Chapter 10.3, Problem 41E

To determine

To find: The slope of the curve $r=2-3\sin \theta$ at points $\left(2,0\right),\left(2,\pi \right),\left(5,\frac{3\pi }{2}\right)$ , and $\left(-1,\frac{\pi }{2}\right)$ .

Answer to Problem 41E

The slope of the curve at $\left(2,0\right)$ is $-\frac{2}{3}$ , at $\left(-1,\frac{\pi }{2}\right)$ is $0$ , at $\left(2,\pi \right)$ is $\frac{2}{3}$ and at $\left(5,\frac{3\pi }{2}\right)$ is $0$ .

Explanation of Solution

Given information: The given equation of curve is $r=2-3\sin \theta$ . The points are $\left(2,0\right),\left(2,\pi \right),\left(5,\frac{3\pi }{2}\right)$ , and $\left(-1,\frac{\pi }{2}\right)$ .

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

Calculation:

Substitute $2-3\sin \theta$ for $r$ in the equation $x=r\cos \theta$ .

  $x=2\cos \theta -3\sin \theta \cos \theta$

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

  $\frac{dx}{d\theta }=-2\sin \theta +3{\sin }^2\theta -3{\cos }^2\theta$

Substitute $2-3\sin \theta$ for $r$ in the equation $y=r\sin \theta$ .

  $y=2\sin \theta -3{\sin }^2\theta$

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

  $\frac{dy}{d\theta }=2\cos \theta -6\sin \theta \cos \theta$

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

  $\begin{array}{c}\frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }\\ =\frac{2\cos \theta -6\sin \theta \cos \theta }{-2\sin \theta +3{\sin }^2\theta -3{\cos }^2\theta }\\ =\frac{2\cos \theta -3\cdot 2\sin \theta \cos \theta }{-2\sin \theta -3\left({\cos }^2\theta -{\sin }^2\theta \right)}\\ =\frac{2\cos \theta -3\sin 2\theta }{-2\sin \theta -3\cos 2\theta }\end{array}$

At the point $\left(2,0\right)$ , $\theta =0$ . Find the value of $\frac{dy}{dx}$ at $\theta =0$ .

  $\begin{array}{c}{\left.\frac{dy}{dx}\right|}_{\theta =0}=\frac{2\cos \left(0\right)-3\sin 2\left(0\right)}{-2\sin \left(0\right)-3\cos 2\left(0\right)}\\ =\frac{2\cdot 1-3\cdot 0}{-2\cdot 0-3\cdot 1}\\ =-\frac{2}{3}\end{array}$

At the point $\left(-1,\frac{\pi }{2}\right)$ , $\theta =\frac{\pi }{2}$ . Find the value of $\frac{dy}{dx}$ at $\theta =\frac{\pi }{2}$ .

  $\begin{array}{c}{\left.\frac{dy}{dx}\right|}_{\theta =\frac{\pi }{2}}=\frac{2\cos \left(\frac{\pi }{2}\right)-3\sin 2\left(\frac{\pi }{2}\right)}{-2\sin \left(\frac{\pi }{2}\right)-3\cos 2\left(\frac{\pi }{2}\right)}\\ =\frac{2\cdot 0-3\cdot \sin \pi }{-2\cdot 1-3\cdot \cos \pi }\\ =\frac{-3\cdot 0}{-2-3\left(-1\right)}\\ =0\end{array}$

At the point $\left(2,\pi \right)$ , $\theta =\pi$ . Find the value of $\frac{dy}{dx}$ at $\theta =\pi$ .

  $\begin{array}{c}{\left.\frac{dy}{dx}\right|}_{\theta =\pi }=\frac{2\cos \left(\pi \right)-3\sin 2\left(\pi \right)}{-2\sin \left(\pi \right)-3\cos 2\left(\pi \right)}\\ =\frac{2\cdot \left(-1\right)-3\cdot \sin 2\pi }{-2\cdot 0-3\cdot \cos 2\pi }\\ =\frac{-2-3\cdot 0}{-3\left(1\right)}\\ =\frac{2}{3}\end{array}$

At the point $\left(5,\frac{3\pi }{2}\right)$ , $\theta =\frac{3\pi }{2}$ . Find the value of $\frac{dy}{dx}$ at $\theta =\frac{3\pi }{2}$ .

  $\begin{array}{c}{\left.\frac{dy}{dx}\right|}_{\theta =\frac{3\pi }{2}}=\frac{2\cos \left(\frac{3\pi }{2}\right)-3\sin 2\left(\frac{3\pi }{2}\right)}{-2\sin \left(\frac{3\pi }{2}\right)-3\cos 2\left(\frac{3\pi }{2}\right)}\\ =\frac{2\cdot \left(0\right)-3\cdot \sin 3\pi }{-2\cdot \left(-1\right)-3\cdot \cos 3\pi }\\ =\frac{-3\cdot 0}{2-3\left(-1\right)}\\ =0\end{array}$

Therefore, the slope of the curve at $\left(2,0\right)$ is $-\frac{2}{3}$ , at $\left(-1,\frac{\pi }{2}\right)$ is $0$ , at $\left(2,\pi \right)$ is $\frac{2}{3}$ and at $\left(5,\frac{3\pi }{2}\right)$ is $0$ .