Question

Volume The radius \(r\) of a sphere is increasing at a rate of 3 inches per minute. (a) Find the rates of change of the volume when \(r=9\) inches and \(r=36\) inches. (b) Explain why the rate of change of the volume of the sphere is not constant even though \(d r / d t\) is constant.

Short answer

The rate of change of volume of the sphere is \(972\pi in^3/min\) when the radius is 9 inches and \(3888\pi in^3/min\) when the radius is 36 inches. The rate of change of volume is not constant because the volume of a sphere is related to the cube of its radius. Therefore, as the radius increases, the increase in volume also increases, despite the rate of change of radius being constant.

Step-by-step solution

Understand the relationship between volume and radius

The volume of a sphere is given by the equation \(V = 4/3 \pi r^3\). This relationship is necessary to link variations in radius to variations in volume.

Differentiate the volume function

Differentiating both sides of the volume function with respect to time \(t\), we get the change in volume over time: \(dV/dt = 4\pi r^2 \cdot dr/dt\).

Calculate \(dV/dt\) at \(r = 9 in\)

Plugging the values \(r = 9 in\) and \(dr/dt = 3 in/min\) into the equation, we get: \(dV/dt = 4\pi \cdot 9^2 \cdot 3 = 972\pi in^3/min\)

Calculate \(dV/dt\) at \(r = 36 in\)

Plugging the values \(r = 36 in\) and \(dr/dt = 3 in/min\) into the equation, we get: \(dV/dt = 4\pi \cdot 36^2 \cdot 3 = 3888\pi in^3/min\)

Explain why the rate of volume change (\(dV/dt\)) isn't constant

Even though the rate of radius change (\(dr/dt\)) is constant, the rate of volume change (\(dV/dt\)) is not constant because the volume of a sphere is proportional to the cube of its radius. As the sphere's radius increases, the increment in volume for each additional unit of radius also increases. Hence, \(dV/dt\) increases as \(r\) increases.