Slope of a polar curve

$\begin{cases} x = r\cos \theta = \cos \theta + \sin \theta \cos \theta \\ y = r\sin \theta = \sin \theta + \sin^2 \theta \end{cases} \quad \small \color{#3D99F6} \text{since }r = 1 + \sin \theta$

$\begin{cases} \dfrac {dx}{d \theta} = - \sin \theta + \cos 2 \theta \\ \dfrac {dy}{d \theta} = \cos \theta + \sin 2 \theta \end{cases}$

$\begin{aligned} \implies \dfrac {dy}{dx} & = \dfrac {\frac {dy}{d\theta}}{\frac {dx}{d\theta}} = \dfrac {\cos \theta + \sin 2 \theta}{- \sin \theta + \cos 2 \theta} \\ \dfrac {dy}{dx}\bigg|_{\theta = \frac \pi 4} & = \dfrac {\cos \frac \pi 4 + \sin \frac \pi 2}{- \sin \frac \pi 4 + \cos \frac \pi 2} = \frac {\frac 1{\sqrt 2}+1}{-\frac 1{\sqrt 2}+0} = - 1 - \sqrt 2 \end{aligned}$

$\implies a+b = 2+1 = \boxed{3}$

In general, with $r = f(\theta)$ , we have that $x = r\cos(\theta)$ and $y = r\sin(\theta)$ and thus

$\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}} = \dfrac{\dfrac{dr}{d\theta}\sin(\theta) + r\cos(\theta)}{\dfrac{dr}{d\theta}\cos(\theta) - r\sin(\theta)}$ .

Now in this case $r = 1 + \sin(\theta) \Longrightarrow \dfrac{dr}{d\theta} = \cos(\theta)$ , and thus

$\dfrac{dy}{dx} = \dfrac{\cos(\theta)\sin(\theta) + (1 + \sin(\theta))\cos(\theta)}{\cos^{2}(\theta) - (1 + \sin(\theta))\sin(\theta)} = \dfrac{2\sin(\theta)\cos(\theta) + \cos(\theta)}{(\cos^{2}(\theta) - \sin^{2}(\theta)) - \sin(\theta)} = \dfrac{\sin(2\theta) + \cos(\theta)}{\cos(2\theta) - \sin(\theta)}$ .

So the slope at $\theta = \dfrac{\theta}{4}$ is $\dfrac{dy}{dx} = \dfrac{1 + \dfrac{\sqrt{2}}{2}}{0 -\dfrac{\sqrt{2}}{2}} = - \dfrac{(2 + \sqrt{2})}{\sqrt{2}} = -\sqrt{2} - 1$ , and so $a + b = 2 + 1 = \boxed{3}$ .