Chapter 10, Problem 26RE

To determine

To Find: The equations for the lines that are tangent to the given cardioid.

Answer to Problem 26RE

The equations are $y=x-1,y=-x-1$

Explanation of Solution

Given information: $r=1+\sin \theta$

Calculation:

The graph for the cardioid is:

  

When $\theta$ is $0$ and $\pi$ then cardioid touches the x-axis. Now, the equation of a line that contains points $\left(x_1,y_1\right)$

with slope m can be written as:

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ y\text{ = }m\left(x-x_1\right)+y_1\end{array}$

So, in order to find slope for a cardioid $\frac{dy}{dx}$ need to be calculated.

In polar coordinates, the point is $\left(r\cos \theta ,rsin\theta \right)$

As, the cardioid equation is $r=1+\sin \theta$ hence,

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

  $\begin{array}{l}\frac{dy}{d\theta }=\frac{dy}{dx}.\frac{dx}{d\theta }\\ \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}\\ \frac{dx}{d\theta }=-\sin \theta +\cos \theta +\sin \theta \left(-\sin \theta \right)\\ \frac{dx}{d\theta }=-\sin \theta +{\cos }^2\theta -{\sin }^2\theta \\ \frac{dx}{d\theta }=\cos 2\theta -\sin \theta \\ \frac{dy}{d\theta }=\cos \theta +2\sin \theta .\cos \theta \\ \frac{dy}{d\theta }=\sin 2\theta +\cos \theta \\ \frac{dy}{dx}=\frac{\sin 2\theta +\cos \theta }{\cos 2\theta -\sin \theta }\end{array}$

Now, at $\theta =0$

  $\begin{array}{l}{\left.\frac{dy}{dx}\right|}_{\theta =0}=\frac{\sin 2\times 0+\cos 0}{\cos 2\times 0-\sin 0}\\ {\left.\frac{dy}{dx}\right|}_{\theta =0}=1\end{array}$

Similarly,

  $\begin{array}{l}{\left.x\right|}_{\theta =0}=\cos 0+\sin 0.\cos 0\\ {\left.x\right|}_{\theta =0}=1\\ {\left.y\right|}_{\theta =0}=\sin 0+{\sin }^{2}0\\ {\left.y\right|}_{\theta =0}=0\end{array}$

This means the point of cardioid $r=1+\sin \theta$ is $\left(1,0\right)\text{ at }\theta =0$

So, the equation of tangent line becomes:

  $\begin{array}{l}y=1.\left(x-1\right)+0\\ y=x-1\end{array}$

Now, calculate at $\theta =\pi$

  $\begin{array}{l}{\left.\frac{dy}{dx}\right|}_{\theta =\pi }=\frac{\sin 2\times \pi +\cos \pi }{\cos 2\times \pi -\sin \pi }\\ {\left.\frac{dy}{dx}\right|}_{\theta =\pi }=-1\end{array}$

  $\begin{array}{l}{\left.x\right|}_{\theta =\pi }=\cos \pi +\sin \pi .\cos \pi \\ {\left.x\right|}_{\theta =\pi }=-1\\ {\left.y\right|}_{\theta =\pi }=\sin \pi +{\sin }^2\pi \\ {\left.y\right|}_{\theta =\pi }=0\end{array}$

So, the point becomes:

  $\left(-1,0\right)$

  $\begin{array}{l}y=-1.\left(x-\left(-1\right)\right)+0\\ y=-1.\left(x+1\right)\\ y=-x-1\end{array}$

Hence, the equations are $y=x-1,y=-x-1$