1. Growing rate

A drop of water is a perfect sphere and by condensation, the droplet picks up moisture at a rate proportional to its surface area. Show that the drops radius increases at a constant rate.

What sort of calculus would I use to tackle this problem? Thanks

2. Originally Posted by RAz
A drop of water is a perfect sphere and by condensation, the droplet picks up moisture at a rate proportional to its surface area. Show that the drops radius increases at a constant rate.

What sort of calculus would I use to tackle this problem? Thanks
Differential Calculus! More specifically, "related rates". Use the formula for volume of a sphere- $V= \frac{4}{3}\pi r^3$- and the formula for area of the surface- $A= 4\pi r^2$. Differentiate the volume formula, with respect to t, using the chain rule, and set it equal to a constant times the area. Solve for dr/dt.

3. Rates of change

Hello RAz
Originally Posted by RAz
A drop of water is a perfect sphere and by condensation, the droplet picks up moisture at a rate proportional to its surface area. Show that the drops radius increases at a constant rate.

What sort of calculus would I use to tackle this problem? Thanks
We can re-write the phrase
the droplet picks up moisture at a rate proportional to its surface area
as "the rate of change of its volume is proportional to its surface area"; or, with the usual notation:

$\frac{dV}{dt}= kS$, for some constant $k$.

So, using $V = \tfrac43\pi r^3$ and $S = 4\pi r^2$, we get:

$\frac{d}{dt}\Big(\tfrac43\pi r^3\Big)=4k\pi r^2$

$\Rightarrow \frac{d}{dr}\Big(\tfrac43\pi r^3\Big)\times\frac{dr}{dt}=4k\pi r^2$

$\Rightarrow 4\pi r^2 \frac{dr}{dt}=4k\pi r^2$

$\Rightarrow \frac{dr}{dt}=k$

$\Rightarrow r$ is increasing at a constant rate.