Can we track this constantly growing patch of fluid?

Can we track this constantly growing patch of fluid?

Question

A viscous liquid is poured on a flat surface. It forms a circular patch which grows at a steady rate of $5\,\text{cm}^{2/s}$. Find, in terms of $\pi$,

1. the radius of the patch 20 seconds after pouring has commenced.

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1. the rate of increase of the radius at this instant.

Solution

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 $t=20$,

$$r=\frac{10}{\sqrt{\pi}}.$$

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.