The radius of a sphere increases at a rate of \( 1 \mathrm{~m} / \mathrm{sec} \). Find the rate at which the volume increases when the radius is 20 m . Type answer with Pi and unit

To find the rate at which the volume of a sphere increases, we use the formula for the volume of a sphere, \( V = \frac{4}{3} \pi r^3 \). First, we differentiate this with respect to time \( t \): \[ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \] Given that \( \frac{dr}{dt} = 1 \) m/s and \( r = 20 \) m, we substitute these values into the equation: \[ \frac{dV}{dt} = 4\pi (20^2)(1) = 4\pi (400) = 1600\pi \] Thus, the rate at which the volume increases when the radius is 20 m is \( 1600\pi \, \text{m}^3/\text{s} \).