Chapter 3.6, Problem 52E

(a)

To determine

To find: the changes in graph of amount of water when the pool is filled at rate of $1360$ gallons per hour.

Answer to Problem 52E

  $u\left(t\right)=1.33v\left(t\right)$ .

Explanation of Solution

Given:

  $v\left(t\right)=1020t$ .

Concept used:

When by either $f\left(x\right)$ or $x$ is multiplied by a number, function can stretch or shrink vertically or horizontally, respectively, when graphed by the equation:

  $y=bf\left(x\right)$ .

Calculation:

When a large hole is found then the pool is filled at a rate of $1360$ gallons per hour so, there is a vertical stretch by a factor $\frac{133}{100}$ of the graph $v\left(t\right)$ .

  $\begin{array}{l}u\left(t\right)=1360t.\\ \Rightarrow \frac{133}{100}\left(v\left(t\right)\right).\\ u\left(t\right)=1.33v\left(t\right).\end{array}$

Hence,

  $u\left(t\right)=1.33v\left(t\right)$ .

(b)

To determine

To find: the changes in graph of amount of water before filling up the pool with and added with $2000$ gallons of water to the pool.

Answer to Problem 52E

vertical translation $2000$ unit up.

Explanation of Solution

Given:

  $v\left(t\right)=1020t$ .

Concept used:

In a geometry vertical translation is a translation of a geometric object in a direction parallel to the vertical axis of the cartesian coordinate system.

  $y=f\left(x\right)+k$ is the vertical translation of the graph $y=f\left(x\right)$ .

Calculation:

There is a vertical translation $2000$ unit up.

Hence, there is vertical translation $2000$ unit up.