Mathematical Immaturity

14.9 Exercises

• A plane curve $C$ in the first quadrant has a negative slope at each of its points and passes through the point $(\frac{3}{2}, 1).$ The position vector $\mathbf{r}$ from the origin to any point $(x,y)$ on $C$ makes an angle $\theta$ with $\mathbf{i},$ and the velocity vector makes an angle $\phi$ with $\mathbf{i},$ where $0 < \theta < \frac{\pi}{2},$ and $0 < \phi < \frac{\pi}{2}.$ If $3 \tan \phi = 4 \cot \theta$ at each point of $C,$ find a Cartesian equation for $C$ and sketch the curve.
• The position vector $\mathbf{r}$ is $(x, y)$ and the velocity vector $\mathbf{v}$ is $\left(\frac{dx}{dt}, \frac{dy}{dt}\right).$ And since velocity is always negative, we instead have $$ \begin{align*} -3\tan\,\phi = 4\cot\,\theta \end{align*} $$ where $\phi$ and $\theta$ are between $0$ and $\pi/2.$
• The position vector $\mathbf{r} = (x, y)$ makes an angle $\theta$ with $\mathbf{i},$ thus $\cos\,\theta$ must be $\frac{x}{\|\mathbf{r}\|},$ implying also that $\sin\,\theta = \frac{y}{\|\mathbf{r}\|}$ since $\|\mathbf{r}\| = \sqrt{x^2 + y^2}.$ Similarly, the velocity vector $\mathbf{v} = \left(\frac{dx}{dt}, \frac{dy}{dt}\right)$ is such that $\cos\,\phi = \frac{dx}{dt}\frac{1}{\|\mathbf{v}\|},$ but since we know that $\frac{dy}{dt} < 0$ for all $t,$ for $0 < \phi < \pi/2,$ it must be the case that $\sin\,\phi = -\frac{dy}{dt}\frac{1}{\|v\|},$ or in other words, $$ \begin{align*} -3\tan\,\phi = 4\cot\,\theta \end{align*} $$ Rewriting the above equation with the values derived for sine and cosine of $\phi$ and $\theta,$ we get: $$ \begin{align*} -3\frac{dy/dt}{dx/dt} = 4\frac{x}{y} \end{align*} $$ Cancelling out the $dt$ factors, we get: $$ \begin{align*} -3\frac{dy}{dx} = 4\frac{x}{y} \end{align*} $$ As we can see, this is a separable differential equation. Separating variables and integrating, we get: $$ \begin{align*} 3\int y\,dy\ +\ 4\int x\,dx &= \frac{3}{2}y^2 + 2x^2\\ &= C \end{align*} $$ To solve for the constant $C,$ recall that when $x = 3/2,$ $y = 1,$ giving us $4x^2 + 3y^2 = 12.$ Putting this into standard form, we get the Cartesian equation $$ \frac{x^{2}}{3} + \frac{y^2}{4} = 1 $$