Chapter 9.4, Problem 11CFU

a.

To determine

To find the polar coordinate of the plane when it comes the closest to the tower.

Answer to Problem 11CFU

The coordinate is, $\left(p,\varphi \right)=\left(5,\frac{5\pi }{6}\right)$ .

Explanation of Solution

Given information:

The path of an airplane can be modelled by the equation $5=r\cos \left(\theta -\frac{5\pi }{6}\right)$ .

Calculation: The required solution is obtained as,

The polar form of the path is given by,

  $5=r\cos \left(\theta -\frac{5\pi }{6}\right)$

Conclude that $p=5$ and $\varphi =\frac{5\pi }{6}$

Since the shortest distance is along the normal,

Therefore $\left(p,\varphi \right)=\left(5,\frac{5\pi }{6}\right)$

Hence, the coordinate is, $\left(p,\varphi \right)=\left(5,\frac{5\pi }{6}\right)$ .

b.

To determine

To graph of the path of the plane .

Explanation of Solution

Given information: The path of an airplane can be modelled by the equation $5=r\cos \left(\theta -\frac{5\pi }{6}\right)$

Graph: Sketch the graph using graphing utility.

  

Interpretation:

  $5=r\cos \left(\theta -\frac{5\pi }{6}\right)$ is a polar equation far a line, so we need just two points to graph the libne.

If $\theta =\frac{5\pi }{6}$ then $r=\frac{5}{\cos 0}=5$

If $\theta =\frac{\pi }{2}$ then $r=\frac{5}{\cos \frac{\pi }{6}}=10$ .