firmezas1

2022-10-03

Find the DE that describes the change in a raindrops volume over time

A spherical raindrop evaporates at a rate proportional to its surface area. Write a differential equation for the volume of the raindrop as a function of time

Volume, $V=\frac{4}{3}\pi {r}^{2}$ but I dont know how to find any way to relate that to an evaporation rate. This course is an introductory differential equations course and they don't give you anything else to work with.

Do they assume we covered this kind of problem in a previous physics course?

IF the question read:

A spherical raindrop evaporates at a rate proportional to its surface area as $\frac{dV}{dt}=\alpha S(t)$. Write a differential equation for the volume of the raindrop as a function of time

I might have fared a bit better.

Is there a way to derive this equation?

A spherical raindrop evaporates at a rate proportional to its surface area. Write a differential equation for the volume of the raindrop as a function of time

Volume, $V=\frac{4}{3}\pi {r}^{2}$ but I dont know how to find any way to relate that to an evaporation rate. This course is an introductory differential equations course and they don't give you anything else to work with.

Do they assume we covered this kind of problem in a previous physics course?

IF the question read:

A spherical raindrop evaporates at a rate proportional to its surface area as $\frac{dV}{dt}=\alpha S(t)$. Write a differential equation for the volume of the raindrop as a function of time

I might have fared a bit better.

Is there a way to derive this equation?

Reagan Tanner

Beginner2022-10-04Added 8 answers

Step 1

Well the evaporation rate is the rate at which water is leaving the raindrop, and if we assume that the drop is incompressible then we can describe this purely as $\frac{dV}{dt}.$.

The surface area of the droplet goes like $S=4\pi {r}^{2}\propto {V}^{2/3}$, so the differential equation that they want you to write is presumably just something as simple as $\frac{dV}{dt}=-\beta {V}^{2/3}$ for some parameter $\beta .$.

Step 2

You can then solve it by dividing both sides by ${V}^{2/3}$ and integrating to find:

${V}^{1/3}-{V}_{0}^{1/3}=-\frac{1}{3}\beta t.$

In other words, had you written the differential equation for the radius it would have been a linear one and quite straightforward, as $\frac{dV}{dt}=4\pi {r}^{2}\frac{dr}{dt}$ and the $4\pi {r}^{2}$ terms cancel directly, leaving just $\dot{r}=-\alpha $.

Well the evaporation rate is the rate at which water is leaving the raindrop, and if we assume that the drop is incompressible then we can describe this purely as $\frac{dV}{dt}.$.

The surface area of the droplet goes like $S=4\pi {r}^{2}\propto {V}^{2/3}$, so the differential equation that they want you to write is presumably just something as simple as $\frac{dV}{dt}=-\beta {V}^{2/3}$ for some parameter $\beta .$.

Step 2

You can then solve it by dividing both sides by ${V}^{2/3}$ and integrating to find:

${V}^{1/3}-{V}_{0}^{1/3}=-\frac{1}{3}\beta t.$

In other words, had you written the differential equation for the radius it would have been a linear one and quite straightforward, as $\frac{dV}{dt}=4\pi {r}^{2}\frac{dr}{dt}$ and the $4\pi {r}^{2}$ terms cancel directly, leaving just $\dot{r}=-\alpha $.

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