Question

A spherical party balloon is being inflated with helium pumped in at a rate of 10 cubic feet per minute. How fast is the radius growing at the instant when the radius has reached 1 ft? HINT [See Example 1.] (The volume of a sphere of radius r is V = 4/3 πr3. Round your answer to two decimal places.)

Answer 1

Answer:

The radius is growing at a rate of   $\frac{dr}{dt} = \frac{10}{4 \pi} = 0.795$ ft per minute.

Step-by-step explanation:

From the information given we know that

$\frac{dV}{dt} = 10$

And we know as well that .

$V = \frac{4}{3} \pi r^3$

Since everything is changing with the time we can compute the implicit  derivative and we would get that

$\frac{dV}{dt} = 10 = 4\pi r^2 \frac{dr}{dt}$

We are told that we are looking for how fast is the radius growing at the instant when the radius has reached 1 ft, therefore  $r = 1.$

And when we solve for  $\frac{dr}{dt} = \frac{10}{4 \pi} = 0.795$