A spherical balloon is inflated at a rate of 10 cm^3/min. How fast does the radius change when the radius is 20 cm?
Need help with set up and work through of problem.
2/25
To solve this problem, we can use the formula for the volume of a sphere:
V = (4/3) * π * r^3
Where V represents the volume and r represents the radius of the sphere.
We are given that the balloon is inflated at a rate of 10 cm^3/min. This means that the rate of change of the volume with respect to time (dV/dt) is 10 cm^3/min. We need to find the rate of change of the radius (dr/dt) when the radius is 20 cm.
To do this, we can differentiate the volume formula with respect to time, using the chain rule:
dV/dt = dV/dr * dr/dt
The derivative of V with respect to r is:
dV/dr = 4πr^2
Now we know:
10 cm^3/min = 4πr^2 * dr/dt
We can plug in the given radius of 20 cm and solve for dr/dt:
10 cm^3/min = 4π(20)^2 * dr/dt
Simplifying:
10 = 4π(400) * dr/dt
10 = 1600π * dr/dt
Dividing both sides by 1600π:
10 / (1600π) = dr/dt
dr/dt ≈ 0.0000397887 cm/min
Therefore, when the radius is 20 cm, the radius is changing at a rate of approximately 0.0000397887 cm/min.