Question

a spherical balloon is inflated and its volume increases at a rate of 23 in.³/min. what is the rate of change of its radius when the radius is 8 in.? the balloons radius is increasing at a rate of at the instant the radius is 8 in. (round to three decimal places as needed.)

Explanation:

Step1: Recall the volume formula for a sphere

The volume formula for a sphere is \(V=\frac{4}{3}\pi r^{3}\).

Step2: Differentiate the volume formula with respect to time \(t\)

Using the chain - rule, \(\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}\).

Step3: Solve for \(\frac{dr}{dt}\)

We know that \(\frac{dV}{dt} = 23\) in³/min and \(r = 8\) in. Rearranging the equation \(\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}\) for \(\frac{dr}{dt}\), we get \(\frac{dr}{dt}=\frac{\frac{dV}{dt}}{4\pi r^{2}}\).
Substitute \(\frac{dV}{dt}=23\) and \(r = 8\) into the formula: \(\frac{dr}{dt}=\frac{23}{4\pi(8)^{2}}\).

Step4: Calculate the value

First, calculate \(4\pi(8)^{2}=4\pi\times64 = 256\pi\). Then \(\frac{dr}{dt}=\frac{23}{256\pi}\approx\frac{23}{256\times3.14159}\approx\frac{23}{804.247}\approx0.029\) in/min.

Answer:

\(0.029\) in/min