Question Write the equation of the line tangent to the graph of \( r=2 \cos \theta \) when \( \theta=\frac{7 \pi}{6} \). Answer Attempt 1 out of 2 \( \sqrt{ } \)

We begin with the polar curve \[ r = 2\cos\theta. \] At \(\theta = \frac{7\pi}{6}\), the radius is \[ r = 2\cos\frac{7\pi}{6} = 2\left(-\frac{\sqrt{3}}{2}\right) = -\sqrt{3}. \] Even though \(r\) is negative, the corresponding Cartesian coordinates are found using \[ x = r\cos\theta \quad \text{and} \quad y = r\sin\theta. \] At \(\theta = \frac{7\pi}{6}\), note that \[ \cos\frac{7\pi}{6} = -\frac{\sqrt{3}}{2}, \quad \sin\frac{7\pi}{6} = -\frac{1}{2}. \] Thus, \[ x = (-\sqrt{3})\left(-\frac{\sqrt{3}}{2}\right) = \frac{3}{2}, \quad y = (-\sqrt{3})\left(-\frac{1}{2}\right) = \frac{\sqrt{3}}{2}. \] Next, we compute the slope of the tangent line. Recall that for a polar curve \(r=f(\theta)\) the derivative \(\frac{dy}{dx}\) is given by \[ \frac{dy}{dx} = \frac{\frac{dr}{d\theta}\sin\theta + r\cos\theta}{\frac{dr}{d\theta}\cos\theta - r\sin\theta}. \] Since \(r = 2\cos\theta\), we differentiate with respect to \(\theta\): \[ \frac{dr}{d\theta} = -2\sin\theta. \] At \(\theta = \frac{7\pi}{6}\): \[ \sin\frac{7\pi}{6} = -\frac{1}{2} \quad \text{and} \quad \cos\frac{7\pi}{6} = -\frac{\sqrt{3}}{2}, \] so that \[ \frac{dr}{d\theta} = -2\left(-\frac{1}{2}\right) = 1. \] Now substitute into the derivative formula: \[ \frac{dy}{dx} = \frac{(1)\left(-\frac{1}{2}\right) + \left(-\sqrt{3}\right)\left(-\frac{\sqrt{3}}{2}\right)}{(1)\left(-\frac{\sqrt{3}}{2}\right) - \left(-\sqrt{3}\right)\left(-\frac{1}{2}\right)}. \] Simplify the numerator: \[ (1)\left(-\frac{1}{2}\right) + \left(-\sqrt{3}\right)\left(-\frac{\sqrt{3}}{2}\right) = -\frac{1}{2} + \frac{3}{2} = \frac{2}{2} = 1. \] And the denominator: \[ \left(-\frac{\sqrt{3}}{2}\right) - \left(\sqrt{3}\cdot\frac{1}{2}\right) = -\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} = -\sqrt{3}. \] Thus, the slope is \[ m = \frac{dy}{dx} = \frac{1}{-\sqrt{3}} = -\frac{1}{\sqrt{3}}. \] We now have the point \(\left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right)\) and slope \(m = -\frac{1}{\sqrt{3}}\). The point-slope form of the tangent line is \[ y - \frac{\sqrt{3}}{2} = -\frac{1}{\sqrt{3}}\left(x - \frac{3}{2}\right). \] This is the equation of the tangent line in point-slope form. For an alternative form, we simplify further. Distribute the right-hand side: \[ y - \frac{\sqrt{3}}{2} = -\frac{1}{\sqrt{3}}x + \frac{3}{2\sqrt{3}}. \] Adding \(\frac{\sqrt{3}}{2}\) to both sides gives \[ y = -\frac{1}{\sqrt{3}}x + \frac{3}{2\sqrt{3}} + \frac{\sqrt{3}}{2}. \] Recognize that \[ \frac{\sqrt{3}}{2} = \frac{3}{2\sqrt{3}}, \] so \[ \frac{3}{2\sqrt{3}} + \frac{3}{2\sqrt{3}} = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}}. \] Thus, the line becomes \[ y = -\frac{1}{\sqrt{3}}x + \frac{3}{\sqrt{3}}. \] Multiplying both sides by \(\sqrt{3}\) to eliminate the radical in the denominator yields \[ \sqrt{3}y = -x + 3, \] which can be rearranged to \[ x + \sqrt{3}y - 3 = 0. \] This is the equation of the tangent line in standard form. The final answer is: \[ y - \frac{\sqrt{3}}{2} = -\frac{1}{\sqrt{3}}\left(x - \frac{3}{2}\right). \]