Chapter 11, Problem 26RE

To determine

To determine: The equations for the lines that are tangent to the cardioids at the points where it crosses the x -axis.

Answer to Problem 26RE

The required equations are $y=x-1$ and $y=-x-1$ .

Explanation of Solution

Given information:

The equation cardioid: $r=1+\sin \theta$

Formula used:

Product rule of differentiation:

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

Calculation:

The given equation is $r=1+\sin \theta$ .

The equation shows that the equation of the cardioids crosses the x -axis at the points where $\theta =0$ and $\theta =\pi$ .

The above are the points where $\theta =\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$ .

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:

  $x=\left(1+\sin \theta \right)\cos \theta$ , $y=\left(1+\sin \theta \right)\sin \theta$

Here, the polar curves lie in the xy -plane. This means the slope of the polar curve and the sloe of the tangent line are same.

So, $\begin{array}{l}\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}\\ \frac{dy}{dx}=\frac{\frac{d}{d\theta }\left(\left(1+\sin \theta \right)\sin \theta \right)}{\frac{d}{d\theta }\left(\left(1+\sin \theta \right)\cos \theta \right)}\\ \frac{dy}{dx}=\frac{\cos \theta \sin \theta +\left(1+\sin \theta \right)\cos \theta }{\cos \theta \cos \theta +\left(1+\sin \theta \right)\sin \theta }\end{array}$

For $\theta =0$ ; $\left(x,y\right)=\left(1,0\right)$ , $m=\frac{dy}{dx}=1$ and the equation of the tangent line is $y=x-1$ .

For $\theta =\pi$ ; $\left(x,y\right)=\left(-1,0\right)$ , $m=\frac{dy}{dx}=-1$ and the equation of the tangent line is $y=-x-1$ .

Hence, the required equations are $y=x-1$ and $y=-x-1$ .