Applications of Differentiation

In short: Differentiation finds rates of change and best values in real problems. Velocity is the derivative of displacement and acceleration is the derivative of velocity, and to find a maximum or minimum you write one formula in one variable, set its derivative to zero and check the endpoints.

Starting point: Years 9 and 10

Formulas you need

Shape	Perimeter or surface area	Area or volume
Rectangle	$P = 2 l + 2 w$	$A = l w$
Circle	$C = 2 π r$	$A = π r^{2}$
Rectangular box	$S A = 2 (l w + l h + w h)$	$V = l w h$
Cylinder	$S A = 2 π r^{2} + 2 π r h$	$V = π r^{2} h$

Speed, distance and time

$speed = \frac{distance}{time}$

On a distance-time graph, the gradient is the speed. On a velocity-time graph, the gradient is the acceleration.

Best values with quadratics

A quadratic with $a < 0$ has a maximum at its vertex, $x = - \frac{b}{2 a}$ .

Applications

Rates of change

dy/dx is a rate
Units: y per x

Motion

x: displacement
v = dx/dt
a = dv/dt

Optimisation

One variable formula
Set derivative to 0
Check it is max or min
Check endpoints

Year 12Motion in a straight line

Displacement $x$ : position relative to the origin.
Velocity $v = \frac{d x}{d t}$ : rate of change of displacement. Its sign shows direction.
Acceleration $a = \frac{d v}{d t} = \frac{d ^{2} x}{d t ^{2}}$ : rate of change of velocity.

Example. $x = t^{3} - 6 t^{2} + 9 t$ metres.

$v = 3 t^{2} - 12 t + 9 = 3 (t - 1) (t - 3)$ , so it is at rest at $t = 1$ and $t = 3$ .

$a = 6 t - 12$ , so acceleration is zero at $t = 2$ .

Speed is the size of velocity. A particle is speeding up when $v$ and $a$ have the same sign.

Going backwards, integration turns acceleration into velocity and velocity into displacement.

Which way does it move? The sign of $v$ gives the direction: $v > 0$ means moving right (in the positive direction), $v < 0$ means moving left.

Total distance is not always the change in displacement, because the particle can turn back. Find when $v = 0$ , then add the distance of each part. For $x = t^{2} - 4 t + 3$ on $0 \leq t \leq 5$ : $v = 2 t - 4 = 0$ at $t = 2$ . The positions are $x (0) = 3$ , $x (2) = - 1$ and $x (5) = 8$ . Distance $= 4 + 9 = 13$ m, even though the displacement only changed by 5.

Year 12Optimisation step by step

Draw a diagram and name the variables.
Write the quantity to optimise in terms of one variable.
Differentiate and set the derivative equal to zero.
Check it is a maximum or minimum (second derivative or table).
Check the endpoints of the allowed domain. Answer the question with units.

The cheapest can. $V = π r^{2} h = 375$ , so $h = \frac{375}{π r ^{2}}$ . The metal used is

$S = 2 π r^{2} + 2 π r h = 2 π r^{2} + \frac{750}{r}$

$\frac{d S}{d r} = 4 π r - \frac{750}{r ^{2}} = 0$ gives $r^{3} = \frac{750}{4 π}$ , so $r \approx 3.91$ cm.

$\frac{d ^{2} S}{d r ^{2}} = 4 π + \frac{1500}{r ^{3}} > 0$ , so this is a minimum. Then $h = \frac{375}{π ( 3.91 ) ^{2}} \approx 7.81$ cm. The cheapest can has its height equal to its diameter.

Related rates. When two quantities change with time, link their rates with the chain rule. A cube's volume is $V = s^{3}$ , so $\frac{d V}{d t} = \frac{d V}{d s} \times \frac{d s}{d t} = 3 s^{2} \frac{d s}{d t}$ . If $s = 4$ cm and the side shrinks at 0.1 cm/min, $\frac{d V}{d t} = 3 (16) (- 0.1) = - 4.8$ cm³/min.

Costs that depend on time. For a trip of 400 km at $v$ km/h, the time is $\frac{400}{v}$ hours. If running costs $C (v)$ dollars per hour, the trip costs $\frac{400}{v} \times C (v)$ . Minimise that.

Shortest distance to a curve. The distance from $(0, 3)$ to a point $(x, x^{2})$ is $x^{2} + (x^{2} - 3)^{2}$ . Minimise the square of the distance instead, $D = x^{2} + (x^{2} - 3)^{2}$ , because it is smallest at the same place and has no root to differentiate. $D^{'} = 2 x + 4 x (x^{2} - 3) = 2 x (2 x^{2} - 5) = 0$ gives $x^{2} = \frac{5}{2}$ , and the shortest distance is $\frac{5}{2} + \frac{1}{4} = \frac{11}{2}$ .

Quick summary

A derivative is a rate of change. Its units are " $y$ per $x$ ".
$v = \frac{d x}{d t}$ and $a = \frac{d v}{d t}$ . At rest means $v = 0$ .
For optimisation, write one formula in one variable, set the derivative to zero, and check the nature and the endpoints.
A quadratic's best value is at its vertex, $x = - \frac{b}{2 a}$ .

Quiz

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Try 3 questions now

A rectangle has perimeter 40 cm and width $x$ . Its length is:
1. A $40 - x$
2. B $20 - x$
3. C $40 - 2 x$
4. D $\frac{40}{x}$
Show answer
B: $20 - x$
$2 l + 2 x = 40$ , so $l = 20 - x$ .
What is the volume of a cylinder with radius 3 and height 10?
1. A $30 π$
2. B $60 π$
3. C $90 π$
4. D $300 π$
Show answer
C: $90 π$
$π r^{2} h = π \times 9 \times 10$ .
On a distance-time graph, the gradient represents:
1. A speed
2. B acceleration
3. C distance
4. D time
Show answer
A: speed
Distance per time.