Problem 70 - Calculus Early Transcendentals

Question

Find an equation of the line tangent to the following curves at the given point. $$r=\frac{1}{1+\sin \theta} ;\left(\frac{2}{3}, \frac{\pi}{6}\right)$$

Step-by-Step Solution

Answer

The equation of the tangent line is \(9y + \sqrt{3}x = 6\).

Step 1: Convert polar coordinates to rectangular coordinates

To convert the point \(\left(\frac{2}{3}, \frac{\pi}{6}\right)\) to rectangular coordinates, we use the formulas \(x = r\cos\theta\) and \(y = r\sin\theta\). For the given point, \(r = \frac{2}{3}\) and \(\theta = \frac{\pi}{6}\). Thus, $$ x = \frac{2}{3}\cos\frac{\pi}{6} = \frac{2}{3}\cdot\frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{3} $$ $$ y = \frac{2}{3}\sin\frac{\pi}{6} = \frac{2}{3}\cdot\frac{1}{2} = \frac{1}{3} $$ So, the rectangular coordinates of the given point are \(\left(\frac{\sqrt{3}}{3},\frac{1}{3}\right)\).

Step 2: Find the derivative of the polar curve

We start by finding \(\frac{dr}{d\theta}\) of the given polar curve \(r = \frac{1}{1+\sin\theta}\).Using the quotient rule, we get: $$ \frac{dr}{d\theta} = \frac{-\frac{d}{d\theta}(1+\sin\theta)}{(1+\sin\theta)^2} = \frac{-\cos\theta}{(1+\sin\theta)^2} $$

Step 3: Evaluate the derivative at the given point

Next, we evaluate \(\frac{dr}{d\theta}\) at \(\theta = \frac{\pi}{6}\). We get: $$ \frac{dr}{d\theta}\Big|_{\frac{\pi}{6}}=\frac{-\cos\frac{\pi}{6}}{(1+\sin\frac{\pi}{6})^2}=\frac{-\frac{\sqrt{3}}{2}}{(1+\frac{1}{2})^2}=-\frac{\sqrt{3}}{9} $$

Step 4: Find the equation of the tangent line

We now use the point-slope form of a line, which is given by: $$ y - y_0 = m(x - x_0) $$ where \((x_0, y_0)\) is the point \(\left(\frac{\sqrt{3}}{3},\frac{1}{3}\right)\), and \(m\) is the slope, which is \(\frac{dr}{d\theta}\Big|_{\frac{\pi}{6}}=-\frac{\sqrt{3}}{9}\). Plugging the values into the equation, we get: $$ y - \frac{1}{3} = -\frac{\sqrt{3}}{9}\left(x - \frac{\sqrt{3}}{3}\right) $$ To find the final equation of the tangent line, we can eliminate fractions and simplify the equation. Doing this, we get: $$ 9y - 3 = -\sqrt{3}(x - \sqrt{3}) $$ $$ 9y + \sqrt{3}x = 6 $$ The equation of the line tangent to the curve at the given point is \(9y + \sqrt{3}x = 6\).