Air is being pumped into a spherical balloon and the volume is increasing at the rate of 100 cubic cm per minute. How fast is the radius of the balloon increasing when the radius is 10 cm?
(V=4/3 times pie times r^3)
V = (4/3)πr^3
dV/dt = 4πr^2 dr/dt
100 = 4π(100) dr/dt
dr/dt = 1/(4π) cm/min
To find how fast the radius of the balloon is increasing, we need to use the relationship between the volume and the radius of a sphere. The formula for the volume of a sphere is V = (4/3)πr^3, where V is the volume and r is the radius.
We are given that the volume is increasing at a rate of 100 cubic cm per minute. Let's denote the rate of change of volume as dV/dt, where dt is the change in time.
We want to find the rate of change of the radius, which is denoted as dr/dt, when the radius is 10 cm.
To solve this problem, we can start by differentiating the volume formula with respect to time (t) using the chain rule:
dV/dt = (dV/dr) * (dr/dt)
Since V = (4/3)πr^3, we can calculate the derivative dV/dr as follows:
dV/dr = 4πr^2
Substituting these values into the differentiation equation, we get:
100 = (4πr^2) * (dr/dt)
Now, we can rearrange the equation to solve for dr/dt:
dr/dt = 100 / (4πr^2)
When the radius is 10 cm, we substitute r = 10 into the equation to find the rate of change of the radius:
dr/dt = 100 / (4π(10^2))
= 100 / (4π(100))
= 100 / (400π)
≈ 0.0796 cm/min
Therefore, when the radius is 10 cm, the radius of the balloon is increasing at a rate of approximately 0.0796 cm per minute.