The problem asks us to find the slope of the tangent line to the polar curves at $\theta = \frac{\pi}{3}$ for the following curves: (a) $r = 2\cos\theta$ (c) $r = \sin(2\theta)$

1. Problem Description

The problem asks us to find the slope of the tangent line to the polar curves at $\theta = \frac{\pi}{3}$ for the following curves:

(a) $r = 2\cos\theta$

(c) $r = \sin(2\theta)$

2. Solution Steps

The slope of the tangent line to a polar curve $r = f(\theta)$ is given by the formula:

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

(a) $r = 2\cos\theta$

$\frac{dr}{d\theta} = -2\sin\theta$

Then we have:

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

At $\theta = \frac{\pi}{3}$, we have $2\theta = \frac{2\pi}{3}$, so

$$\frac{dy}{dx} = -\cot\left(\frac{2\pi}{3}\right) = -\left(-\frac{1}{\sqrt{3}}\right) = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$$

(c) $r = \sin(2\theta)$

$\frac{dr}{d\theta} = 2\cos(2\theta)$

Then we have:

$$\frac{dy}{dx} = \frac{2\cos(2\theta)\sin\theta + \sin(2\theta)\cos\theta}{2\cos(2\theta)\cos\theta - \sin(2\theta)\sin\theta}$$

At $\theta = \frac{\pi}{3}$, we have $2\theta = \frac{2\pi}{3}$. Thus $\sin\theta = \frac{\sqrt{3}}{2}$, $\cos\theta = \frac{1}{2}$, $\sin(2\theta) = \sin(\frac{2\pi}{3}) = \frac{\sqrt{3}}{2}$, and $\cos(2\theta) = \cos(\frac{2\pi}{3}) = -\frac{1}{2}$.

$$\frac{dy}{dx} = \frac{2(-\frac{1}{2})(\frac{\sqrt{3}}{2}) + \frac{\sqrt{3}}{2}(\frac{1}{2})}{2(-\frac{1}{2})(\frac{1}{2}) - \frac{\sqrt{3}}{2}(\frac{\sqrt{3}}{2})} = \frac{-\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{4}}{-\frac{1}{2} - \frac{3}{4}} = \frac{-\frac{\sqrt{3}}{4}}{-\frac{5}{4}} = \frac{\sqrt{3}}{5}$$

3. Final Answer

(a) $\frac{\sqrt{3}}{3}$

(c) $\frac{\sqrt{3}}{5}$