Chapter 10, Problem 23RE

To determine

To state: The equation for the horizontal and vertical tangent lines to the curves if $r=1-\cos \left(\frac{\theta }{2}\right),0\le \theta \le 4\pi$ .

Answer to Problem 23RE

The equation for the horizontal and vertical tangent lines are:

  $-\cos \theta \cos \left(\frac{\theta }{2}\right)+\frac{\sin \theta \sin \left(\frac{\theta }{2}\right)}{2}+\cos \theta =0$ and $\sin \theta \cos \left(\frac{\theta }{2}\right)+\frac{\cos \theta \sin \left(\frac{\theta }{2}\right)}{2}-\sin \theta =0$

Explanation of Solution

Given information:

The given equation in polar coordinate system is $r=1-\cos \left(\frac{\theta }{2}\right),0\le \theta \le 4\pi$ .

Consider the given equation:

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

If $r=f\left(\theta \right)$ is a curve in polar coordinate system, then in rectangular coordinate system:

  $\begin{array}{l}x=r\cos \theta ,x=f\left(\theta \right)\cos \theta \\ y=r\sin \theta ,y=f\left(\theta \right)\sin \theta \end{array}$

Since the polar curves are shown in the $xy-$ plane, the slope of a polar curve is still the slope of tangent line, which is $\frac{dy}{dx}$ .

Now find the $\frac{dy}{dx}$ .

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

Now consider,

  $\begin{array}{l}x=\left(1-\cos \left(\frac{\theta }{2}\right)\right)\cos \theta \\ y=\left(1-\cos \left(\frac{\theta }{2}\right)\right)\sin \theta \end{array}$

And the slope is:

  $\frac{dy}{dx}=\frac{\frac{d}{d\theta }\left(\left(1-\cos \left(\frac{\theta }{2}\right)\right)\sin \theta \right)}{\frac{d}{d\theta }\left(\left(1-\cos \left(\frac{\theta }{2}\right)\right)\cos \theta \right)}$

For horizontal line:

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

Now use a trigonometric equation solver to get the solution of the above equation and for those values of $\theta$ , the value of $y$ at that point and $y=y_1$ is the equation of horizontal tangent line at that point.

These tangents are:

  $y=0\text{, }y\approx \pm 0.443,y\approx \pm 1.739$

For vertical tangent line:

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

Now use a trigonometric equation solver to get the solution of the above equation and for those values of $\theta$ , the value of $x$ at that point and $x=x_1$ is the equation of vertical tangent line at that point.

These tangents are:

  $x=0\text{, x}\approx 0.067,y\approx -1.104$