Find the derivative of the polar curve $r = 4\theta + \sin\theta$

Question

I have been trying to solve this equation: Find the derivative $\frac{dy}{dx}$ of the polar curve $r = 4\theta + \sin\theta$ and I'm not sure where to go with this. Anyone mind giving a hand?

Answer 1

HINT:

$x = r \cos \theta = (4 \theta + \sin \theta) \cos \theta$

$y = r \sin \theta = (4 \theta + \sin \theta) \sin \theta$

$\displaystyle \frac{dy}{dx} = \frac{dy}{d\theta} / \frac{dx}{d\theta}$

Can you take it from here?

Answer 2

By the Chain Rule, we know that:

$$\frac{dy}{dx}\frac{dx}{d\theta} = \frac{dy}{d\theta}$$

Thus:

$$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$$

However, we know that $x = r\cos(\theta)$ and $y = r\sin(\theta)$. Thus, we have:

$$\frac{dy}{dx} = \frac{\frac{d}{d\theta}(r\sin(\theta))}{\frac{d}{d\theta}(r\cos(\theta))}$$

Doing some straightforward product and chain rules (remember that $r$ is also a function of $\theta$):

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

Can you take it from here? If you're still stuck:

Taking derivatives, we find:

$$\frac{dr}{d\theta} = 4 + \cos(\theta)$$

Thus, we have:

$$\frac{dy}{dx} = \frac{(4 + \cos(\theta))\sin(\theta) + (4\theta + \sin(\theta))\cos(\theta)}{(4 + \cos(\theta))\cos(\theta) - (4\theta + \sin(\theta))\sin(\theta)}$$ $$\frac{dy}{dx} = \frac{4\sin(\theta) + 4\sin(\theta)\cos(\theta) + 4\theta\cos(\theta) + \sin(\theta)\cos(\theta)}{4\cos(\theta) + \cos^{2}(\theta) - 4\theta\sin(\theta) - \sin^{2}(\theta)}$$ $$\frac{dy}{dx} = \boxed{\frac{8(\sin(\theta) + \theta\cos(\theta)) + 5\sin(2\theta)}{8(\cos(\theta) - \theta\sin(\theta)) + 2\cos(2\theta)}}$$