# Applications of Differentiation

## Applications of Differentiation Revision

**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.

**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}}.

**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}.

**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]**

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}.