Questions: A spherical balloon is inflated and its volume increases at a rate of 16 in^3 / min. What is the rate of change of its radius when the radius is 5 in? The balloon's radius is increasing at a rate of (Round to three decimal places as needed.) at the instant the radius is 5 in.

Solution Steps

We are given that the volume of a spherical balloon is increasing at a rate of \(16 \, \text{in}^3/\text{min}\). We need to find the rate of change of the radius when the radius is 5 inches.

The volume \(V\) of a sphere with radius \(r\) is given by the formula: \[ V = \frac{4}{3} \pi r^3 \]

To find the rate of change of the radius, we need to differentiate the volume with respect to time \(t\). Using the chain rule, we have: \[ \frac{dV}{dt} = \frac{d}{dt} \left(\frac{4}{3} \pi r^3\right) = 4 \pi r^2 \frac{dr}{dt} \]

We know \(\frac{dV}{dt} = 16 \, \text{in}^3/\text{min}\) and \(r = 5 \, \text{in}\). Substitute these values into the differentiated equation: \[ 16 = 4 \pi (5)^2 \frac{dr}{dt} \]

Simplify and solve for \(\frac{dr}{dt}\): \[ 16 = 100 \pi \frac{dr}{dt} \] \[ \frac{dr}{dt} = \frac{16}{100 \pi} = \frac{4}{25 \pi} \]

Calculate the numerical value of \(\frac{dr}{dt}\) and round to three decimal places: \[ \frac{dr}{dt} \approx \frac{4}{25 \times 3.1416} \approx 0.0509 \]

Final Answer