Chapter 8.3, Problem 65PS

Solution

(a)

To calculate : The number of miles for which the plane will be in the top-most layer of Class B airspace.

The plane will be in the top-most layer of Class B airspace for a distance of $28.9\text{ miles}$ .

Given information :

"Class B" airspace sometimes consists of a stack of cylindrical layers as shown. Seen from above, the airspace forms circle whose origin is the control tower. A plane in straight and level flight flies through the top layer along the line $y=-4$ ,as shown.

  

Calculation :

The top layer of Class B airspace is represented by $x^2+y^2={15}^2$ .

When the plane enters the top layer of Class B airspace, it will have a negative $x-$ coordinate and on leaving the top layer it will have a positive $x-$ coordinate. In both the cases, the $y-$ coordinate will be $-4$ .

Represent the situation using a figure as below:

  

Substitute $y=-4$ in $x^2+y^2={15}^2$ and solve for $x$ .

  $\begin{array}{c}{x}^2+{\left(-4\right)}^2={15}^2\\ x^2+16=225\\ x^2=209\\ x=\pm 14.45\end{array}$

So, when the plane enters the top layer of Class B airspace, its $x-$ coordinate will be $-14.45$ and on leaving the top layer, its $x-$ coordinate will be $14.45$ .

So, the plane will be in top layer at the distance of

  $\begin{array}{c}\text{Distance}=14.45-\left(-14.5\right)\\ =28.9\text{ miles}\end{array}$

Therefore, the plane will be in the top-most layer of Class B airspace for a distance of $28.9\text{ miles}$ .

(b)

To calculate : The number of miles for which the plane will be above the middle layer of Class B airspace.

The plane will be above the middle layer of Class B airspace for a distance of $18.32\text{ miles}$ .

Given information :

"Class B" airspace sometimes consists of a stack of cylindrical layers as shown. Seen from above, the airspace forms circle whose origin is the control tower. A plane in straight and level flight flies through the top layer along the line $y=-4$ ,as shown.

  

Calculation :

The middle layer of Class B airspace is represented by $x^2+y^2={10}^2$ .

When the plane enters the middle layer of Class B airspace, it will have a negative $x-$ coordinate and on leaving the middle layer it will have a positive $x-$ coordinate. In both the cases, the $y-$ coordinate will be $-4$ .

Represent the situation using a figure as below:

  

Substitute $y=-4$ in $x^2+y^2={10}^2$ and solve for $x$ .

  $\begin{array}{c}{x}^2+{\left(-4\right)}^2={10}^2\\ x^2+16=100\\ x^2=84\\ x=\pm 9.16\end{array}$

So, when the plane enters the middle layer of Class B airspace, its $x-$ coordinate will be $-9.16$ and on leaving the middle layer, its $x-$ coordinate will be $9.16$ .

So, the plane will be above middle layer at the distance of,

  $\begin{array}{c}\text{Distance}=9.16-\left(-9.16\right)\\ =18.32\text{ miles}\end{array}$

Therefore, the plane will be above the middle layer of Class B airspace for a distance of $18.32\text{ miles}$ .

(c)

To calculate : The number of miles for which the plane will be above the lowest layer of Class B airspace.

The plane will be above the lowest layer of Class B airspace for a distance of $6\text{ miles}$ .

Given information :

"Class B" airspace sometimes consists of a stack of cylindrical layers as shown. Seen from above, the airspace forms circles whose origin is the control tower. A plane in straight and level flight flies through the top layer along the line $y=-4$ ,as shown.

  

Calculation :

The lowest layer of Class B airspace is represented by $x^2+y^2=5^2$ .

When the plane enters the lowest layer of Class B airspace, it will have a negative $x-$ coordinate and on leaving the lowest layer it will have a positive $x-$ coordinate. In both the cases, the $y-$ coordinate will be $-4$ .

Represent the situation using a figure as below:

  

Substitute $y=-4$ in $x^2+y^2=5^2$ and solve for $x$ .

  $\begin{array}{c}{x}^2+{\left(-4\right)}^2=5^2\\ x^2+16=25\\ x^2=9\\ x=\pm 3\end{array}$

So, when the plane enters the lowest layer of Class B airspace, its $x-$ coordinate will be $-3$ and on leaving the lowest layer, its $x-$ coordinate will be $3$ .

So, the plane will be above lowest layer at the distance of

  $\begin{array}{c}\text{Distance}=3-\left(-3\right)\\ =6\text{ miles}\end{array}$

Therefore, the plane will be above the lowest layer of Class B airspace for a distance of $6\text{ miles}$ .