Chapter 10, Problem 21RE

To determine

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

Answer to Problem 21RE

The slope of the tangent line of the polar curve $r=\cos 2\theta$ at $\theta =\frac{\pi }{3}$ is $4.04$ .

Explanation of Solution

Given information:

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

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

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

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

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

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

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

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

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

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

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

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

  $\begin{array}{c}\frac{dy}{dx}=\frac{\cos \frac{\pi }{3}\cos 2\left(\frac{\pi }{3}\right)-2\sin \frac{\pi }{3}\sin 2\frac{\pi }{3}}{-\sin \frac{\pi }{3}\cos 2\left(\frac{\pi }{3}\right)-2\cos \frac{\pi }{3}\sin 2\left(\frac{\pi }{3}\right)}\\ =\frac{\left(1/2\right)\left(-1/2\right)-2\left(\sqrt{3}/2\right)\left(\sqrt{3}/2\right)}{-\left(\sqrt{3}/2\right)\left(-1/2\right)-2\left(1/2\right)\left(\sqrt{3}/2\right)}\\ =\frac{-\frac{1}{4}-\frac{3}{2}}{\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{2}}\\ =\frac{-\frac{7}{4}}{\frac{-\sqrt{3}}{4}}\\ =\frac{7}{\sqrt{3}}\\ \approx 4.04\end{array}$

The slope of the tangent line at $\theta =\frac{\pi }{3}$ is $4.04$ .