A spherical party balloon is being inflated with helium pumped in at a rate of 8 cubic feet per minute. How fast is the radius growing at the instant when the radius has reached 4 ft? Round your answer to two decimal places. (Hint: The volume of a sphere of radius \( r \) is \( V=\frac{4}{3} \pi r^{3} \).)

To find how fast the radius is growing when the radius is 4 ft, we first need to relate the volume of the sphere to its radius. The formula for the volume of a sphere is given by \( V = \frac{4}{3} \pi r^3 \). We can differentiate both sides with respect to time \( t \): \[ \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \] We know that \( \frac{dV}{dt} = 8 \) cubic feet per minute (the rate at which helium is being pumped in) and we want to determine \( \frac{dr}{dt} \) when \( r = 4 \) ft. First, we can substitute these values into the differentiated equation: \[ 8 = 4 \pi (4^2) \frac{dr}{dt} \] Calculating \( 4^2 \) gives us 16, so: \[ 8 = 4 \pi (16) \frac{dr}{dt} \] This simplifies to: \[ 8 = 64 \pi \frac{dr}{dt} \] Now, divide both sides by \( 64 \pi \): \[ \frac{dr}{dt} = \frac{8}{64\pi} = \frac{1}{8\pi} \] Now let's calculate \( \frac{dr}{dt} \) to two decimal places: \[ \frac{1}{8\pi} \approx \frac{1}{25.13} \approx 0.0397 \] Rounding to two decimal places, we get: \[ \frac{dr}{dt} \approx 0.04 \text{ ft/min} \] So, the radius is growing at approximately 0.04 feet per minute when the radius reaches 4 feet!