Question

The volume of a sphere is decreasing at a constant rate of 6031 cubic centimeters per second. At the instant when the volume of the sphere is 3292 cubic centimeters, what is the rate of change of the radius? The volume of a sphere can be found with the equation V=4/3 π r^3. Round your answer to three decimal places.

Answer 1

The rate of change of the radius at the instant when the volume of the sphere is 3292 cubic centimeters is approximately -0.075 cm/s.

Explanation:

Let's begin by differentiating the volume formula with respect to time:

dV/dt = 4πr^2(dr/dt)

We know that dV/dt = -6031 cm^3/s (because the volume is decreasing at a constant rate), and we're trying to find dr/dt when V = 3292 cm^3.

Substituting these values into the above equation, we get:

-6031 = 4πr^2(dr/dt)

We can solve for dr/dt by plugging in V = 3292 cm^3 into the volume formula to find the corresponding radius:

3292 = (4/3)πr^3

r = (3V/4π)^(1/3) = (3(3292)/(4π))^(1/3) ≈ 6.76 cm

Now we can plug in this value for r and solve for dr/dt:

-6031 = 4π(6.76)^2(dr/dt)

dr/dt = -6031 / (4π(6.76)^2) ≈ -0.075 cm/s

Therefore, the rate of change of the radius at the instant when the volume of the sphere is 3292 cubic centimeters is approximately -0.075 cm/s.

Answer 2

differentiating the equation for the volume of a sphere with respect to time:

V = 4/3 π r^3

dV/dt = 4π r^2 (dr/dt)

dV/dt = -6031

V = 3292

V = 4/3 π r^3

3292 = 4/3 π r^3

r^3 = 3/4 (3292/π)

r = 9.96

-6031 = 4π (9.96)^2 (dr/dt)

dr/dt = -6031 / (4π (9.96)^2)

dr/dt = -0.609 cm/s

the rate of change of the radius is approximately -0.609 cm/s when the volume of the sphere is 3292 cm^2.