Question

a spherical balloon is inflated and its volume increases at a rate of 19 in.³/min. what is the rate of change of its radius when the radius is 12 in.?
the balloons radius is increasing at a rate of at the instant the radius is 12 in.
(round to three decimal places as needed.)

Explanation:

Step1: Write the volume formula for a sphere

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

Step2: Differentiate both sides 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 \(\frac{dV}{dt} = 19\) and \(r = 12\). Rearranging the equation \(\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}\) for \(\frac{dr}{dt}\), we get \(\frac{dr}{dt}=\frac{\frac{dV}{dt}}{4\pi r^{2}}\).

Step4: Substitute the given values

Substitute \(\frac{dV}{dt}=19\) and \(r = 12\) into the formula: \(\frac{dr}{dt}=\frac{19}{4\pi\times(12)^{2}}\).
Calculate \(4\pi\times(12)^{2}=4\pi\times144 = 576\pi\). Then \(\frac{dr}{dt}=\frac{19}{576\pi}\).
Using a calculator, \(\frac{19}{576\pi}\approx\frac{19}{576\times3.14159}\approx\frac{19}{1809.557}\approx0.011\).

Answer:

\(0.011\) in/min