Question

Writing to Learn Recall that the volume \(V\) of a sphere of radius \(r\) is \((4 / 3) \pi r^{3}\) and that the surface area \(A\) is 4\(\pi r^{2}\) . Notice that \(d V / d r=A .\) Explain in terms of geometry why the instantaneous rate of change of the volume with respect to the radius should equal the surface area.

Short answer

The instantaneous rate of change of the volume of a sphere with respect to the radius equals the surface area because when the radius increases, the volume increases by the amount of a new 'shell' whose area is equal to the surface area of the original sphere.

Step-by-step solution

Understand the underlying concept

First, remember that the derivative measures rate of change. In this situation, \(dV / dr\) signifies how fast the volume of the sphere changes as the radius \(r\) changes.

Relate radius change with the volume change

This change in volume can be visualized as if we are adding a small 'shell' of thickness \(dr\) to the sphere's radius. This 'shell', from a 3D perspective, covers the entire surface of the sphere.

Relate volume change with the surface area

The volume of this small 'shell' is the amount by which the volume of the sphere increases, \(dV\). If we flatten the shell, the area of this flat shape is the surface area \(A\) of the sphere. Thus, we can conclude that the volume increase \(dV\) due to the increase in the radius \(dr\) is equal to the surface area \(A\) times \(dr\). Hence, \(dV = A * dr\).

Derive the Instantaneous rate of change of volume - The Derivative

The previous equation tells us that for small changes in radius, the change in volume is proportional to the surface area of the sphere. And if we divide both sides of the equation by \(dr\) we get \(dV / dr = A\), which indeed is the formula provided. Therefore, the instantaneous rate of change of the volume, which is the derivative of the volume with respect to the radius, is equal to the surface area of the sphere.