How does your rectangle grow?

Excerpt: This two-part task looks at how the rate of change of the area of a rectangle depends on the side lengths. It offers a visual representation of the Product Rule.

Sideways

This white rectangle has a fixed height, $v$ , and its width $u$ 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, $A=$
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, $\delta u =$
The area increases by the shaded area, $\delta A =$
The rate of change of the area of the rectangle, $\dfrac{dA}{dt} \approx \dfrac{\delta A}{\delta t} =$

Example

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

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

$t=0$ ?

$t=3$ ?

Both ways

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

Example

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

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

$t=0$ ?

$t=3$ ?

Solution

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 $v=3\,\text{cm}$ and width $u=\left(\frac{1}{2}t^2+1\right)\,\text{cm}$ at time $t\,\text{s}$ .

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

$t=0$ ?

$t=3$ ?

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

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

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

$t=0$ ?

$t=3$ ?

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

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