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 determine how fast the radius of the balloon is growing at the instant it pops, we'll use the relationship between the volume of a sphere and its radius. **Given:** - Volume of a sphere: \( V = \frac{4}{3} \pi r^3 \) - Rate at which helium is added: \( \frac{dV}{dt} = 10 \) cm³/s - Radius at the instant of popping: \( r = 8 \) cm **Steps:** 1. **Differentiate the volume with respect to time:** \[ \frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right) = 4 \pi r^2 \frac{dr}{dt} \] 2. **Plug in the known values:** \[ 10 = 4 \pi (8)^2 \frac{dr}{dt} \] \[ 10 = 4 \pi \times 64 \frac{dr}{dt} \] \[ 10 = 256 \pi \frac{dr}{dt} \] 3. **Solve for \(\frac{dr}{dt}\):** \[ \frac{dr}{dt} = \frac{10}{256 \pi} = \frac{5}{128 \pi} \text{ cm/s} \] 4. **Calculate the numerical value:** \[ \frac{5}{128 \pi} \approx \frac{5}{402.123} \approx 0.01243 \text{ cm/s} \] 5. **Round to two decimal places:** \[ \frac{dr}{dt} \approx 0.01 \text{ cm/s} \] **Answer:** At the instant the balloon pops, the radius is increasing at approximately 0.01 cm per second.