Chapter 10.3, Problem 39E

To determine

To find: The slope of the curve $r=-1+\sin \theta$ at $\theta =0,\pi$ .

Answer to Problem 39E

The slope at $\theta =0$ is $-1$ and the slope at $\theta =\pi$ is $1$ .

Explanation of Solution

Given information: The given curve is $r=-1+\sin \theta$ and the point is $\theta =0,\pi$ .

Formula used: The rectangular coordinates in polar form are $\left(x,y\right)=\left(r\cos \theta ,r\sin \theta \right)$ .

Calculation:

Substitute $-1+\sin \theta$ for $r$ in the equation $x=r\cos \theta$ .

  $\begin{array}{c}x=\left(-1+\sin \theta \right)\cos \theta \\ =\sin \theta \cos \theta -\cos \theta \end{array}$

Differentiate the above equation with respect to $\theta$ .

  $\begin{array}{c}\frac{dx}{d\theta }=\cos \theta \cdot \cos \theta -\sin \theta \cdot \sin \theta +\sin \theta \\ ={\cos }^2\theta -{\sin }^2\theta +\sin \theta \\ =\cos 2\theta +\sin \theta \end{array}$

Substitute $-1+\sin \theta$ for $r$ in the equation $y=r\sin \theta$ .

  $\begin{array}{c}y=\left(-1+\sin \theta \right)\sin \theta \\ ={\sin }^2\theta -\sin \theta \end{array}$

Differentiate the above equation with respect to $\theta$ .

  $\begin{array}{c}\frac{dy}{d\theta }=2\sin \theta \cos \theta -\cos \theta \\ =\sin 2\theta -\cos \theta \end{array}$

To find the slope, find the value of $\frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }$ .

  $\begin{array}{c}\frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }\\ =\frac{\sin 2\theta -\cos \theta }{\cos 2\theta +\sin \theta }\end{array}$

Find the value of $\frac{dy}{dx}$ at $\theta =0$ .

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

Find the value of $\frac{dy}{dx}$ at $\theta =\pi$ .

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

Therefore, the slope at $\theta =0$ is $-1$ and the slope at $\theta =\pi$ is $1$ .