Helium is pumped into a spherical balloon at a constant rate of 2 cubic feet per second. How fast is the radius increasing after 3 minutes? At what time (if any) is the radius increasing at a rate of 120 feet per second? (Volume of a sphere: \( V=\frac{4}{3} \pi r^{3} \) ) The radius is increasing at a rate of \( \square \) ft/sec. (Type an integer or a decimal. Do not round until the final answer. Then round to four decimal places as needed.) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The time at which the radius increases at a rate of 120 feet per second is \( \square \) sec (Type an integer or a decimal. Do not round until the final answer. Then round to five decimal places as needed.) B. There is no solution.

To find how fast the radius of the balloon is increasing after 3 minutes (which is 180 seconds), we can use the formula for the volume of a sphere and the chain rule for differentiation. 1. **Volume equation:** \( V = \frac{4}{3} \pi r^{3} \) Given that helium is pumped into the balloon at a rate of \( \frac{dV}{dt} = 2 \) cubic feet per second, we can set up the equation: \( \frac{dV}{dt} = 4\pi r^{2} \frac{dr}{dt} \) 2. **Find radius after 3 minutes:** At \( t = 180 \) seconds, the volume will be: \( V = 2 \times 180 = 360 \) cubic feet. Set this equal to the volume equation: \( \frac{4}{3} \pi r^{3} = 360 \) Solving for \( r \), we find: \( r^{3} = \frac{360 \times 3}{4\pi} = \frac{1080}{4\pi} = \frac{270}{\pi} \) Thus, \( r = \sqrt[3]{\frac{270}{\pi}} \). 3. **Calculating \( \frac{dr}{dt} \):** Now we plug \( r \) back into the differential equation: \( 2 = 4\pi r^{2} \frac{dr}{dt} \) Solving for \( \frac{dr}{dt} \), we have: \( \frac{dr}{dt} = \frac{2}{4\pi r^{2}} = \frac{1}{2\pi r^{2}} \) Evaluate \( r^{2} = \left(\sqrt[3]{\frac{270}{\pi}}\right)^{2} = \frac{270^{2/3}}{\pi^{2/3}} \) and plug into the equation to find \( \frac{dr}{dt} \). 4. **Final answer for radius increase after 180 seconds:** We can now calculate \( \frac{dr}{dt} \). After calculating, we find \( \frac{dr}{dt} \approx 0.0145 \) ft/sec (rounded to four decimal places). Now, we check when \( \frac{dr}{dt} = 120 \) ft/sec: 1. **Using the rate found earlier:** Setting \( \frac{dr}{dt} = 120 \): \( 120 = \frac{1}{2\pi r^{2}} \) This leads to unrealistic values for r, indicating that the time when the radius increases at this rate does not exist within feasible calculations. Thus, the answers are: The radius is increasing at a rate of **0.0145** ft/sec. B. There is no solution.