Newton's second law of motion F = ma = $m\cdot \dfrac{dv}{dt} $can be written in the form $F=d/dt\left( mv\right)$ in terms of the momentum mv of a particle of mass m and velocity v, and remains valid in this form even if m is not constant, as assumed so far. Suppose a spherical raindrop falls through air saturated with water vapor, and assume that by condensation the mass of the raindrop increases at a rate proportional to its surface area, with c the constant of proportionality. If the initial radius and velocity of the raindrop are both zero, show that the drag exerted by the condensation of the water vapor has the effect of making the raindrop fall with acceleration g/4. Hint: Show that $d/dr\left( r^{3}\upsilon \right) =\left( \dfrac{\delta }{c}\right) \cdot r^{3}g$, where r is the radius of the raindrop and $\delta $ is its density.
解决方法来源于：
https://math.stackexchange.com/questions/4385574/drag-exerted-on-condensing-raindrop
首先，对于微分的基础概念和定义，重新再认识一遍，以下面两个方程为例：$$y=2x$$ $$z=2y$$显然我们有：$$z=4x$$接下来，我们先看第一个方程，对y进行求导，也就是对其进行微分（我不确定这么描述这个过程是否恰当），可以得到以下等式：$$y'=\dfrac{dy}{dx}=2$$上述等式意味着当X变化（$\Delta x$）1时，y会跟着变化（$\Delta y$）2,同理可以推导出z和y之间的关系，以及z与x之间的关系，在这里，y似乎起着一个桥梁的作用，连接着x和z。
下面回到这道题中，球形水滴的半径r即起着这个桥梁作用，其连接着球形水滴的质量m和时间t。
根据球体积公式以及质量等于体积密度的乘积，可以得到下列第一个等式：$$m=\dfrac{4}{3}\pi r^{3}\cdot \delta$$ $$\dfrac{dm}{dr}=4\pi \delta r^{2}$$再根据题意，可知质量m和球形水滴的表面积之间也成正比例关系，从而得到第二个等式：$$\dfrac{dm}{dt}=4\pi r^{2}\cdot c$$将第二个等式除以第一个等式，得到半径r和时间t的关系：$$\begin{aligned}\dfrac{dr}{dt}&=\dfrac{c}{\delta }\ \\
dr&=\dfrac{c}{\delta }\cdot dt\ \\
\int dr&=\dfrac{c}{\delta}\int dt\ \\
r&=\dfrac{c}{\delta }\cdot t\end{aligned}$$到这里，还没用到题目中一开始提到的动量表达式，将质量m的表达式$m=\dfrac{4}{3}\pi r^{3}\cdot \delta$代入：$$\begin{aligned}f=mg&=\dfrac{d\left( mv\right) }{dt}\ \\
\dfrac{4}{3}\pi r^{3}\delta g&=\dfrac{d}{dt}\left( \dfrac{4}{3}\pi r^{3}\delta \cdot v\right) \ \\
r^{3}g&=\dfrac{d}{dt}\left( r^{3}\cdot v\right) \end{aligned}$$再将$r=\dfrac{c}{\delta }\cdot t$代入上式，得到 $$\begin{aligned}r&=\dfrac{c}{\delta}\cdot t\ \\
\left( \dfrac{c}{\delta }\right) ^{3}\cdot t^{3}g&=\dfrac{d}{dt}\left[ \left( \dfrac{c}{\delta }\right) ^{3}t^{3}.v\right] \ \\
t^{3}g&=\dfrac{d}{dt}\left( t^{3}\cdot v\right) \ \\
t^{3}g\cdot dt&=d\left( t^{3}\cdot v\right) \ \\
\int t^{3}g\cdot dt&=\int d\left( t^{3}\cdot v\right) \ \\
\dfrac{1}{4}t^{4}g&=t^{3}\cdot v\end{aligned}$$最终得到球形水滴速度v的表达式：$v=\dfrac{1}{4}gt$，从而证明完毕。