Applications of Differentiation

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

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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 $\textcolor{red}{V}_{max} = \dfrac{20000}{3\sqrt{50}}$ .

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.

Applications of Differentiation Example Questions

Question 1: Find the maximum volume possible when we create a hollow box of length $5x$ , width $3x$ and height $h$ , and we have a limit of $30000\text{ cm}^2$ of wood to use.

[5 marks]

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 $V_{max} = \dfrac{6250000\sqrt{3}}{\sqrt{1000}}$ .

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Question 2: A firework is fired directly upwards and has a height of $h = 1000\left( \dfrac{t^2}{40} - \dfrac{t^3}{600}\right)$ . Find the time $t$ when the firework finds its maximum height, and state this height.

[4 marks]

$h = 1000\left( \dfrac{t^2}{40} - \dfrac{t^3}{600}\right)$ 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 $h=833\text{ m}$ .

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Question 3: Let a bath be modelled as a trapezium with the dimensions shown in the diagram, and width $1\text{ m}$ . Assuming that water occupies a volume of $\dfrac{1}{2}t^2 + 4t\text{ cm}^3$ , find the amount of time it takes for the bath to fill completely, and find the rate that the water is filling the bath at this point.

[5 marks]