Applications of Differentiation

A LevelAQAEdexcelOCR

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.

A LevelAQAEdexcelOCR

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 LevelAQAEdexcelOCR

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 LevelAQAEdexcelOCR

Applications of Differentiation 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}.

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