Question

The radius of a sphere is increasing at the rate 7.7 cm/s. When the surface area of the sphere becomes 520 cm², what is the rate at which the volume is increasing?

a. 2,002 cm³/s
b. 4,004 cm³/s
c. 6,006 cm³/s
d. 12,012 cm³/s

Answer 1

The rate at which the volume of the sphere is increasing when the surface area becomes 520 cm² is calculated using related rates and differentiation. By substituting the radius we find from the surface area into the volume's rate of change formula, the answer is determined to be 4004 cm³/s, corresponding to option b.

Step-by-step explanation:

To determine the rate at which the volume is increasing when the surface area of the sphere becomes 520 cm², we will use related rates calculus, specifically formulas for surface area and volume of a sphere.

The formula for the surface area of a sphere is S = 4πr², where r is the radius of the sphere. Given the surface area S is 520 cm², we can solve for r:

520 = 4πr²

r² = 520 / (4π)

r = √(520 / (4π)) = √(130 / π)

Now, to find the rate of change of volume (dV/dt), we use the formula V = (4/3)πr³.

To find dV/dt when r is changing, we differentiate both sides of the volume formula with respect to time t:

dV/dt = (4/3)π(3r²)(dr/dt)

Since the radius is increasing at 7.7 cm/s (dr/dt = 7.7), we substitute r and dr/dt into the equation to find dV/dt:

dV/dt = (4/3)π(3(√(130 / π))²)(7.7)

dV/dt = 4π(√(130 / π))²(7.7)

We now calculate this expression to determine the rate of volume increase:

dV/dt = 4π((130 / π))(7.7) = 4(130)(7.7)

dV/dt = 4004 cm³/s

So, the rate at which the volume is increasing when the surface area is 520 cm² is 4004 cm³/s, which is option b.