Chapter 10.8, Problem 20E

To determine

Find the maximum value and any zeros of r.

Answer to Problem 20E

The maximum value of r is 18

  $\theta =\frac{2\pi }{3}$ is the zero of $r=6+12\cos \theta$

Explanation of Solution

Given:

  $r=6+12\cos \theta$

Consider a following curve

  $r=6+12\cos \theta$

Sum Rule: If $f(x)$ and $g(x)$ are differentiable functions, then the derivative of the sum of two functions is the sum of their derivatives. That is

  $\frac{d}{dx}(f(x)+g(x))=\frac{d}{dx}f(x)+\frac{d}{dx}g(x)$

Constant Multiple Rule: Let $f(x)$ be a differentiable function and c being a constant. Then, $c\cdot f(x)$ is also differentiable such that

  $\frac{d}{dx}(c\cdot f(x))=c\cdot \frac{d}{dx}(f(x))$

It is known that,

  $\frac{d}{dx}(\sin x)=\cos x$

  $\frac{d}{dx}(\cos x)=-\sin x$

  $\frac{d}{dx}(a)=0,$ a being a constant

For maximum value of r, find derivative of r and equate to zero

  $\begin{array}{r}\hfill \frac{dr}{d\theta }=0\\ \hfill \frac{d}{d\theta }(6+12\cos \theta )=0\\ \hfill \frac{d}{d\theta }(6)+\frac{d}{d\theta }(12\cos \theta )=0\\ \hfill 12\frac{d}{d\theta }(\cos \theta )=0\\ \hfill -12\sin \theta =0\\ \hfill \sin \theta =0\end{array}$

This implies that $\theta =0$

Find second derivative of $r$ ,

  $\begin{array}{cc}\frac{d^{2}r}{d{\theta }^2}& =\frac{d}{d\theta }(-12\sin \theta )\\ & =-12\cos \theta \end{array}$

Now,

  ${\left(\frac{d^{2}r}{d{\theta }^2}\right)}_{\theta =0}=-12\cos 0$

  $=-12<0$

Thus maximum is value of $r$ is at $\theta =0$ , and

  $r_{\max }=6+12\cos 0$

  $\begin{array}{l}=6+12\hfill \\ =18\hfill \end{array}$

Therefore, the maximum value of r is 18

For zeros of $r$ , set $r=0,$ , that is

  $\begin{array}{cc}6+12\cos \theta & =0\\ \cos \theta & =\frac{-1}{2}\\ \theta & =\frac{2\pi }{3}\end{array}$

Therefore, $\theta =\frac{2\pi }{3}$ is the zero of $r=6+12\cos \theta$