Chapter 1.3, Problem 54E

(a)

To determine

To express: The radius of the balloon as a function of time t.

Answer to Problem 54E

The function of the radius of the balloon in terms of time t is $r\left(t\right)=2t$.

Explanation of Solution

It is given that the balloon is being inflated and therefore the radius is increasing at a rate of 2 cm/s.

Let the radius of the balloon be r.

Recall the formula, Distance = Time × Speed.

Note that the distance is same as radius.

Substitute t for time and 2 for speed in a distance formula.

Thus, the function is $r\left(t\right)=2t$ where r(t) is measured in cm.

Therefore, the radius of the spherical balloon after t seconds is $r\left(t\right)=2t$.

(b)

To determine

To find: The expression for $\left(V\circ r\right)\left(t\right)$ and interpret where V is the volume of the function of radius.

Answer to Problem 54E

The value of $\left(V\circ r\right)\left(t\right)=\frac{32}{3}\pi t^3$, which represents the area of the circular ripple at a time t.

Explanation of Solution

Area of the circle is, $V\left(r\right)=\frac{4}{3}\pi {\left(r\left(t\right)\right)}^3$ (1)

The composite function $\left(V\circ r\right)\left(t\right)$ is defined as $\left(V\circ r\right)\left(t\right)=V\left(r\left(t\right)\right)$.

From part (a), the value of $r\left(t\right)=2t$.

Thus, $\left(V\circ r\right)\left(t\right)=V\left(2t\right)$.

Substitute $2t$ in equation (1),

$\begin{array}{c}V\left(r\right)=\frac{4}{3}\pi {\left(2t\right)}^3\\ =\frac{4}{3}\pi \left(8t^3\right)\\ =\frac{32}{3}\pi t^3\end{array}$

Thus, $\left(V\circ r\right)\left(t\right)=\frac{32}{3}\pi t^3$, which represents the area of the circular ripple with respect to the radius of time t.