A rather flimsy spherical balloon is designed to pop at the instant its radius has reached 3 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. (The volume of a sphere of radius r is V = 4/3 πr3. Round your answer to two decimal places.)

Answer:The radius is growing at the rate of 0.09 centimeter per second at the instant it pops.            Step-by-step explanation: We are given the following information in the question:A spherical balloon is designed to pop at the instant its radius has reached 3 centimeters.The balloon is filled with helium at a rate of 10 cubic centimeters per second .[tex]\displaystyle\frac{dV}{dt} = 10~cm^3/sec.[/tex]We have to find fast the radius is growing at the instant it pops.Volume of spherical balloons = [tex]\displaystyle\frac{4}{3}\pi r^3[/tex]Differentiating, we get,[tex]\displaystyle\frac{dV}{dt} = \frac{4}{3}\pi 3r^2 \displaystyle\frac{dr}{dt} = 4\pi r^2 \displaystyle\frac{dr}{dt}[/tex]Putting the values, we get,[tex]10 = 4\times 3.14\times (3)^2\times \displaystyle\frac{dr}{dt}\\\\\frac{dr}{dt} = \frac{10}{4\times 3.14\times (3)^2} = 0.0884 \approx 0.09\\\\\frac{dr}{dt} = 0.09\text{ centimeter per second}[/tex]The radius is growing at the rate of 0.09 centimeter per second at the instant it pops.