A calculus problem by Rocco Dalto

$\: \dfrac{dx}{dt} = - y + x (1 - x^2 - y^2)$ $\: \dfrac{dy}{dt} = x + y (1 - x^2 - y^2)$ $$

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: $$

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

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

$x * (\dfrac{dx}{dt} = - y + x (1 - x^2 - y^2) )$ $y * (\dfrac{dy}{dt} = x + y (1 - x^2 - y^2) )$

Adding the above equations we obtain: $r \dfrac{dr}{dt} = r^2 (1 - r^2) \implies \dfrac{dr}{dt} = r (1 - r^2)$ $$

and, $-y * (\dfrac{dx}{dt} = - y + x (1 - x^2 - y^2) )$ $x * (\dfrac{dy}{dt} = x + y (1 - x^2 - y^2) )$ $$

Adding the above equations we obtain:

$r^2 \dfrac{d\theta}{dt} = r^2 \implies \dfrac{d\theta}{dt} = 1$ $$

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

$\dfrac{dr}{dt} = r (1 - r^2)$ $\dfrac{d\theta}{dt} = 1$

$\implies \int \dfrac{dr}{r (1 - r^2)} = \int dt \implies$ $$

$\int \dfrac{1}{r} + \dfrac{1}{2 (1 - r)} - \dfrac{1}{2 (1 + r)} \: dr = \int dt \implies$ $$

$\ln(\dfrac{r}{\sqrt{1 - r^2}}) = t + K \implies \dfrac{r^2}{1 - r^2} = C e^{2t} \implies$ $$

$r^2 (1 + C e^{2t}) = C e^{2t} \implies r = \sqrt{\dfrac{C e^{2t}}{1 + C e^{2t}}}$ $\implies r = \dfrac{1}{\sqrt{1 + C e^{-2t}}}$ and, $\theta = t + t_{0}$ $$

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

$x(t) = \dfrac{cos(t + t_{0})}{\sqrt{1 + C e^{-2t}}}$ $y(t) = \dfrac{sin(t + t_{0})}{\sqrt{1 + C e^{-2t}}}$

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

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

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

$\implies \dfrac{1}{2} (1 + C) = 1 \implies C = 1 \implies$

$x(t) = \dfrac{cos(t + t_{0})}{\sqrt{1 + e^{-2t}}}$ $y(t) = \dfrac{sin(t + t_{0})}{\sqrt{1 + e^{-2t}}}$

Note: Using $C = 1 \implies t_{0} = \dfrac{\pi}{4}$ here, but it's not needed for the integral below. $$

The integral $I = \int_{0}^{\ln(\sqrt{3})} \sqrt{x(t)^2 + y(t)^2} \: dt =$ $$

$\int_{0}^{\ln(\sqrt{3})} \dfrac{e^{t}}{\sqrt{e^{2t} + 1}} \: dt$ $$

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

$I = \int_{0}^{\ln(\sqrt{3})} \dfrac{e^{t}}{\sqrt{e^{2t} + 1}} \: dt = \int_{\dfrac{\pi}{4}}^{\dfrac{\pi}{3}} sec(\theta) \: d\theta =$ $\ln(sec(\theta) + tan(\theta))|_{\dfrac{\pi}{4}}^{\dfrac{\pi}{3}} =$ $$

$\ln(\dfrac{2 + \sqrt{3}}{\sqrt{2} + 1}) = \ln(\dfrac{\alpha + \sqrt{\beta}}{\sqrt{\alpha} + \lambda})$ $$

$\implies \alpha + \beta + \lambda = \fbox{6}$