Chapter 8.4, Problem 2MRPS

Solution

(a)

To determine : An inequality that describes the region covered by the radar if the fishing boat is anchored at the origin.

The required inequality is $x^2+y^2\le {16}^2$ .

Given information :

A fishing boat's radar has a range of $16$ miles. A second boat is $12$ miles west and $12$ miles south of the fishing boat.

Explanation :

Draw the figure on the basis of the given information.

  

Therefore, the required inequality is $x^2+y^2\le {16}^2$ .

(b)

To calculate : The second boat is in range of the radar or not.

The second boat is not in radar.

Given information :

A fishing boat's radar has a range of $16$ miles. A second boat is $12$ miles west and $12$ miles south of the fishing boat.

Calculation :

The coordinates of the fishing boat anchored at the origin is $\left(0,0\right)$ . The coordinates of the second boat that is $12$ miles west and $12$ miles south of the fishing boat will be $\left(12,12\right)$ .

The inequality that describes the region covered by the radar if the fishing boat is anchored at the origin is $x^2+y^2\le {16}^2$ .

The distance of the second boat from the fishing boat is,

  $\begin{array}{c}\text{Distance}=\sqrt{{\left(12\right)}^2+{\left(12\right)}^2}\\ =\sqrt{144+144}\\ =\sqrt{288}\\ =16.9\text{ miles}\end{array}$

The distance of the second boat from the fishing boat is more than $16$ .

Therefore, the second boat is not in radar.

(c)

To calculate : The distance for which the third boat will be in radar range of the fishing boat.

The third boat will be in radar for $8.8\text{ miles}$ more.

Given information :

A fishing boat's radar has a range of $16$ miles. A second boat is $12$ miles west and $12$ miles south of the fishing boat. A third boat $6$ miles north and $4$ miles east of the fishing boat begins moving westward.

Calculation :

The coordinates of the fishing boat anchored at the origin is $\left(0,0\right)$ . The coordinates of the second boat that is $12$ miles west and $12$ miles south of the fishing boat will be $\left(4,6\right)$ .

The inequality that describes the region covered by the radar if the fishing boat is anchored at the origin is $x^2+y^2\le {16}^2$ .

The distance of the third boat from the fishing boat is,

  $\begin{array}{c}\text{Distance}=\sqrt{{\left(4\right)}^2+{\left(6\right)}^2}\\ =\sqrt{16+36}\\ =\sqrt{52}\\ =7.2\text{ miles}\end{array}$

The distance of the third boat from the fishing boat is less than $16$ so it is presently in radar but as it has begun moving westward, so it will be in radar for,

  $16-7.2=8.8\text{ miles}$

Therefore, the third boat will be in radar for $8.8\text{ miles}$ more.