Question

1. a spherical balloon is expanding at a rate of 60π in³/sec. how fast is the surface area of the balloon expanding when the radius of the balloon is 4 inches? v = 4/3πr³ and a = 4πr². answer: 30π in²/sec

Explanation:

Step1: Differentiate volume formula

Given \(V=\frac{4}{3}\pi r^{3}\), differentiate with respect to \(t\) (time).
Using the chain - rule \(\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}\).
We know \(\frac{dV}{dt} = 60\pi\) in³/sec.
So, \(60\pi=4\pi r^{2}\frac{dr}{dt}\).

Step2: Solve for \(\frac{dr}{dt}\)

Divide both sides of \(60\pi = 4\pi r^{2}\frac{dr}{dt}\) by \(4\pi r^{2}\).
\(\frac{dr}{dt}=\frac{60\pi}{4\pi r^{2}}=\frac{15}{r^{2}}\).
When \(r = 4\) inches, \(\frac{dr}{dt}=\frac{15}{4^{2}}=\frac{15}{16}\) in/sec.

Step3: Differentiate surface - area formula

Given \(A = 4\pi r^{2}\), differentiate with respect to \(t\) (time).
Using the chain - rule \(\frac{dA}{dt}=8\pi r\frac{dr}{dt}\).

Step4: Substitute \(r = 4\) and \(\frac{dr}{dt}=\frac{15}{16}\)

\(\frac{dA}{dt}=8\pi\times4\times\frac{15}{16}\).
First, \(8\times4\times\frac{15}{16}=\frac{8\times4\times15}{16}=\frac{480}{16} = 30\).

Answer:

\(30\pi\) in²/sec