### Home > APCALC > Chapter 7 > Lesson 7.1.5 > Problem7-49

7-49.

As Khalid inflates a spherical balloon, Kareem wonders about its different rates. He knows that the rate at which Khalid blows is equal to the rate at which the volume changes $( \frac { d V } { d t } )$. As the balloon inflates, other aspects are changing as well, such as the radius and the surface area.

1. If  $\frac { d V } { d t }=10 \ \frac { \text{cm} ^ { 3 } } { \text{sec} }$, calculate the rate of change of the radius, $\frac { d r } { d t }$, when $r = 3$ cm.

Using just geometry (no Calculus), find the equation for a generic spherical balloon.

$V=\frac{4}{3}\pi r^{3}$

$\text{Implicitly differentiate:}\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}$

Substitute the given information and solve.

2. If $\frac { d V } { d t }= 12 \ \frac { \text{cm} ^ { 3 } } { \text{sec} }$, calculate the rate of change of the surface area, $\frac { d A } { d t }$, when $r = 5$ cm.

$\text{Repeat the steps you used in part (a), this time evaluating at }r=5\text{ and }\frac{dV}{dt}=12.$

$\text{Find }\frac{dr}{dt}, \text{ you will need that information later.}$

Using just geometry, write the equation for the Surface Area of a generic spherical balloon: $A = 4πr^2$

$\text{Implicitly differentiate, then evaluate to find }\frac{dA}{dt} \text{ when }r=5\text{ cm.}$

3. Describe is happening to the balloon when $\frac { d V } { d t }$ is negative.

$\frac{dV}{dt}\text{ represents the RATE that the volume of the balloon changes (grows) with respect to time. }$

What does a negative rate represent?