# Applications of Differentiation

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

## Applications of Differentiation

In this section, we’ll look at how to use differentiation in mechanics and practical problems.

Make sure you are happy with the following topics before continuing.

A Level   ## Finding Maximum and Minimum Values of Volume and Area

We can use differentiation to find optimal values of dimensions of objects.

So, let’s say wish to create a hollow box of length $2\textcolor{blue}{x}$, width $\textcolor{blue}{x}$ and height $\textcolor{purple}{h}$, and we have a limit of $600\text{ cm}^2$ of wood to use.

To find the maximum volume possible, we must find two equations in volume and area:

$\textcolor{red}{V} = 2\textcolor{blue}{x} \times \textcolor{blue}{x} \times \textcolor{purple}{h} = 2\textcolor{blue}{x}^2\textcolor{purple}{h}$

and

$\textcolor{limegreen}{A} = 2(2\textcolor{blue}{x} \times \textcolor{blue}{x}) + 2(2\textcolor{blue}{x} \times \textcolor{purple}{h}) + 2(\textcolor{blue}{x} \times \textcolor{purple}{h}) = 4\textcolor{blue}{x}^2 + 6\textcolor{blue}{x}\textcolor{purple}{h} = 600$

Of course, we only want to differentiate in terms of one variable, so we’ll transform the second equation to $\textcolor{purple}{h} = \dfrac{300 - 2\textcolor{blue}{x}^2}{3\textcolor{blue}{x}}$

Plugging this back into our equation for $\textcolor{red}{V}$, we have

$\textcolor{red}{V} = \dfrac{2\textcolor{blue}{x}^2(300 - 2\textcolor{blue}{x}^2)}{3\textcolor{blue}{x}} = 200\textcolor{blue}{x} - \dfrac{4\textcolor{blue}{x}^3}{3}$

We can then differentiate with respect to $\textcolor{blue}{x}$ to find the maximum volume of the box:

$\dfrac{d\textcolor{red}{V}}{d\textcolor{blue}{x}} = 200 - 4\textcolor{blue}{x}^2 = 0$, giving $\textcolor{blue}{x} = \sqrt{50}$

We must also verify that this is a maximum, rather than a minimum, so we find $\dfrac{d^2\textcolor{red}{V}}{d\textcolor{blue}{x}^2}$:

$\dfrac{d^2\textcolor{red}{V}}{d\textcolor{blue}{x}^2} = - 8\textcolor{blue}{x} = -8\sqrt{50} < 0$, so this is definitely a maximum point.

Therefore, we have $\textcolor{blue}{x} = \sqrt{50}$ and $\textcolor{purple}{h} = \dfrac{200}{3\sqrt{50}}$, giving $\textcolor{red}{V}_{max} = \dfrac{20000}{3\sqrt{50}}$.

A Level   ## Finding Rates of Change in Mechanics

In Mechanics, we’ll talk about the derivation of acceleration and velocity from displacement.

From the displacement $\textcolor{limegreen}{s}$, we can differentiate with respect to $\textcolor{purple}{t}$ to find $\textcolor{red}{v}$, and differentiate again to find $\textcolor{blue}{a}$.

So,

$\dfrac{d\textcolor{limegreen}{s}}{d\textcolor{purple}{t}} = \textcolor{red}{v}$

and

$\dfrac{d^2\textcolor{limegreen}{s}}{d\textcolor{purple}{t}^2} = \dfrac{d\textcolor{red}{v}}{d\textcolor{purple}{t}} = \textcolor{blue}{a}$

So, for example, let’s say we have an equation for the displacement $\textcolor{limegreen}{s} = 2\textcolor{purple}{t}^3 - 3\textcolor{purple}{t}^2 -4\textcolor{purple}{t} + 1$.

We can find an equation for the velocity:

$\textcolor{red}{v} = 6\textcolor{purple}{t}^2 - 6\textcolor{purple}{t} - 4$

and an equation for the acceleration:

$\textcolor{blue}{a} = 12\textcolor{purple}{t} - 6$

Therefore, we can find the values of $\textcolor{purple}{t}$ such that $\textcolor{red}{v} = 0$:

$6\textcolor{purple}{t}^2 - 6\textcolor{purple}{t} - 4 = 0$ has solutions at $\textcolor{purple}{t} = \dfrac{1}{2} \pm \sqrt{\dfrac{11}{12}}$, but we cannot have a negative time, so we have $\textcolor{purple}{t} = \dfrac{1}{2} + \sqrt{\dfrac{11}{12}}$.

$\textcolor{blue}{a} = 12\left( \dfrac{1}{2} + \sqrt{\dfrac{11}{12}}\right) - 6 = 11.49$, so we can conclude that this is a point of minimum displacement.

A Level   ## Example Questions

To find the maximum volume possible, we must find two equations in volume and area:

$V = 5x \times 3x \times h = 15x^2h$

and

$A = 2(5x \times 3x) + 2(5x \times h) + 2(3x \times h) = 30x^2 + 16xh = 30000$

Of course, we only want to differentiate in terms of one variable, so we’ll transform the second equation to

$h = \dfrac{30000 - 30x^2}{16x}$

Plugging this back into our equation for $V$, we have

$V = \dfrac{15x^2(30000 - 30x^2)}{16x} = 28125x - 28.125x^3$

We can then differentiate with respect to $x$ to find the maximum volume of the box:

$\dfrac{dV}{dx} = 28125 - 84.375x^2 = 0$, giving $x = \sqrt{\dfrac{1000}{3}}$

We must also verify that this is a maximum, rather than a minimum, so we find $\dfrac{d^2V}{dx^2}$:

$\dfrac{d^2V}{dx^2} = - 168.75x = -168.75\sqrt{\dfrac{1000}{3}} < 0$, so this is a maximum point.

Therefore, we have $x = \sqrt{\dfrac{1000}{3}}$ and $h = \dfrac{20000}{16\sqrt{\dfrac{1000}{3}}}$, giving $V_{max} = \dfrac{6250000\sqrt{3}}{\sqrt{1000}}$.

$h = 1000\left( \dfrac{t^2}{40} - \dfrac{t^3}{600}\right)$ gives

$\dfrac{dh}{dt} = 1000 \left( \dfrac{t}{20} - \dfrac{t^2}{200}\right)$

and

$\dfrac{d^2h}{dt^2} = 1000 \left( \dfrac{1}{20} - \dfrac{t}{100}\right)$

Setting $\dfrac{dh}{dt} = 0$ gives $1000\left( \dfrac{t}{20}\left( 1 - \dfrac{t}{10}\right) \right) = 0$, meaning that $t = 0$ or $t = 10$.

Using $t=10$ gives $h=833\text{ m}$.

Volume of bath, $V$:

$V = \dfrac{1}{2}(2.5 + 1.5) \times 0.5 \times 1 = 1\text{ m}^3 = 1 \times 10^6 \text{ cm}^3$

Then, we must find $t$ such that $\dfrac{1}{2}t^2 + 4t - 1000000 = 0$.

Using the quadratic formula, we have

$t = \dfrac{-4 \pm \sqrt{16 - \left( 4 \times \dfrac{1}{2} \times -1000000 \right)}}{1} = -1418.22, 1410.22\text{ s}$

So, we know that the bath takes approximately $23.5\text{ minutes}$ to fill.

When $t = 1410.22$, $\dfrac{dV_{filled}}{dt} = t + 4$, so the water is filling the bath at a rate of $1414.22\text{ cm}^3\text{ s}^{-1}$.

A Level

A Level

A Level

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