The radius of a sphere is increasing at a rate of $3 mm / s$ . How fast is the volume increasing (in ${mm}^{3} / s$ ) when the diameter is $40 mm$ ? (Round your answer to two decimal places.)
${mm}^{3} / s$

Answer

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Answer

$15079.64$ is the rate at which the volume is increasing when the diameter is 40 mm.

Steps

Step 1 ：Given the rate of change of the radius (dr/dt) is 3 mm/s and the diameter is 40 mm, so the radius is 20 mm.

Step 2 ：Using the formula dV/dt = (4 * pi * r^2) * dr/dt, we can plug in the values: dV/dt = (4 * pi * (20)^2) * 3.

Step 3 ：Calculate the final answer: $d V / d t = 15079.64 {mm}^{3} / s$

Step 4 ： $15079.64$ is the rate at which the volume is increasing when the diameter is 40 mm.