Final answer:
To find the radius r(t) of a spherical water droplet shrinking by evaporation, we use the rate of volume change proportional to surface area. The resulting radius as a function of time is r(t) = r_0 - k * t, given the initial radius r_0 and the proportionality constant k.
Step-by-step explanation:
The student has asked us to find the radius r(t) as a function of time for a spherical water droplet that loses volume by evaporation at a rate proportional to its surface area. Let's denote the proportionality constant as k. We're given the initial radius r0.
Firstly, we can express the rate of change in volume as:
dV/dt = -k * A
where A is the surface area of the sphere. Since the surface area of a sphere is given by A = 4πr2 and the volume by V = (4/3)πr3, we substitute these expressions into our equation:
dV/dt = -k * 4πr2
We then take the derivative of the volume with respect to the radius:
dV/dr = 4πr2
Now, we use the chain rule to replace dV/dt with dV/dr * dr/dt and solve for dr/dt:
dr/dt = (dV/dt) / (dV/dr) = (-k * 4πr2) / (4πr2) = -k
The negative sign indicates the radius is shrinking. Integrating both sides with respect to time, we get:
r(t) = r0 - k * t
This equation allows us to calculate the radius at any time t, given the initial radius r0 and proportionality constant k.