Chapter 10, Problem 24RE

To determine

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

Answer to Problem 24RE

The equation for the horizontal and vertical tangent lines are $\left(2-4\sin \theta \right)\cos \theta =0$ and $2\sin \theta \left(\sin \theta -1\right)-2{\cos }^2\theta =0$ respectively.

Explanation of Solution

Given information:

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

Consider the given equation:

  $r=2\left(1-\sin \theta \right),0\le \theta \le 2\pi$

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,

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

And the slope is:

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

For horizontal line:

  $\begin{array}{c}\frac{dy}{dx}=0\\ \frac{d}{d\theta }\left(2\left(1-\sin \theta \right)\sin \theta \right)=0\\ \left(2-4\sin \theta \right)\cos \theta =0\end{array}$

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:

  $y=\frac{1}{2}\text{, }y=-4$

For vertical tangent line:

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

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:

  $x=0,x\approx \pm 2.598$