Chapter 8.7, Problem 4MRPS

Solution

a.

To find: To find the shortest possible horizontal distance from the jet when hearing first sonic boom.

The horizontal distance from the jet is $\text{6 mi}$ .

Given information : The jet makes a sonic boom heard along $\frac{x^2}{36}-\frac{y^2}{100}=1$ .

Calculation:

When the observer is positioned at the hyperbola's vertex, the horizontal distance between them and the plane is the shortest.

As a result of the plane being at the cone's vertex, the distance between them is equal to the horizontal half-axis of the hyperbola.

  $\frac{x^2}{36}-\frac{y^2}{100}=1$

Therefore,

  $\text{a = }\sqrt{\text{36}}\text{ = 6 mi}$

Conclusion:

The horizontal distance from the jet when hearing the first sonic boom is $\text{6 mi}$ .

b.

To find: To find the shortest possible horizontal distance from the jet when hearing second sonic boom.

The horizontal distance from the jet is $\text{3 mi}$ .

Given information : The jet makes a sonic boom heard along $\frac{x^2}{9}-\frac{y^2}{25}=1$ .

Calculation:

Similarly comparing to the part (a), when the observer is positioned at the hyperbola's vertex, the horizontal distance between them and the plane is the shortest.

As a result of the plane being at the cone's vertex, the distance between them is equal to the horizontal half-axis of the hyperbola.

  $\frac{x^2}{9}-\frac{y^2}{25}=1$

Therefore,

  $\begin{array}{c}\text{a = }\sqrt{\text{9}}\\ \text{= 3 miles}\end{array}$

Conclusion:

The horizontal distance from the jet when hearing the first sonic boom is $\text{3 mi}$ .

c.

To find: To describe the relationship between the two hyperbolas.

The first hyperbola is simply the second one scaled in both directions by a factor of $2$ .

Given information : The two hyperbola’s equations are $\frac{x^2}{36}-\frac{y^2}{100}=1$ and $\frac{x^2}{9}-\frac{y^2}{25}=1$ .

Calculation:

The hyperbola equation from the first sonic boom is,

  $\frac{x^2}{36}-\frac{y^2}{100}=1$

The hyperbola equation from the second sonic boom is,

  $\frac{x^2}{9}-\frac{y^2}{25}=1$

Comparing the above two hyperbolae’s, to get $9\cdot 4=36\text{ and }25.4=100$

As a result, the first hyperbola is half-axes double the length of the second hyperbola.

Conclusion:

The first hyperbola is simply the second one scaled in both directions by a factor of $2$ .