Question

1. a spherical balloon is expanding at a rate of ( 60pi \text{ in}^3/\text{sec} ). how fast is the surface area of the balloon expanding when the radius of the balloon is 4 inches? ( v=\frac{4}{3}pi r^3 ) and ( a = 4pi r^2 ).

Explanation:

Step1: Differentiate volume formula

Given \(V=\frac{4}{3}\pi r^{3}\), differentiate with respect to \(t\):
\(\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}\).
We know \(\frac{dV}{dt} = 60\pi\) and \(r = 4\). Substitute values:
\(60\pi=4\pi(4)^{2}\frac{dr}{dt}\).
Solve for \(\frac{dr}{dt}\):
\(\frac{dr}{dt}=\frac{60\pi}{16\pi}=\frac{15}{4}\).

Step2: Differentiate surface - area formula

Given \(A = 4\pi r^{2}\), differentiate with respect to \(t\):
\(\frac{dA}{dt}=8\pi r\frac{dr}{dt}\).
Substitute \(r = 4\) and \(\frac{dr}{dt}=\frac{15}{4}\):
\(\frac{dA}{dt}=8\pi\times4\times\frac{15}{4}\).
Calculate:
\(\frac{dA}{dt}=120\pi\).

Answer:

The surface area of the balloon is expanding at a rate of \(120\pi\space in^{2}/sec\).