Problem 13 - Calculus of a Single Variable

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.

Step-by-Step Solution

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 1: 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.

Step 2: Differentiate the volume function

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

Step 3: Calculate \(dV/dt\) at \(r = 9 in\)

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

Step 4: Calculate \(dV/dt\) at \(r = 36 in\)

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

Step 5: 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.