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 $2textcolor{blue}{x}$ , width $textcolor{blue}{x}$ and height $textcolor{purple}{h}$ , and we have a limit of $600text{ cm}^2$ of wood to use.

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

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

and

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

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

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

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

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

$dfrac{dtextcolor{red}{V}}{dtextcolor{blue}{x}} = 200 – 4textcolor{blue}{x}^2 = 0$ , giving $textcolor{blue}{x} = sqrt{50}$

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

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

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

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,

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

and

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

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

We can find an equation for the velocity:

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

and an equation for the acceleration:

$textcolor{blue}{a} = 12textcolor{purple}{t} – 6$

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

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

$textcolor{blue}{a} = 12left( dfrac{1}{2} + sqrt{dfrac{11}{12}}right) – 6 = 11.49$ , 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 $30000text{ cm}^2$ 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 $dfrac{d^2V}{dx^2}$ :

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

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

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

[4 marks]

$h = 1000left( dfrac{t^2}{40} – dfrac{t^3}{600}right)$ 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 $1000left( dfrac{t}{20}left( 1 – dfrac{t}{10}right) right) = 0$ , meaning that $t = 0$ or $t = 10$ .

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

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Question 3: Let a bath be modelled as a trapezium with the dimensions shown in the diagram, and width $1text{ m}$ . Assuming that water occupies a volume of $dfrac{1}{2}t^2 + 4ttext{ cm}^3$ , 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.