Respuesta :
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.
