# How does your rectangle grow? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

## Solution

This white rectangle has a fixed height, $v$, and width $u$ which varies as a function of time, $t$. We can write the rate of change of the width as $\dfrac{du}{dt}$ or $\dot{u}$. You can think of this as the speed at which the right hand edge is moving.

 The area of the white rectangle, $A=$ $u\times v$ In a short time interval, $\delta t$, we can assume that $\dot{u}$ is roughly constant, so thinking about speed and distance… The width increases by a small amount, $\delta u =$ $\dot{u}\,\delta t$ The area increases by the shaded area, $\delta A =$ $v\times\dot{u}\,\delta t$ The rate of change of the area of the rectangle, $\dfrac{dA}{dt} \approx \dfrac{\delta A}{\delta t} =$ $v\times\dot{u} = v\dfrac{du}{dt}$

Note that $\frac{\delta A}{\delta t}$ is a fraction whereas $\frac{dA}{dt}$ is a derivative. They become identical in the limit as $\delta t\to0$.

Our rectangle has height $v=\quantity{3}{cm}$ and width $u=\quantity{\left(\frac{1}{2}t^2+1\right)}{cm}$ at time $\quantity{t}{s}$.

What is the rate of change of area, $\dfrac{dA}{dt}$, at

1. $t=0$?

2. $t=3$?

Differentiating, we have $\frac{du}{dt}=\frac{1}{2}\times2t = t$ so $\frac{dA}{dt}=v\frac{du}{dt}=3t.$

At $t=0$, $\frac{dA}{dt}=0$.

At $t=3$, $\frac{dA}{dt}=\quantity{9}{cm^2\,s^{-1}}$.

This white rectangle has width $u$ and height $v$, both of which vary as functions of time, $t$. We can write the rates of change as $\dot{u}$ and $\dot{v}$.

 The area of the white rectangle, $A=$ $u\times v$ In a short time interval, $\delta t$, we can assume that $\dot{u}$ and $\dot{v}$ are roughly constant, so… The width increases by $\delta u =$ $\dot{u}\,\delta t$ The height increases by $\delta v =$ $\dot{v}\,\delta t$ The area increases by approximately the shaded area, so $\delta A \approx$ $v\times\dot{u}\,\delta t+u\times\dot{v}\,\delta t$ The rate of change of the area of the rectangle, $\dfrac{dA}{dt} \approx \dfrac{\delta A}{\delta t} \approx$ $v\times\dot{u}+u\times\dot{v}$ $= v\dfrac{du}{dt}+u\dfrac{dv}{dt}$

Note that in writing down $\delta A$ we ignored the small white rectangle in the top right-hand corner. Why is it OK to do this?

This result is known as the Product Rule which is often expressed as $\frac{d}{dx}\big(uv\big)=u\frac{dv}{dx}+v\frac{du}{dx}.$

Our rectangle has $v=\quantity{\sqrt{t+1}}{cm}$ and $u=\quantity{\left(\frac{1}{2}t^2+1\right)}{cm}$ at time $\quantity{t}{s}$.

What is the rate of change of area, $\dfrac{dA}{dt}$, at

1. $t=0$?

2. $t=3$?

Differentiating, we have \begin{align*} \frac{dv}{dt} &= \frac{1}{2}(t+1)^{-\frac{1}{2}} \\ \frac{du}{dt} &= t \end{align*}

and so $\frac{dA}{dt} = \left(\frac{1}{2}t^2+1\right)\times\frac{1}{2}(t+1)^{-\frac{1}{2}} + \sqrt{t+1}\times t.$

At $t=0$, $\frac{dA}{dt} = 1\times\frac{1}{2}\times(1)^{-\frac{1}{2}}+0 = \quantity{\frac{1}{2}}{cm^2\,s^{-1}}$.

At $t=3$, $\frac{dA}{dt} = \left(\frac{9}{2}+1\right)\times\frac{1}{2}\times(4)^{-\frac{1}{2}}+\sqrt{4}\times 3 = \quantity{7\frac{3}{8}}{cm^2\,s^{-1}}$.