Chapter 10, Problem 22RE

To determine

To find:the slope of the tangent of the polar curve $r=2+\cos 2\theta$ at the point where $\theta =\frac{\pi }{3}$ .

Answer to Problem 22RE

The slope of the tangent line $r=2+\cos 2\theta$ at $\theta =\frac{\pi }{3}$ is $0.364$ .

Explanation of Solution

Given information:

Equation of the polar curve: $r=2+\cos 2\theta$ .

Formula used:

  $x=r\cos \theta$ and $y=r\sin \theta$ .

The formula to convert polar equation to cartesian equation: $r^2=x^2+y^2$ .

  $\frac{d}{d\theta }\left(\sin \theta \right)=\cos \theta$

  $\frac{d}{d\theta }\left(\cos \theta \right)=-\sin \theta$

Calculation:

Substitute $2+\cos 2\theta$ for $r$ in $x=r\cos \theta$ .

  $x=\left(2+\cos 2\theta \right)\cos \theta$

Substitute $2+\cos 2\theta$ for $r$ in $y=r\sin \theta$ .

  $y=\left(2+\cos 2\theta \right)\sin \theta$

Find the derivative $\frac{dx}{d\theta }$ .

  $\begin{array}{c}\frac{dx}{d\theta }=\left(2+\cos 2\theta \right)\left(-\sin \theta \right)+\left(-2\sin 2\theta \right)\left(\cos \theta \right)\\ =-2\sin \theta -\sin \theta \cos 2\theta -2\cos \theta \sin 2\theta \end{array}$

Find the derivative $\frac{dy}{d\theta }$ .

  $\begin{array}{c}\frac{dy}{d\theta }=\left(2+\cos 2\theta \right)\cos \theta +\left(-2\sin 2\theta \right)\left(\sin \theta \right)\\ =2\cos \theta +\cos \theta \cos 2\theta -2\sin \theta \sin 2\theta \end{array}$

Find $\frac{dy}{dx}$ , by substituting $-\sin \theta \cos 2\theta -2\cos \theta \sin 2\theta$ for $\frac{dx}{d\theta }$ and $2\cos \theta +\cos \theta \cos 2\theta -2\sin \theta \sin 2\theta$ for $\frac{dy}{d\theta }$ in $\frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }$

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

Substitute $\frac{\pi }{3}$ for $\theta$ in $\frac{dy}{dx}$ .

  $\begin{array}{c}\frac{dy}{dx}=\frac{2\cos \left(\frac{\pi }{3}\right)+\cos \frac{\pi }{3}\cos 2\left(\frac{\pi }{3}\right)-2\sin \frac{\pi }{3}\sin 2\left(\frac{\pi }{3}\right)}{-2\sin \frac{\pi }{3}-\sin \frac{\pi }{3}\cos 2\left(\frac{\pi }{3}\right)-2\cos \frac{\pi }{3}\sin 2\left(\frac{\pi }{3}\right)}\\ =\frac{2\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)-2\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)}{-2\left(\frac{\sqrt{3}}{2}\right)-\left(\frac{\sqrt{3}}{2}\right)\left(-\frac{1}{2}\right)-2\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)}\\ =\frac{1+\left(-\frac{1}{4}\right)-\left(\frac{3}{2}\right)}{-\left(\sqrt{3}\right)+\left(\frac{\sqrt{3}}{4}\right)-\left(\frac{\sqrt{3}}{2}\right)}\\ =\frac{\frac{4-1-6}{4}}{\frac{-4\sqrt{3}+\sqrt{3}-2\sqrt{3}}{4}}\end{array}$

Further, simplify.

  $\begin{array}{c}\frac{dy}{dx}=\frac{-3}{-5\sqrt{3}}\\ =\frac{\sqrt{3}}{5}\\ \approx 0.364\end{array}$

The slope of the tangent line at $\theta =\frac{\pi }{3}$ is $0.364$ .