A calculus problem by Rocco Dalto

$\dfrac{dy}{dx} = \dfrac{x + \alpha y}{\alpha x - y}.$ $$

Converting to polar coordinates, let $x = r cos\theta, y = r sin\theta$ $(1): \: x^2 + y^2 = r^2, (2) \: \theta = \arctan(\dfrac{y}{x})$ $$

Taking the derivatives with respect to $x$ of $(1)$ and $(2)$ we obtain: $$

$r \dfrac{dr}{dx} = x + y \dfrac{dy}{dx}$ $r^2 \dfrac{d\theta}{dx} = x \dfrac{dy}{dx} - y$
$\implies$ $\dfrac{1}{r} \dfrac{dr}{d\theta} =\dfrac{\alpha x^2- xy + xy + \alpha y^2}{x^2 + \alpha xy - \alpha xy + y^2} = \alpha \implies$ $$

$\dfrac{dr}{d\theta} = \alpha r \implies \ln(r) = \alpha \theta + K \implies$ $$

$r(\theta) = C e^{\alpha \theta}$ , $r(0) = 1 \implies r(\theta) = e^{\alpha \theta}.$ $$

So, we have a spiral, and since $\alpha > 0$ , it spirals outward in a counter close-wise direction.

To obtain the arc length of $r(\theta)$ on ( $0 <= \theta <= \dfrac{\ln(\alpha)}{\alpha})$ we note that: $$

$\vec{r(t)} = r \vec{i_{r}},$ where the unit vector $\vec{i_{r}} = cos(\theta(t)) \vec{i} + sin(\theta(t)) \vec{j} \implies \vec{i_{r}'} = (-sin(\theta(t)) \vec{i} + cos(\theta(t)) \vec{j}) \dfrac{d\theta}{dt} = \dfrac{d\theta}{dt} \vec{i_\theta}$ $$ and the dot product of the unit vectors $\vec{i_{r}} \cdot \vec{i_\theta} = 0,$ so that $\vec{i_{r}}$ and $\vec{i_\theta}$ are perpendicular vectors.

$\dfrac{d\vec{r}}{dt} = r(t) \vec{i_{r}'} + \vec{i_{r}} \dfrac{dr}{dt} \implies$ $\dfrac{d\vec{r}}{dt} = \dfrac{dr}{dt} \vec{i_{r}}+ r(t) \dfrac{d\theta}{dt} \vec{i_{\theta}}$

$\implies \vert\dfrac{d\vec{r}}{dt}\vert = \sqrt{(\dfrac{dr}{dt})^2 + r^2(t) (\dfrac{d\theta}{dt})^2} \implies$

Arc length $L = \int_{t_{1}}^{t_{2}} \vert\dfrac{d\vec{r}}{dt}\vert \: dt =$ $\int_{0}^{\dfrac{\ln(\alpha)}{\alpha}} \sqrt{r^2 + (\dfrac{dr}{d\theta})^2} \: d\theta =$ $$ $\sqrt{1 + \alpha^2} \int_{0}^{\dfrac{\ln(\alpha)}{\alpha}} e^{\alpha \theta} \ d\theta =$ $$

$\dfrac{\sqrt{1 + \alpha^2}}{\alpha} \: e^{\alpha \theta} |_{0}^{\dfrac{\ln(\alpha)}{\alpha}} =$ $\dfrac{\sqrt{1 + \alpha^2}}{\alpha} (\alpha - 1) = \dfrac{\sqrt{1 + \alpha^2}}{\alpha} \implies$ $$

$\dfrac{\sqrt{1 + \alpha^2}}{\alpha} (\alpha - 2) = 0 \implies \alpha = 2.$

$$

Note: You can get the same result solving the homogeneous differential equation $$

$\dfrac{dy}{dx} = \dfrac{x + \alpha y}{\alpha x - y},$ by making the substitution $z = \dfrac{y}{x}$ which results in $\int \dfrac{\alpha - z}{1 + z^2} \: dz = \int \dfrac{dx}{x}$ $$

Letting $z = tan\theta \implies dz = sec^2\theta \: d\theta \implies$

$\alpha \theta = ln(\dfrac{x}{c \: cos\theta}) = \ln(\dfrac{r}{c}),$ where $z = tan\theta$ and $x = r \cos\theta \implies r = C e^{\alpha \theta}$ , which is the same solution as above.