A spherical balloon is designed to pop at the instant its radius has reached 8 centimeters. Assuming the balloon is filled with helium at a rate of 10 cubic centimeters per second, calculate how fast the radius is growing at the instant it pops. (Round your answer to two decimal places.) (The volume of a sphere with radius \( r \) is given by \( V=\frac{4}{3} \pi r^{3} \).)

To find out how fast the radius of the balloon is growing when its radius is 8 centimeters, we first need to relate the volume of the sphere to its radius and the given rate of volume increase. Given the volume \( V = \frac{4}{3} \pi r^3 \), we can differentiate this with respect to time \( t \): \[ \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \] We know that \( \frac{dV}{dt} = 10 \) cubic centimeters per second (the rate at which the helium is filling the balloon) and we need to find \( \frac{dr}{dt} \) when \( r = 8 \) cm. Substituting in the values we have: \[ 10 = 4 \pi (8^2) \frac{dr}{dt} \] Calculating \( 8^2 \): \[ 10 = 4 \pi (64) \frac{dr}{dt} \] \[ 10 = 256 \pi \frac{dr}{dt} \] Now we solve for \( \frac{dr}{dt} \): \[ \frac{dr}{dt} = \frac{10}{256 \pi} \] \[ \frac{dr}{dt} = \frac{10}{256 \cdot 3.14159} \approx \frac{10}{804.247} \approx 0.01243 \] Rounding this to two decimal places, we have: \[ \frac{dr}{dt} \approx 0.01 \text{ cm/s} \] Therefore, the radius of the balloon is growing at approximately **0.01 cm/s** at the instant it pops.