consider the polar curve $r = 1 + 2sin(\theta)$. what is the equation of the tangent line to the curve $r$…

Explanation:

Step1: Convert polar to Cartesian coordinates

We know that $x = r\cos\theta=(1 + 2\sin\theta)\cos\theta=\cos\theta+2\sin\theta\cos\theta=\cos\theta+\sin2\theta$ and $y = r\sin\theta=(1 + 2\sin\theta)\sin\theta=\sin\theta+2\sin^{2}\theta$.

Step2: Find $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$

$\frac{dx}{d\theta}=-\sin\theta + 2\cos2\theta$ and $\frac{dy}{d\theta}=\cos\theta+4\sin\theta\cos\theta=\cos\theta + 2\sin2\theta$.

Step3: Evaluate at $\theta=\frac{5\pi}{6}$

First, find $r$ at $\theta=\frac{5\pi}{6}$: $r = 1+2\sin\frac{5\pi}{6}=1 + 2\times\frac{1}{2}=2$. $x=r\cos\theta=2\times(-\frac{\sqrt{3}}{2})=-\sqrt{3}$, $y=r\sin\theta=2\times\frac{1}{2}=1$. $\frac{dx}{d\theta}\big|{\theta=\frac{5\pi}{6}}=-\sin\frac{5\pi}{6}+2\cos\frac{5\pi}{3}=-\frac{1}{2}+2\times\frac{1}{2}=\frac{1}{2}$. $\frac{dy}{d\theta}\big|{\theta=\frac{5\pi}{6}}=\cos\frac{5\pi}{6}+2\sin\frac{5\pi}{3}=-\frac{\sqrt{3}}{2}+2\times(-\frac{\sqrt{3}}{2})=-\frac{3\sqrt{3}}{2}$.

Step4: Calculate $\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$. At $\theta = \frac{5\pi}{6}$, $\frac{dy}{dx}=\frac{-\frac{3\sqrt{3}}{2}}{\frac{1}{2}}=- 3\sqrt{3}$.

Step5: Use point - slope form

The point - slope form of a line is $y - y_0=m(x - x_0)$. Here $x_0=-\sqrt{3}$, $y_0 = 1$ and $m=-3\sqrt{3}$, so the equation of the tangent line is $y - 1=-3\sqrt{3}(x+\sqrt{3})$.

Answer:

B. $y - 1=-3\sqrt{3}(x+\sqrt{3})$