Chapter 9.4, Problem 32E

a.

To determine

To write that how close did the fly come to the origin.

Answer to Problem 32E

Closest the fly came to origin was 6 cm.

Explanation of Solution

Given information:

The equation of the path of the fly is $r=6\sec \left(\theta -15°\right)$

Write the equation and rearrange it in polar form.

  $\begin{array}{l}r=6\sec \left(\theta -15°\right)\\ \frac{r}{\sec \left(\theta -15°\right)}=6\left[\text{divide each side by}\sec \left(\theta -15°\right)\right]\\ r\cos \left(\theta -15°\right)=6\left[\frac{1}{\sec x}=\cos x\right]\\ 6=r\cos \left(\theta -15°\right)\end{array}$

Here $p=6$ , since the shortest distance is along the normal.

Therefore,

Closest the fly came to origin was 6 cm.

b.

To determine

To write that polar coordinates of the fly when it was closest to origin.

Answer to Problem 32E

The polar coordinates of the fly when it was closest to origin was $\left(6,16\right)$

Explanation of Solution

Given information:

The equation of the path of the fly is $r=6\sec \left(\theta -15°\right)$

Write the polar form of the equation.

  $6=r\cos \left(\theta -15°\right)$

Since the shortest distance occurs along the normal. Therefore the polar coordinates are

  $\left(p,\phi \right)=\left(6,16\right)$