A calculus problem by Rocco Dalto

$\frac{dx}{dt} = a^2 x +a^2 y - b^2 x ( x^2 - y^2) \\ \frac{dy}{dt} = -a^2 x + a^2 y -b^2 (x^2 - y^2)$

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 $t$ of $(1)$ and $(2)$ we obtain: $$

$r \dfrac{dr}{dt} = x \dfrac{dx}{dt} + y \dfrac{dy}{dt}$

$r^2 \dfrac{d\theta}{dt} = x \dfrac{dy}{dt} - y \dfrac{dx}{dt}$ $$

$x * (\frac{dx}{dt} = a^2 x +a^2 y - b^2 x ( x^2 - y^2))$ $y * (\frac{dy}{dt} = -a^2 x + a^2 y -b^2 (x^2 - y^2))$

Adding the above equations we obtain: $r \dfrac{dr}{dt} = r^2 (a^2 - b^2 r^2) \implies \dfrac{dr}{dt} = r (a^2 - b^2 r^2)$

$-y * (\frac{dx}{dt} = a^2 x +a^2 y - b^2 x ( x^2 - y^2))$ $x * (\frac{dy}{dt} = -a^2 x + a^2 y -b^2 (x^2 - y^2))$

Adding the above equations we obtain: $$

$\dfrac{d\theta}{dt} = -a^2$ $$

The system of differentials above has a single critical point at $r = 0$ . Assuming $r > 0$ , the system above reduces to:

$\dfrac{dr}{dt} = r (a^2 - b^2 r^2)$ $\dfrac{d\theta}{dt} = -a^2$

$\implies \dfrac{1}{a^2} \int (\dfrac{1}{r} + \dfrac{b}{2 (a - b r)} - \dfrac{b}{2 (a + b r)}) \: dr = \dfrac{1}{a^2} \ln(\dfrac{r}{\sqrt{a^2 - b^2 r^2}}) = t + C \implies$ $$

$\dfrac{r^2}{a^2 - b^2 r^2} = C e^{2 a^2 t} \implies$ $r = \dfrac{a}{\sqrt{b^2 + C e^{-2 a^2 t}}}$ and $\theta = -a^2 t + t_{0}$ . $$

$x = r cos(-a^2 t + t_{0}), y = r sin(-a^2 t + t_{0}) \implies$

$x(t) = \dfrac{a * cos(-a^2 t + t_{0})}{\sqrt{b^2 + C e^{-2 a^2 t}}}$ $y(t) = \dfrac{a * sin(-a^2 t + t_{0})}{\sqrt{b^2 + C e^{-2 a^2 t}}}$

Using initial conditions $x(t = 0) = y(t = 0) = \dfrac{1}{2}$ we obtain: $$

$\dfrac{1}{4} (b^2 + C) = a^2 cos^2(t_{0})$ $$

$\dfrac{1}{4} (b^2 + C) = a^2 sin^2(t_{0})$ $$

$\implies C = 2 a^2 - b^2 \implies t_{0} = \dfrac{\pi}{4},$ although $t_{0} = \dfrac{\pi}{4}$ is not needed here. $$

$\int_{ \frac{1}{a^2} (\ln(\frac{\sqrt{2 a^2 - b}}{b})) }^{\frac{1}{a^2} (\ln(\frac{\sqrt{3(2 a^2 - b)}}{b}))} \sqrt{x(t)^2 + y(t)^2} \: dt =$ $$

$\int_{ \frac{1}{a^2} (\ln(\frac{\sqrt{2 a^2 - b}}{b})) }^{\frac{1}{a^2} (\ln(\frac{\sqrt{3(2 a^2 - b)}}{b}))} \dfrac{e^{a^2 t}}{\sqrt{(b e^{a^2 t})^2 + 2 a^2 - b^2}}$ $$

Let $b e^{a^2 t} = \sqrt{2 a^2 - b^2} tan\theta \implies$ $dt = \dfrac{sec^2(\theta)}{a^2 tan\theta} \: d\theta \implies$ $$

$\dfrac{1}{a^2 * b} \int_{\dfrac{\pi}{4}}^{\dfrac{\pi}{3}} sec\theta \: d\theta =$ $$

$\dfrac{1}{a^2 * b} ln(sec\theta + tan\theta)|_{ \frac{1}{a^2} (\ln(\frac{\sqrt{2 a^2 - b}}{b})) }^{\frac{1}{a^2} (\ln(\frac{\sqrt{3(2 a^2 - b)}}{b}))} =$ $$

$\dfrac{1}{a^2 * b} ln(\dfrac{2+ \sqrt{3}}{\sqrt{2} + 1}) =$ $\dfrac{1}{a^2 * b} \ln \left(\frac{\alpha + \sqrt{\beta}}{\sqrt{\alpha} + \lambda}\right)$ $\implies \alpha + \beta + \lambda = \fbox{6}$