^2 = x,$ let $\alpha$ be the angle that $\mathbf{r}$ makes with the tangent line, $0 \leq \alpha \leq \pi.$ Express $\alpha$ in terms of $\theta,$ where $\theta$ is the angle that $\mathbf{r}$ makes with the positive $x$-axis, $0 \leq \theta \leq \pi.$
• Recall that if $\alpha$ is the angle between two vectors $A$ and $B,$ we have \begin{align*} \\ \sin\alpha &= \frac{\|A \times B\|}{\|A\|\|B\|}, \quad \cos\alpha = \frac{A \cdot B}{\|A\|\|B\|}. \end{align*} We can also express $\alpha$ in terms of its tangent to eliminate the norm calculations. For $A \cdot B \neq 0$ we have: \begin{align*} \\ \tan\alpha &= \frac{\|A \times B\|}{A \cdot B}. \end{align*} Putting $\mathbf{r}$ in rectangular coordinates gives us $\mathbf{r} = (x, y).$ But, every point on the parabola must satisfy $x = y^2,$ giving us $\mathbf{r} = (y^2, y).$ Taking its derivative with respect to $y$ gives us $\mathbf{r}' = (2y, 1).$ But $\mathbf{r}'$ is parallel to the line tangent to the curve at $(x, y).$ Thus, $\alpha$ can be expressed as the angle between $\mathbf{r}$ and $\mathbf{r}'.$
• We know that the line tangent to $\mathbf{r}$ must be parallel to $\mathbf{r}',$ so we can use the dot and cross products of $\mathbf{r}$ and $\mathbf{r}'$ to give us: \begin{align*} \\ \tan \alpha &= \frac{\sin \alpha}{\cos \alpha} \\ \\ &= \frac{\|\mathbf{r} \times \mathbf{r}'\|}{\|\mathbf{r}\|\|\mathbf{r}'\|}\frac{\|\mathbf{r}\|\|\mathbf{r}'\|}{\mathbf{r} \cdot \mathbf{r}'} \\ \\ &= \frac{\|\mathbf{r} \times \mathbf{r}'\|}{\mathbf{r} \cdot \mathbf{r}'} \end{align*} And since $x = y^2$ for all $(x, y)$ on the parabola, we get $\mathbf{r} = (y^2, y)$ and $\mathbf{r}' = (2y, 1),$ giving us: \begin{align*} \\ \tan \alpha &= \frac{\|\mathbf{r} \times \mathbf{r}'\|}{\mathbf{r} \cdot \mathbf{r}'} \\ \\ &= \frac{\|-y^2\,\mathbf{k}\|}{2y^3 + y} \\ \\ &= \frac{y^2}{2y^3 + y} \\ \\ &= \frac{y}{2y^2 + 1} \end{align*} Now, to express $\tan\alpha$ in terms of the angle $\theta$ between $\mathbf{r}$ and the positive $x$-axis, we first express $\mathbf{r}$ in terms of polar coordinates to give us $x = r\cos\theta$ and $y = r\sin\theta.$ This gives us $y/x = \tan\theta$ for $x \neq 0.$ But since $x = y^2,\,\tan\theta = 1/y$ for $y \neq 0.$ We can then express $\tan \alpha$ in terms of $\theta$ as: \begin{align*} \\ \tan\alpha &= \frac{1/\tan\theta}{2/\tan^2\theta + 1} \\ \\ &= \frac{\tan\theta}{2 + \tan^2\theta} \quad \blacksquare \end{align*}