Question

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

Short answer

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

Step-by-step solution

Convert polar coordinates to rectangular coordinates

To convert the point $(\frac{2}{3},\frac{\pi }{6})$ 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 $(\frac{\sqrt{3}}{3},\frac{1}{3})$.

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: $$\frac{dr}{d\theta }=\frac{-\frac{d}{d\theta }(1+\sin \theta )}{(1+\sin \theta {)}^2}=\frac{-\cos \theta }{(1+\sin \theta {)}^2}$$

Evaluate the derivative at the given point

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

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 $(\frac{\sqrt{3}}{3},\frac{1}{3})$, and $m$ is the slope, which is $\frac{dr}{d\theta }{|}_{\frac{\pi }{6}}=-\frac{\sqrt{3}}{9}$. Plugging the values into the equation, we get: $$y-\frac{1}{3}=-\frac{\sqrt{3}}{9}(x-\frac{\sqrt{3}}{3})$$ 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 $9y+\sqrt{3}x=6$.