A viscous liquid is poured on a flat surface. It forms a circular patch which grows at a steady rate of \(\quantity{5}{cm^2/s}\). Find, in terms of \(\pi\),
- the radius of the patch 20 seconds after pouring has commenced.
The area \(A\) of the patch is a function of time such that\[ \dfrac{dA}{dt}=5 \implies A = 5t,\]
since \(A = 0\) when \(t = 0\).
We also know that at any given time, the area of a circle is connected to its radius via \[ A=\pi r^2.\]
If we combine both equations, eliminating A and solving for \(r\), we find\[ r=\sqrt{\frac{5t}{\pi}}, \text{or} \: r=\left(\sqrt{\frac{5}{\pi}}\right)t^{\frac {1}{2}}\]
and hence at \(t=20\),
\[r=\frac{10}{\sqrt{\pi}}.\]
- the rate of increase of the radius at this instant.
Differentiating our expression for \(r\) gives \[\frac{dr}{dt} = \frac{1}{2}\sqrt{\frac{5}{\pi}}t^{-\frac{1}{2}}=\frac{1}{2}\sqrt{\frac{5}{\pi t}},\] and hence we find at a time of 20 seconds an increase of
\[\frac{dr}{dt} = \frac{1}{4\sqrt{\pi}}.\]
Alternatively here we could use the Chain Rule to say \(\dfrac{dA}{dt}=\dfrac{dA}{dr}\times\dfrac{dr}{dt}\). We know \(\dfrac{dA}{dt}=5\) and can easily find \(\dfrac{dA}{dr} = 2\pi r\).
Thus \(\dfrac{dr}{dt} = \dfrac{5}{2\pi r}\). So when \(t = 20\), \(\dfrac{dr}{dt}= \dfrac{5}{2\pi \frac{10}{\sqrt{\pi}}} = \dfrac{1}{4\sqrt{\pi}}\), as before.
