A spherical balloon is being filled with air in such a way that its radius is increasing at a rate of 2 centimeters per second. At what rate is the volume of the balloon changing at the instant when its surface has an area of 4 pi square centimeters.
To solve this problem, we need to use the formulas for the surface area and volume of a sphere.
The surface area (A) of a sphere with radius (r) is given by the formula: A = 4πr².
The volume (V) of a sphere with radius (r) is given by the formula: V = (4/3)πr³.
Given that the rate at which the radius is increasing is 2 centimeters per second (dr/dt = 2 cm/s), we want to find the rate at which the volume is changing (dV/dt) when the surface area is 4π square centimeters (A = 4π).
To find dV/dt, we will first differentiate the volume formula with respect to time (t):
dV/dt = d/dt [(4/3)πr³]
To differentiate with respect to t, we need to apply the chain rule. Since the radius is changing with respect to time (dr/dt = 2 cm/s), we can write r = r(t). Applying the chain rule, we have:
dV/dt = d/dt [(4/3)πr(t)³]
= (4/3)π d/dt [r(t)³]
= (4/3)π [(d/dr) (r(t)³)] (dr/dt)
= (4/3)π [3r(t)²] (dr/dt)
Now, we plug in the given values to find dV/dt when A = 4π:
A = 4πr²
4π = 4πr²
r² = 1
r = 1 cm
Substituting r = 1 cm and dr/dt = 2 cm/s, we get:
dV/dt = (4/3)π [3(1 cm)²] (2 cm/s)
= (4/3)π (3 cm²) (2 cm/s)
= 8π cm³/s
Therefore, the volume of the balloon is changing at a rate of 8π cubic centimeters per second when its surface area is 4π square centimeters.