# Given the rate of which a circular area grows, find the radius and its rate of change

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

(a) the radius of the patch 20 seconds after pouring has commenced

(b) the rate of increase of the radius at this instant. • It should have been a cat. – David Mitra May 7 '12 at 2:23

Let $r(t)$ be the radius of the patch at time $t$, measured in cm, with $t$ measure in seconds (why cm and seconds? Because the information we are given is in cm and in seconds, so there is no point in introducing unnecessary complications by changing units).

The area of the patch at time $t$ is then $A(t) = \pi \Bigl(r(t)\Bigr)^2$. Note that $A(t)$ is measured in $\text{cm}^2$.

We are told the rate of change of the Area over time; this is the derivative with respect to $t$; so we are told that $$\frac{dA}{dt} = 5\frac{\text{cm}^2}{\text{s}}.$$ Why are the units $\text{cm}^2/\text{s}$? Because $\frac{dA}{dt}$ is measured in units of $A$ over units of $t$, in this case, square cm over seconds.

We are asked for: $r(20)$, and $r'(20)$.

How do we find them?

Well, we have an equation that relates $A$ and $r$. By taking derivatives, we obtain an equation that relates $A'$, $A$, $r'$ and $r$. We have: \begin{align*} A &= \pi r^2\\ \frac{d}{dt}(A) &= \frac{d}{dt}(\pi r^2)\\ \frac{dA}{dt} &= \pi\left( 2r\frac{dr}{dt}\right). \end{align*} Now, the area was $0$ when $t=0$, and grows at $5\ \text{cm}^2/\text{s}$, so after $20$ seconds the area will be $5(20) = 100$. Solving for $r(20)$ we get: \begin{align*} A(t) &= \pi(r(t))^2\\ A(20) &= \pi (r(20))^2\\ 100 &=\pi (r(20))^2\\ \frac{100}{\pi} &= (r(20))^2\\ \sqrt{\frac{100}{\pi}} &= |r(20)|\\ \frac{10}{\sqrt{\pi}} &= r(20)\quad\text{(radius is positive)}\\ \frac{10\sqrt{\pi}}{\pi} &= r(20). \end{align*} So the radius at $t=20$ is $\frac{10\sqrt{\pi}}{\pi}$.

To find the rate of change of $r$ at $20$, we use the formula we found $$\frac{dA}{dt} = \pi\left(2r\frac{dr}{dt}\right).$$ We know the value of $\frac{dA}{dt}$, and now we know the value of $r$ at $t=20$. So we just need to plug in and solve for $\frac{dr}{dt}\Bigm|_{t=20}$.

• Thanks! I was just gonna say you should've been a teacher, but apparently you already are :) cheers – Deniz May 7 '12 at 3:39

A viscous liquid is poured on to a flat surface. It forms a circular patch whose area grows at a steady rate of 5cm^2/s. Find in terms of π,

(a) the radius of the patch 20 seconds after pouring has commenced

(b) the rate of increase of the radius at this instant.


The growth of the area in function of time can be expressed as

$$A(t) = 5 \frac{cm^2}{s}t$$

Since we know that $A = \pi r^2$, then we can relate radius and area by

$$R(t) = \sqrt{\frac {5t} {\pi\cdot s}} cm$$

The rate of increase of the radius is $\dfrac {dR}{dt}$.

Can you try an answer the questions now?