A spherical balloon is inflated at the rate of 5 cubic feet/minute. What is the volume of the balloon if when the radius is increasing at the rate of 3 feet/minute?

dV = surface area * dr

for a sphere
dV = 4 pi r^2 dr

dV/dt = 4 pi r^2 dr/dt

5 = 4 pi r^2 (3)

r^2 = 5/(12 pi)

Volume=4/3 PI r^3

dV/dt=5=4 PI r^2 dr/dt

you are given dr/dt = 3ft/min
and dv/dt=5ft^3/min

solve for r, and then calculate V

To find the volume of the balloon, we need to use the formula for the volume of a sphere, which is V = (4/3)πr³, where V represents the volume and r represents the radius.

We know that the rate of change of the radius (dr/dt) is given as 3 feet/minute. Since we are interested in finding the rate of change of the volume (dV/dt), we need to differentiate the volume formula with respect to time.

Differentiating both sides of V = (4/3)πr³ with respect to time (t), we get:

dV/dt = d/dt((4/3)πr³)
= (4/3)π * d/dt(r³)

Now, we need to use the chain rule to differentiate r³ with respect to time. The chain rule states that if y = f(g(x)), then dy/dx = f'(g(x)) * g'(x).

In this case, y = r³ and g(x) = r. So, applying the chain rule, we have:

dV/dt = (4/3)π * d/dt(r³)
= (4/3)π * 3r² * (dr/dt)

We can substitute the given values: dr/dt = 3 feet/minute and solve for dV/dt:

dV/dt = (4/3)π * 3r² * 3
= 36πr²

Now, we need to find the volume of the balloon when the rate of change of the radius is 3 feet/minute. Let's assume the radius is increasing from an initial value of r₀ to r₁. Then, the volume will be:

V = ∫(dV/dt) dt
= ∫(36πr²) dt
= 36π∫r² dt
= 36π * ∫r² dt
= 36π * (r₁² - r₀²)

However, we are not given the specific values of r₀ and r₁, so we cannot calculate the exact volume without that information.