A spherical party balloon is being inflated with helium pumped in at a rate of 11 cubic feet per minute. How fast is the radius growing at the instant when the radius has reached 2 ft? HINT [See Example 1.] (The volume of a sphere of radius r is
V = 4/3𝜋r^3
Round your answer to two decimal places.)
____ ft/min
The CHANGE in volume is the surface area times the change in radius.
V = (4/3) pi r^3
dV/dr = 4 pi r^2 = surface area of of sphere. (here 16 pi = 50.2 ft^2)
dV/dt = 4 pi r^2 dr/dt = 50.2 dr/dt
so
dr/dt = dV/dt / (4 pi r^2)
= 11 /(4 pi *4) = 11/ (16 pi ) = 0.219 ft/minute
To find out how fast the radius is growing, we can use the formula for the volume of a sphere:
V = (4/3)πr^3
Where V is the volume and r is the radius.
Taking the derivative of this equation with respect to time, we can find how the volume is changing as the radius changes:
dV/dt = (d/dt)[(4/3)πr^3]
Since we are given that the helium is being pumped in at a rate of 11 cubic feet per minute, we can substitute dV/dt with 11:
11 = (d/dt)[(4/3)πr^3]
Next, we need to find the derivative of (4/3)πr^3 with respect to time. To do this, we can bring the derivative inside the brackets:
11 = (4/3)π(d/dt)[r^3]
Now, let's take the derivative of r^3. Since we are interested in the rate when the radius has reached 2 ft, we can substitute r as 2:
11 = (4/3)π(d/dt)[(2)^3]
11 = (4/3)π(d/dt)(8)
11 = (4/3)π(0) [Since d/dt of a constant is always zero]
Therefore, the rate at which the radius is growing when the radius has reached 2 ft is 0 ft/min.