Review question

Can we track this constantly growing patch of fluid? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6959

Solution

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 $\pi$,

1. 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 $t=20$,

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

1. 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.