Chapter 11, Problem 23RE

To determine

To determine: The equations for the horizontal and vertical tangent lines to the curves.

Answer to Problem 23RE

The required tangents are $x=2,0.067,-1.104$ and $y=0,0.443,1.739$ .

Explanation of Solution

Given information:

The equation: $r=1-\cos \left(\frac{\theta }{2}\right)$ , $0\le \theta \le 4\pi$

Formula used:

Product rule of differentiation:

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

Calculation:

The given equation is $r=1-\cos \left(\frac{\theta }{2}\right)$ .

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:

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

Rewrite the coordinates as:

  $x=\left(1-\cos \left(\frac{\theta }{2}\right)\right)\cos \theta$ , $y=\left(1-\cos \left(\frac{\theta }{2}\right)\right)\sin \theta$

Here, the slope of the tangent is,

  $\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}$

It is known that for a tangent line to be horizontal, the slope of the line is 0.

  $\begin{array}{c}\frac{dy}{dx}=0\\ \frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=0\\ \frac{dy}{d\theta }=0\end{array}$

Simplify the above expression further.

  $\begin{array}{c}\frac{d}{d\theta }\left(\left(1-\cos \left(\frac{\theta }{2}\right)\right)\sin \theta \right)=0\\ -\cos \theta \cos \left(\frac{\theta }{2}\right)+\frac{\sin \theta \sin \left(\frac{\theta }{2}\right)}{2}+\cos \theta =0\end{array}$

From the above equation we get the tangents as:

  $y=0$ , $y\approx \pm 0.443$ and $y\approx \pm 1.739$

Now, it is known that for a tangent line to be vertical, the slope of the line is undefined.

  $\begin{array}{c}\frac{dy}{dx}=\infty \\ \frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\infty \\ \frac{dy}{d\theta }=\infty \end{array}$

Simplify the above expression further.

  $\begin{array}{c}\frac{d}{d\theta }\left(\left(1-\cos \left(\frac{\theta }{2}\right)\right)\cos \theta \right)=0\\ \sin \theta \cos \left(\frac{\theta }{2}\right)+\frac{\cos \theta \sin \left(\frac{\theta }{2}\right)}{2}-\sin \theta =0\end{array}$

From the above equation we get the tangents as:

  $x=2$ , $x\approx 0.067$ and $x\approx -1.104$

Hence, the required tangents are $x=2,0.067,-1.104$ and $y=0,0.443,1.739$ .