Chapter 11, Problem 22RE

To determine

To determine: The slope of the tangent at the point where $\theta =\frac{\pi }{3}$ .

Answer to Problem 22RE

The required slope of the tangent is 0.346.

Explanation of Solution

Given information:

The equation: $r=2+\cos 2\theta$

Formula used:

Product rule of differentiation:

  $\frac{d}{dx}\left(uv\right)=u^{\prime }v+v^{\prime }u$

Calculation:

The given equation is $r=2+\cos 2\theta$ .

It is seen that in polar coordinate system if $r=f\left(\theta \right)$ is a curve, then the coordinates of the curve in rectangular coordinate system is:

  $x=r\cos \theta =f\left(x\right)\cos \theta$ , $y=r\sin \theta =f\left(\theta \right)\sin \theta$

Rewrite the coordinates as:

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

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

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

Find the slope of the tangent.

  $\begin{array}{c}\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}\\ \frac{dy}{dx}=\frac{\cos \theta \cos 2\theta -2\sin \theta \sin 2\theta +2\cos \theta }{-\sin \theta \cos 2\theta -2\cos \theta \sin 2\theta -2\sin \theta }\end{array}$

Substitute $\frac{\pi }{3}$ for $\theta$ in the above expression.

  $\begin{array}{c}\frac{dy}{dx}=\frac{\cos \frac{\pi }{3}\cos \frac{2\pi }{3}-2\sin \frac{\pi }{3}\sin \frac{2\pi }{3}+2\cos \frac{\pi }{3}}{-\sin \frac{\pi }{3}\cos \frac{2\pi }{3}-2\cos \frac{\pi }{3}\sin \frac{2\pi }{3}-2\sin \frac{\pi }{3}}\\ =\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)-2\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)+2\left(\frac{1}{2}\right)}{-\left(\frac{\sqrt{3}}{2}\right)\left(-\frac{1}{2}\right)-2\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)-2\left(\frac{\sqrt{3}}{2}\right)}\\ =\frac{\left(-\frac{1}{4}-\frac{3}{2}+1\right)}{\left(\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{2}-\sqrt{3}\right)}\\ =\frac{3}{5\sqrt{3}}\end{array}$

Simplify the above expression.

  $\frac{3}{5\sqrt{3}}\approx 0.346$

Hence, the required slope of the tangent is 0.346.