Question

Air is being pumped into a spherical balloon at a rate of 1 (cm)^3/s. Determine the area's rate of change when the volume is 90 (cm)^3.

Answer 1

\[\frac{dv}{dt}=1\]

Answer 2

\[A= 4 \pi r^2 \]

Answer 3

\[V= \frac{4}{3} \pi r^3\]

Answer 4

we want dA/dt.

\[\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt} = \frac{dA}{dr} \times [ \frac{dr}{dv} \times \frac{dv}{dt}] = \frac{dA}{dr} \times [ \frac{1}{\frac{dv}{dr}} \times \frac{dv}{dt}] \]

Answer 5

\[\frac{dA}{dr} = 8 \pi r \]
\[\frac{dv}{dr} = 4 \pi r^2 \]

and given

\[\frac{dv}{dt} =1 \]

Answer 6

\[\frac{dA}{dt} = 8 \pi r \times \frac{1}{4\pi r^2} \times 1 = \frac{2}{r}\]

Answer 7

Now, we just need to find the value of r which corresponds to volume of 90

Answer 8

\[r= \sqrt[3]{\frac{3V}{4 \pi}} \]

Answer 9

The rest is putting values into a calculator , you can do that .