Question

Suppose, instead, that the liquid is known to dissolve the material at a rate of 1 m3/hour, at what rate will its radius r change with respect to time when r = 1 meter? when r = 1/2 meter?

Answer 1

The rate of change of the radius r of the dissolving material when r = 1 meter is 1/(4π) meters per hour, and when r = 1/2 meter, it is 1/(π) meters per hour.

Step-by-step explanation:

A student asked about the rate at which the radius r of a spherical material will change over time given that the material dissolves at a rate of 1 m³/hour. To find the rate at which the radius changes when r = 1 meter and when r = 1/2 meter, we need to apply the formula for the volume of a sphere, which is V = (4/3)πr³. By differentiating both sides with respect to time (dt), we get dV/dt = 4πr²dr/dt. We can rearrange this to solve for dr/dt, the rate of radius change with respect to time.

The next step is to plug in the given values. With the rate of dissolution dV/dt given as 1 m³/hour, when r = 1 meter, we can find dr/dt by solving 4π(1)²dr/dt = 1. This gives us dr/dt = 1/(4π) meters per hour. Similarly, when r = 1/2 meter, we solve 4π(1/2)²dr/dt = 1, which gives us dr/dt = 1/(π) meters per hour.