Leaving Certificate · Higher Level · 2023 specification

Applied Maths formula sheet

Every formula on the course — 190 of them across all 12 strands — with what each symbol means and when the formula actually applies. Free to print, free to hand out, no account needed.

See the practice questions

Units & Vectors

13 formulas · pages 5-28

The newton

Dimensional Analysis

1\,\mathrm{N} = 1\,\mathrm{kg\,m\,s^{-2}}

The force needed to accelerate one kilogram at one metre per second squared.

$\mathrm{N}$: newton, the SI unit of force $\mathrm{kg\,m\,s^{-2}}$

Used whenever a force must be reduced to m, kg and s for a dimensional analysis.

p. 6

The joule

Dimensional Analysis

1\,\mathrm{J} = 1\,\mathrm{kg\,m^2\,s^{-2}}

The work done when a force of one newton moves through one metre.

$\mathrm{J}$: joule, the SI unit of work and of energy $\mathrm{kg\,m^2\,s^{-2}}$

Work and energy share this unit, so both reduce the same way.

p. 6

The watt

Dimensional Analysis

1\,\mathrm{W} = 1\,\mathrm{kg\,m^2\,s^{-3}}

One joule of work done per second.

$\mathrm{W}$: watt, the SI unit of power $\mathrm{kg\,m^2\,s^{-3}}$

Power is work per unit time, which is where the third power of seconds comes from.

p. 6

Magnitude of a vector

The i - j plane

\lvert a\vec{i} + b\vec{j} \rvert = \sqrt{a^2 + b^2}

Length of a vector given in component form, straight from Pythagoras' Theorem.

$a$: component along the $\vec{i}$ -axis
$b$: component along the $\vec{j}$ -axis

Signs do not matter here because both components are squared.

p. 13

Direction of a vector

The i - j plane

\tan\theta = \frac{b}{a}

Angle a vector makes with the $\vec{i}$ -axis.

$\theta$: angle measured from the positive $\vec{i}$ -axis $^\circ$
$a$: $\vec{i}$ component
$b$: $\vec{j}$ component

The $\vec{j}$ component is on top - inverting the fraction gives the complement of the correct angle.
Check the quadrant against a sketch; $\tan^{-1}$ alone cannot distinguish opposite directions.

p. 13

Unit vector

The i - j plane

\hat{a} = \frac{\vec{a}}{\lvert\vec{a}\rvert}

A vector one unit long in the same direction as $\vec{a}$ .

$\hat{a}$: unit vector in the direction of $\vec{a}$
$\lvert\vec{a}\rvert$: magnitude of $\vec{a}$

Undefined for the zero vector, which has no direction.

p. 14

Dot product

Dot products

(a\vec{i} + b\vec{j}) \cdot (c\vec{i} + d\vec{j}) = ac + bd

Multiply matching components and add; the result is a real number, not a vector.

$a, b$: components of the first vector
$c, d$: components of the second vector

Commutative: $\vec{p} \cdot \vec{q} = \vec{q} \cdot \vec{p}$ .
Distributive over addition: $\vec{x} \cdot (\vec{y} + \vec{z}) = \vec{x} \cdot \vec{y} + \vec{x} \cdot \vec{z}$ .

p. 15

Perpendicularity test

Dot products

\vec{p} \perp \vec{q} \iff \vec{p} \cdot \vec{q} = 0

Two non-zero vectors are perpendicular exactly when their dot product vanishes.

$\vec{p}, \vec{q}$: the two vectors being tested

Both vectors must be non-zero.
This is an 'if and only if', so it works as a test in either direction.

p. 15

Angle between two vectors

Dot products

\cos\theta = \frac{\vec{p} \cdot \vec{q}}{\lvert\vec{p}\rvert \lvert\vec{q}\rvert}

The smaller angle between two vectors.

$\theta$: smaller angle between the vectors $^\circ$
$\vec{p} \cdot \vec{q}$: their dot product
$\lvert\vec{p}\rvert, \lvert\vec{q}\rvert$: their magnitudes

Given indirectly on page 17 of Formulae and Tables.
A negative $\cos\theta$ means the angle is obtuse.

p. 16

Adjacent side

Writing vectors in terms of i and j

\text{Adjacent} = \text{Hypotenuse} \times \cos\theta

Component of a vector along the axis the angle is measured from.

$\theta$: angle at the origin between the vector and the axis $^\circ$
Hypotenuse: magnitude of the vector

Which side counts as adjacent depends on where the angle is marked - sketch it first.

p. 17

Opposite side

Writing vectors in terms of i and j

\text{Opposite} = \text{Hypotenuse} \times \sin\theta

Component of a vector perpendicular to the axis the angle is measured from.

$\theta$: angle at the origin between the vector and the axis $^\circ$
Hypotenuse: magnitude of the vector

The sign of the finished component comes from the sketch, not from this formula.

p. 17

Polar form to component form

Vectors in Polar Form

\langle r \angle A \rangle = r\cos A\,\vec{i} + r\sin A\,\vec{j}

Converts a magnitude-and-argument description into $\vec{i}$ and $\vec{j}$ components.

$r$: magnitude of the vector
$A$: argument, measured anti-clockwise from the positive $x$ -axis $^\circ$

The formula produces the correct signs on its own for any argument, including those past $90^\circ$ .
The argument may be in degrees or radians - set the calculator to match.

p. 21

Component form to polar form

Vectors in Polar Form

x\vec{i} + y\vec{j} = \left\langle \sqrt{x^2 + y^2} \, \angle \, \tan^{-1}\frac{y}{x} \right\rangle

Converts components into magnitude and argument.

$x$: $\vec{i}$ component
$y$: $\vec{j}$ component

$\tan^{-1}$ returns an angle between $-90^\circ$ and $90^\circ$ only - adjust for the quadrant using a sketch.

p. 21

Uniform Acceleration

14 formulas · pages 29-46

Definition of acceleration

Uniform acceleration

a = \frac{v - u}{t}

Acceleration is the rate at which velocity changes: the change in velocity divided by the time it took.

$a$: acceleration $\mathrm{m\,s^{-2}}$
$u$: initial velocity $\mathrm{m\,s^{-1}}$
$v$: final velocity $\mathrm{m\,s^{-1}}$
$t$: time taken $\mathrm{s}$

The acceleration must be uniform, otherwise this gives only the average acceleration.
A negative result means the body is slowing; quote it as a deceleration or retardation of that magnitude.

p. 30

First equation of motion

Uniform acceleration

v = u + at

The velocity after time $t$ , when the acceleration is constant. Use it when displacement is neither known nor wanted.

$v$: final velocity $\mathrm{m\,s^{-1}}$
$u$: initial velocity $\mathrm{m\,s^{-1}}$
$a$: acceleration $\mathrm{m\,s^{-2}}$
$t$: time $\mathrm{s}$

Uniform acceleration only.
Equation 2.1 in the book; printed on page 50 of Formulae and Tables.
The one equation with no $s$ in it.

p. 30

Second equation of motion

Uniform acceleration

s = \left(\frac{u + v}{2}\right)t

Distance as average velocity times time. Under uniform acceleration the average velocity really is the mean of the first and last velocities.

$s$: displacement $\mathrm{m}$
$u$: initial velocity $\mathrm{m\,s^{-1}}$
$v$: final velocity $\mathrm{m\,s^{-1}}$
$t$: time $\mathrm{s}$

Uniform acceleration only - the mean-of-the-endpoints shortcut fails otherwise.
Equation 2.2 in the book.
The one equation with no $a$ in it, so reach for it when the acceleration is unknown.

p. 30

Third equation of motion

Uniform acceleration

s = ut + \frac{1}{2}at^2

Displacement from the initial velocity, the acceleration and the time. It is the area under the time-velocity graph: a rectangle plus a triangle.

$s$: displacement $\mathrm{m}$
$u$: initial velocity $\mathrm{m\,s^{-1}}$
$a$: acceleration $\mathrm{m\,s^{-2}}$
$t$: time $\mathrm{s}$

Uniform acceleration only.
Equation 2.3 in the book, obtained by substituting $v = u + at$ into Equation 2.2.
The one equation with no $v$ in it.
Quadratic in $t$ , so solving for time can give two roots - discard any that is negative.

p. 30

Fourth equation of motion

Uniform acceleration

v^2 = u^2 + 2as

Relates the two velocities to the distance covered, with time eliminated entirely.

$v$: final velocity $\mathrm{m\,s^{-1}}$
$u$: initial velocity $\mathrm{m\,s^{-1}}$
$a$: acceleration $\mathrm{m\,s^{-2}}$
$s$: displacement $\mathrm{m}$

Uniform acceleration only.
Equation 2.4 in the book, obtained by eliminating $t$ between Equations 2.1 and 2.2.
The one equation with no $t$ in it - the first choice whenever time is neither given nor asked for.

p. 30

Area under a time-velocity graph

Using Time-Velocity graphs

s = \text{area under the velocity-time curve}

The distance travelled equals the area between the graph and the time axis. Unlike the four equations, this holds even when the acceleration is not uniform.

$s$: distance travelled $\mathrm{m}$

Holds for any motion, uniform or not - this is what makes it so useful.
Split a multi-stage journey into triangles, rectangles and trapezia and add the areas.
The slope of the same graph is the acceleration; do not confuse the two.

p. 33

Average speed

Using Time-Velocity graphs

\text{average speed} = \frac{\text{total distance}}{\text{total time}}

Computed once across the whole journey. It is not the average of the individual stage speeds.

total distance: sum of all stage areas $\mathrm{m}$
total time: sum of all stage durations $\mathrm{s}$

Add every stage, including short final ones, before dividing.
For a rest-to-rest triangular journey the average speed is exactly half the top speed.

p. 34

Top speed from acceleration and deceleration

Problems with just Acceleration and Deceleration

\frac{v}{a} + \frac{v}{d} = T

For a journey that accelerates from rest to $v$ and immediately decelerates to rest, the two stage times must add to the total time.

$v$: top speed reached $\mathrm{m\,s^{-1}}$
$a$: acceleration $\mathrm{m\,s^{-2}}$
$d$: deceleration, entered as a positive magnitude $\mathrm{m\,s^{-2}}$
$T$: total time from rest to rest $\mathrm{s}$

Requires rest to rest with no constant-speed stage in between.
The two stage times come out in the ratio $d : a$ , so the larger rate takes the shorter time.

p. 37

Distance for a rest-to-rest triangular journey

Problems with just Acceleration and Deceleration

s = \frac{1}{2}Tv

The area of the triangle: half the total time times the top speed.

$s$: total distance $\mathrm{m}$
$T$: total time $\mathrm{s}$
$v$: top speed $\mathrm{m\,s^{-1}}$

Only for rest to rest with no constant-speed stage; with one the shape is a trapezium instead.

p. 37

Overtaking condition

Problems involving two objects

s_P = s_Q

Two bodies that set out from the same point at the same instant are level when they have covered the same distance.

$s_P$: distance travelled by the first body $\mathrm{m}$
$s_Q$: distance travelled by the second body $\mathrm{m}$

Both must be measured from the same point and the same instant.
If $P$ started a distance $k$ behind $Q$ , the condition becomes $s_P = s_Q + k$ .
If $B$ sets out $n$ seconds after $A$ , and $t$ is $A$ 's time on the road, then $B$ 's time is $t - n$ .

p. 39

Meeting condition for bodies approaching each other

Problems involving two objects

s_P + s_Q = l

Two bodies moving towards each other meet when the distances they have covered add up to the gap that separated them.

$s_P$: distance travelled by the first body $\mathrm{m}$
$s_Q$: distance travelled by the second body $\mathrm{m}$
$l$: initial separation $\mathrm{m}$

Applies only when they move towards each other along the same line.
The greatest gap between two bodies instead occurs when their speeds are equal.

p. 46

Equations of motion under gravity

Motion under Gravity

s = ut + \frac{1}{2}gt^2

The third equation of motion with the acceleration replaced by $g$ , the constant rate at which gravity accelerates a freely moving body.

$s$: displacement $\mathrm{m}$
$u$: initial velocity $\mathrm{m\,s^{-1}}$
$g$: acceleration due to gravity, $9.8$ at Higher Level and $10$ at Ordinary Level $\mathrm{m\,s^{-2}}$
$t$: time $\mathrm{s}$

Higher Level uses $g = 9.8\,\mathrm{m\,s^{-2}}$ .
Fix a positive direction first; if up is positive then $g$ enters as $-9.8$ .
Applies to free motion under gravity only, with air resistance ignored.

p. 42

Greatest height of a body projected upwards

Motion under Gravity

h_{\max} = \frac{u^2}{2g}

The highest point reached, found by setting the velocity to zero in $v^2 = u^2 - 2gh$ .

$h_{\max}$: greatest height above the launch point $\mathrm{m}$
$u$: launch speed, vertically upwards $\mathrm{m\,s^{-1}}$
$g$: acceleration due to gravity, $9.8$ at Higher Level $\mathrm{m\,s^{-2}}$

The velocity is zero at the highest point, but the acceleration is still $g$ .
Measured from the point of projection, so add the launch height if the body was thrown from above ground.
For a launch and return to the same level, the total flight time is $\frac{2u}{g}$ .

p. 42

Average speed over an interval equals the midpoint speed

Motion under Gravity

v_{\text{avg over } [t_1,\,t_2]} = v\left(\frac{t_1 + t_2}{2}\right)

Under uniform acceleration the average speed across any interval is exactly the instantaneous speed at the middle of that interval.

$t_1$: start of the interval $\mathrm{s}$
$t_2$: end of the interval $\mathrm{s}$

Requires uniform acceleration; it fails if the acceleration changes.
The phrase 'during the $n$ th second' means the interval from $t = n-1$ to $t = n$ , whose midpoint is $t = n - 0.5$ .

p. 42

Projectiles

17 formulas · pages 47-63

Acceleration of a projectile

Why study projectiles?

\vec{a} = 0\vec{i} - g\vec{j}

The only force acting is gravity, so the acceleration is $g$ downwards and there is none horizontally.

$\vec{a}$: acceleration of the projectile $\mathrm{m\,s^{-2}}$
$g$: acceleration due to gravity, $9.8$ at Higher Level and $10$ at Ordinary Level $\mathrm{m\,s^{-2}}$

Air resistance is ignored.
The flight is short enough that $g$ can be taken as constant.

p. 54

Horizontal velocity is constant

Why study projectiles?

v_x = u_x

Gravity has no horizontal component, so the horizontal velocity never changes during the flight.

$u_x$: initial horizontal component of velocity $\mathrm{m\,s^{-1}}$
$v_x$: horizontal component of velocity at time $t$ $\mathrm{m\,s^{-1}}$

No air resistance and no horizontal force.

p. 54

Horizontal displacement

When the speed and angle are known

s_x = u_xt

Horizontal distance travelled after time $t$ . There is no $\tfrac{1}{2}at^2$ term because $a_x = 0$ .

$s_x$: horizontal displacement from the point of projection $\mathrm{m}$
$u_x$: horizontal component of the initial velocity $\mathrm{m\,s^{-1}}$
$t$: time since projection $\mathrm{s}$

Valid for the whole flight, since the horizontal motion is uniform.

p. 54

Vertical velocity

When the speed and angle are known

v_y = u_y - gt

The vertical component of velocity falls by $g$ every second.

$v_y$: vertical component of velocity at time $t$ $\mathrm{m\,s^{-1}}$
$u_y$: initial vertical component of velocity $\mathrm{m\,s^{-1}}$
$g$: acceleration due to gravity $\mathrm{m\,s^{-2}}$
$t$: time since projection $\mathrm{s}$

Upwards taken as positive.

p. 54

Vertical displacement

When the speed and angle are known

s_y = u_yt - \frac{1}{2}gt^2

Height above the point of projection after time $t$ . It is negative when the particle is below its launch point.

$s_y$: vertical displacement from the point of projection $\mathrm{m}$
$u_y$: initial vertical component of velocity $\mathrm{m\,s^{-1}}$
$g$: acceleration due to gravity $\mathrm{m\,s^{-2}}$
$t$: time since projection $\mathrm{s}$

Upwards taken as positive, with the point of projection as the origin.
For a projectile fired from a cliff of height $h$ , the sea is at $s_y = -h$ .

p. 54

Resolving speed and angle into components

When the speed and angle are known

\vec{u} = u\cos\alpha\,\vec{i} + u\sin\alpha\,\vec{j}

Turns a speed and an angle of projection into the $\vec{i}$ and $\vec{j}$ components every other formula needs.

$u$: speed of projection $\mathrm{m\,s^{-1}}$
$\alpha$: angle of projection above the horizontal

$\alpha$ measured from the horizontal, not from the vertical.

p. 52

Speed at any instant

Projectiles fired at an angle

v = \sqrt{v_x^2 + v_y^2}

The speed is the magnitude of the velocity vector, never the sum of its components.

$v$: speed $\mathrm{m\,s^{-1}}$
$v_x$: horizontal component of velocity $\mathrm{m\,s^{-1}}$
$v_y$: vertical component of velocity $\mathrm{m\,s^{-1}}$

p. 63

Direction of travel

Projectiles fired at an angle

\tan\theta = \frac{v_y}{v_x}

The angle the velocity makes with the horizontal. It is negative when the particle is descending.

$\theta$: angle of the velocity to the horizontal
$v_x$: horizontal component of velocity $\mathrm{m\,s^{-1}}$
$v_y$: vertical component of velocity $\mathrm{m\,s^{-1}}$

p. 55

Velocity and position vectors

Projectiles fired at an angle

\vec{v} = v_x\vec{i} + v_y\vec{j}, \qquad \vec{r} = s_x\vec{i} + s_y\vec{j}

The velocity and the position vector of the projectile at time $t$ , both measured relative to the point of projection.

$\vec{v}$: velocity vector at time $t$ $\mathrm{m\,s^{-1}}$
$\vec{r}$: position vector relative to the point of projection $\mathrm{m}$

The point of projection is taken as the origin.

p. 54

Time to the greatest height

Projectiles launched vertically

t = \frac{u_y}{g}

At the top of the flight $v_y = 0$ , so the rise takes $u_y/g$ seconds.

$t$: time to reach the greatest height $\mathrm{s}$
$u_y$: initial vertical component of velocity $\mathrm{m\,s^{-1}}$
$g$: acceleration due to gravity $\mathrm{m\,s^{-2}}$

p. 48

Greatest height from the vertical components

Projectiles launched vertically

H = \frac{u_y^2}{2g}

Greatest height above the point of projection, from $v^2 = u^2 + 2as$ applied in the $y$ -direction with $v_y = 0$ .

$H$: greatest height above the point of projection $\mathrm{m}$
$u_y$: initial vertical component of velocity $\mathrm{m\,s^{-1}}$
$g$: acceleration due to gravity $\mathrm{m\,s^{-2}}$

Measured from the point of projection, not from the ground, if the two differ.

p. 55

Time of flight on a horizontal plane

Projectiles fired at an angle

T = \frac{2u_y}{g} = \frac{2u\sin\alpha}{g}

Total time in the air for a projectile that lands at the same level it was fired from.

$T$: time of flight $\mathrm{s}$
$u_y$: initial vertical component of velocity $\mathrm{m\,s^{-1}}$
$u$: speed of projection $\mathrm{m\,s^{-1}}$
$\alpha$: angle of projection

Landing point level with the point of projection. Not valid from a cliff.

p. 57

Greatest height in terms of speed and angle

Maximum Range

H = \frac{u^2\sin^2\alpha}{2g}

Greatest height reached by a particle projected with speed $u$ at angle $\alpha$ .

$H$: greatest height above the point of projection $\mathrm{m}$
$u$: speed of projection $\mathrm{m\,s^{-1}}$
$\alpha$: angle of projection above the horizontal
$g$: acceleration due to gravity $\mathrm{m\,s^{-2}}$

Must be derived on paper in the exam - it is not in the Formulae and Tables booklet.

p. 57

Range on a horizontal plane

Maximum Range

R = \frac{u^2\sin 2\alpha}{g}

Horizontal distance from the point of projection to the landing point, on level ground.

$R$: range $\mathrm{m}$
$u$: speed of projection $\mathrm{m\,s^{-1}}$
$\alpha$: angle of projection above the horizontal
$g$: acceleration due to gravity $\mathrm{m\,s^{-2}}$

Landing point level with the point of projection.
Uses the identity $\sin 2\alpha = 2\sin\alpha\cos\alpha$ .

p. 58

Maximum range

Maximum Range

R_{\max} = \frac{u^2}{g} \quad \text{at} \quad \alpha = 45^\circ

$\sin 2\alpha$ is greatest when it equals $1$ , which happens at $2\alpha = 90^\circ$ .

$R_{\max}$: greatest range obtainable with a fixed speed $\mathrm{m}$
$u$: speed of projection $\mathrm{m\,s^{-1}}$
$g$: acceleration due to gravity $\mathrm{m\,s^{-2}}$

Level ground only.
At this angle the ratio of greatest height to range is $1 : 4$ .

p. 58

Double angle identity

Maximum Range

\sin 2A = 2\sin A\cos A

The identity that collapses $2u^2\sin\alpha\cos\alpha$ into $u^2\sin 2\alpha$ when deriving the range.

$A$: any angle

Page 14 of the Formulae and Tables booklet.

p. 58

Secant identity for target problems

Maximum Range

\frac{1}{\cos^2 A} = \sec^2 A = 1 + \tan^2 A

Used with $\tan A = \frac{\sin A}{\cos A}$ to turn a target-practice equation into a quadratic in $\tan A$ .

$A$: angle of projection

Page 13 of the Formulae and Tables booklet.

p. 59

Newton's laws & connected particles

14 formulas · pages 64-88

Newton's second law

Newton's Laws of Motion

F = ma

The resultant of all the forces acting on a body equals its mass times its acceleration.

$F$: resultant force, the sum of every force acting $\mathrm{N}$
$m$: mass of the body $\mathrm{kg}$
$a$: acceleration produced $\mathrm{m\,s^{-2}}$

$F$ must be the resultant, not any single named force.
Mass in kilograms - convert grams first.

p. 66

Momentum

Newton's Laws of Motion

\vec{p} = m\vec{v}

Momentum is mass times velocity, and is a vector in the direction of the velocity.

$\vec{p}$: momentum $\mathrm{kg\,m\,s^{-1}}$
$m$: mass $\mathrm{kg}$
$\vec{v}$: velocity $\mathrm{m\,s^{-1}}$

Momentum has no unit of its own; it is measured in $\mathrm{kg\,m\,s^{-1}}$ .

p. 65

Weight

Newton's Laws of Motion

W = mg

Weight is the force of gravity on a body. It is a force in newtons, not a mass in kilograms.

$W$: weight $\mathrm{N}$
$m$: mass $\mathrm{kg}$
$g$: acceleration due to gravity, $9.8$ at Higher Level $\mathrm{m\,s^{-2}}$

Mass is the same everywhere; weight depends on the local value of $g$ .

p. 66

Equation of motion for a hanging particle

Tension

T - mg = ma

For a particle hanging from a string and accelerating upwards, with up taken as positive.

$T$: tension in the string $\mathrm{N}$
$m$: mass of the particle $\mathrm{kg}$
$a$: acceleration, upwards positive $\mathrm{m\,s^{-2}}$

Reverse to $mg - T = ma$ when the particle accelerates downwards.
In equilibrium $a = 0$ and the equation reduces to $T = mg$ .

p. 75

Normal reaction on a horizontal surface

Two More Kinds of Force

R = mg

The perpendicular push of a horizontal surface on a body resting on it, when no other vertical force acts.

$R$: normal reaction $\mathrm{N}$
$m$: mass $\mathrm{kg}$
$g$: acceleration due to gravity $\mathrm{m\,s^{-2}}$

No vertical component from any applied force.
The body moves neither up nor down.

p. 71

Normal reaction with an inclined pull

Two More Kinds of Force

R = mg - T\sin A

A rope pulling at angle $A$ above the horizontal lifts part of the weight, reducing the reaction and therefore the friction.

$R$: normal reaction $\mathrm{N}$
$T$: tension in the rope $\mathrm{N}$
$A$: angle of the rope above the horizontal

Add instead of subtract when the applied force pushes downwards at an angle.

p. 73

Coefficient of friction

The Laws of Friction

\mu = \frac{F}{R}

The fixed ratio of limiting friction to normal reaction for two given surfaces.

$\mu$: coefficient of friction
$F$: friction force at the point of slipping $\mathrm{N}$
$R$: normal reaction $\mathrm{N}$

Depends on the nature of the two surfaces, not on the area, size or shape of the body.
Valid when the body is moving or just on the point of moving.

p. 71

Limiting friction

The Laws of Friction

F_{\text{limiting}} = \mu R

The greatest friction the surfaces can supply. Below this, friction equals whatever force is trying to move the body.

$F_{\text{limiting}}$: limiting friction $\mathrm{N}$
$\mu$: coefficient of friction
$R$: normal reaction $\mathrm{N}$

A driving force greater than $\mu R$ moves the body; anything less and it stays put.
Once moving, the friction stays at $\mu R$ and is independent of the speed.

p. 72

Motion along a rough horizontal surface

The Laws of Friction

P - \mu R = ma

Equation of motion once the driving force exceeds the limiting friction.

$P$: driving force along the surface $\mathrm{N}$
$\mu R$: friction, opposing the motion $\mathrm{N}$
$m$: mass $\mathrm{kg}$
$a$: acceleration $\mathrm{m\,s^{-2}}$

Use the horizontal component of $P$ if the driving force is inclined.

p. 72

Weight resolved on a slope

Particles on slopes

mg\sin A \text{ along the slope}, \quad mg\cos A \text{ perpendicular to it}

On an inclined plane the weight must be split into a component down the line of greatest slope and one at right angles to the surface.

$m$: mass of the particle $\mathrm{kg}$
$A$: angle of the slope to the horizontal
$g$: acceleration due to gravity $\mathrm{m\,s^{-2}}$

A slope of '1 in $n$ ' means $\sin A = \frac{1}{n}$ .
Do not leave the original $mg$ on the diagram once it has been replaced by its components.

p. 73

Reaction and acceleration on a smooth slope

Particles on slopes

R = mg\cos A, \qquad a = g\sin A

Nothing moves perpendicular to the plane, so $R$ balances $mg\cos A$ ; the unbalanced $mg\sin A$ then gives an acceleration independent of the mass.

$R$: normal reaction $\mathrm{N}$
$a$: acceleration down the line of greatest slope $\mathrm{m\,s^{-2}}$
$A$: angle of the slope

Smooth plane, no string and no applied force along the slope.
On a rough slope subtract $\mu mg\cos A$ before dividing by $m$ .

p. 74

Two masses over a fixed smooth pulley

Systems of connected particles

a = \frac{(M - m)g}{M + m}

Common acceleration when two particles hang freely from either side of a fixed smooth pulley.

$M$: the heavier mass, which descends $\mathrm{kg}$
$m$: the lighter mass, which rises $\mathrm{kg}$
$a$: common acceleration $\mathrm{m\,s^{-2}}$

Light inextensible string, smooth light pulley.
Obtained by adding $Mg - T = Ma$ and $T - mg = ma$ .

p. 77

Block on a table with masses over both edges

Systems of connected particles

a = \frac{(m' - m)g}{m' + m + M}

Common acceleration of a mass $M$ on a smooth horizontal table pulled by hanging masses $m$ and $m'$ over opposite edges.

$M$: mass on the table $\mathrm{kg}$
$m'$: the heavier hanging mass $\mathrm{kg}$
$m$: the lighter hanging mass $\mathrm{kg}$

Smooth table. If the table is rough, subtract $\mu Mg$ from the numerator.
The two strings are separate, so their tensions $T$ and $S$ differ.

p. 79

Movable pulley

Systems of connected particles

2T - Mg = Ma

A movable pulley of mass $M$ is held up by two segments of the same string, so the upward force on it is $2T$ , not $T$ .

$T$: tension in the string $\mathrm{N}$
$M$: mass of the movable pulley $\mathrm{kg}$
$a$: acceleration of the pulley $\mathrm{m\,s^{-2}}$

If the pulley moves $x$ , the free end of the string moves $2x$ , so the free particle has acceleration $2a$ .
To decide which way the system moves, compare $\frac{M}{2}$ with the hanging mass.

p. 79

Work, Power, Energy & Momentum

16 formulas · pages 89-105

Work done by a constant force

Work and Power

W = Fs

Work is the force along the direction of motion multiplied by the distance moved. It is a scalar in joules.

$W$: work done $\mathrm{J}$
$F$: component of the force along the motion $\mathrm{N}$
$s$: distance the point of application moves $\mathrm{m}$

The force must be constant.
$1\,\mathrm{J} = 1\,\mathrm{N\,m} = 1\,\mathrm{kg\,m^2\,s^{-2}}$ .

p. 90

Work done by an inclined force

Work and Power

W = Fs\cos A

Only the component along the motion does work; the perpendicular component does none.

$W$: work done $\mathrm{J}$
$F$: magnitude of the applied force $\mathrm{N}$
$A$: angle between the force and the direction of motion
$s$: distance moved $\mathrm{m}$

A force at right angles to the motion, such as the tension in a swing rope, does zero work.

p. 90

Power as a rate of work

Work and Power

P = \frac{W}{t}

Power is the work done per unit time, measured in watts.

$P$: power output $\mathrm{W}$
$W$: work done $\mathrm{J}$
$t$: time taken $\mathrm{s}$

$1\,\mathrm{W} = 1\,\mathrm{J\,s^{-1}}$ and $1000\,\mathrm{W} = 1\,\mathrm{kW}$ .

p. 90

Power, tractive effort and speed

Work and Power

P = Tv

The power output of an engine equals the tractive effort it produces multiplied by the speed at that instant.

$P$: power output of the engine $\mathrm{W}$
$T$: tractive effort, the driving force $\mathrm{N}$
$v$: speed $\mathrm{m\,s^{-1}}$

At a steady speed the tractive effort balances the resistance plus any component of weight down the slope.
With a fixed power output, $T$ falls as $v$ rises.

p. 91

Drag force model

Drag forces

D = kv^n

Resistance from a fluid increases with speed; the exponent $n$ is always stated in the question.

$D$: drag force $\mathrm{N}$
$k$: constant of proportionality, found from the data given
$v$: speed through the fluid $\mathrm{m\,s^{-1}}$
$n$: a stated positive number, commonly 1 or 2

Drag always opposes the motion, so it is subtracted from the driving force.

p. 92

Maximum speed under drag

Drag forces

\frac{P}{v_{\max}} = kv_{\max}^{\,n}

At maximum speed the acceleration is zero, so the tractive effort exactly equals the drag.

$v_{\max}$: maximum speed attainable $\mathrm{m\,s^{-1}}$
$P$: maximum power output $\mathrm{W}$
$k, n$: drag constants

Level ground. On a hill, add the component of weight down the slope to the drag.
Rearranges to $P = kv_{\max}^{\,n+1}$ .

p. 92

Terminal velocity

Drag forces

kv_{\text{term}}^{\,n} = mg

A falling body reaches terminal velocity when the drag has grown enough to balance its weight.

$v_{\text{term}}$: terminal velocity $\mathrm{m\,s^{-1}}$
$m$: mass of the falling body $\mathrm{kg}$
$g$: acceleration due to gravity $\mathrm{m\,s^{-2}}$

At terminal velocity the acceleration is zero, so the body falls at a steady speed.

p. 93

Potential energy

Energy

\text{P.E.} = mgh

Energy a body has because of its height above the standard position, equal to the work it can do falling to that position.

$m$: mass $\mathrm{kg}$
$g$: acceleration due to gravity $\mathrm{m\,s^{-2}}$
$h$: height above the standard position $\mathrm{m}$

Only changes in height matter, so the choice of standard position never affects the answer.

p. 94

Kinetic energy

Energy

\text{K.E.} = \frac{1}{2}mv^2

Energy a body has because of its speed, equal to the work it can do in coming to rest.

$m$: mass $\mathrm{kg}$
$v$: speed $\mathrm{m\,s^{-1}}$

Depends on the square of the speed, so doubling the speed quadruples the energy.
A change in kinetic energy is $\frac{1}{2}mv^2 - \frac{1}{2}mu^2$ , never $\frac{1}{2}m(v-u)^2$ .

p. 94

Speed gained in a free fall

Conservation of Energy

v = \sqrt{2gh}

All the potential energy $mgh$ converts to kinetic energy $\frac{1}{2}mv^2$ , and the mass cancels.

$v$: speed on reaching the lower level $\mathrm{m\,s^{-1}}$
$h$: height fallen $\mathrm{m}$
$g$: acceleration due to gravity $\mathrm{m\,s^{-2}}$

Started from rest, with gravity the only force doing work.
Independent of the mass and of the path taken.

p. 95

Principle of Conservation of Energy

mgh + \frac{1}{2}mv^2 = \text{constant}

If gravitational forces are the only forces doing work on a body, the sum of its potential and kinetic energy is the same at every point of its path.

$m$: mass $\mathrm{kg}$
$h$: height above the standard position $\mathrm{m}$
$v$: speed at that point $\mathrm{m\,s^{-1}}$

Only valid when no non-gravitational force does work - friction and air resistance break it.
A force perpendicular to the motion, such as a swing's tension, does no work and so does not break it.

p. 95

Momentum

Conservation of Momentum

\vec{p} = m\vec{v}

Momentum is mass times velocity, a vector in the direction of the velocity.

$\vec{p}$: momentum $\mathrm{kg\,m\,s^{-1}}$
$m$: mass $\mathrm{kg}$
$\vec{v}$: velocity $\mathrm{m\,s^{-1}}$

A vector: fix a positive direction before using it.

p. 98

Impulse

Conservation of Momentum

\vec{I} = \vec{F}t = m\vec{v} - m\vec{u}

The impulse imparted to a body is its change in momentum - measurable even when the force and the contact time are not.

$\vec{I}$: impulse $\mathrm{N\,s}$
$\vec{F}$: average force during contact $\mathrm{N}$
$t$: contact time $\mathrm{s}$
$\vec{u}, \vec{v}$: velocities before and after the blow $\mathrm{m\,s^{-1}}$

A vector in newton-seconds.
By Newton's third law the impulses on the two bodies are equal and opposite.

p. 98

Conservation of momentum

Conservation of Momentum

m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2

In the absence of an external force in a given direction, the total momentum of the system in that direction is constant.

$m_1, m_2$: the two masses $\mathrm{kg}$
$u_1, u_2$: velocities before the collision $\mathrm{m\,s^{-1}}$
$v_1, v_2$: velocities after the collision $\mathrm{m\,s^{-1}}$

No external force acting in the direction concerned.
Kinetic energy is generally NOT conserved in the same collision.

p. 105

Coalescing bodies

Conservation of Momentum in 2 dimensions

m_1\vec{u}_1 + m_2\vec{u}_2 = (m_1 + m_2)\vec{v}

When two bodies stick together or become entangled they move off with one common velocity.

$\vec{v}$: common velocity of the joint mass $\mathrm{m\,s^{-1}}$
$m_1 + m_2$: total mass after the collision $\mathrm{kg}$

Holds component by component: one equation for $\vec{i}$ and one for $\vec{j}$ .
The speed afterwards is $\lvert \vec{v} \rvert = \sqrt{v_x^2 + v_y^2}$ .

p. 100

Momentum through a jolt on a string

Conservation of Momentum as It Applies to Strings

m_{\text{before}}\,u = m_{\text{after}}\,v

When a system on an inextensible string picks up extra mass, momentum is applied to the whole string-particle system as one body.

$m_{\text{before}}$: total mass of the system before the jolt $\mathrm{kg}$
$u$: common speed just before the jolt $\mathrm{m\,s^{-1}}$
$m_{\text{after}}$: total mass after the jolt $\mathrm{kg}$
$v$: common speed just after the jolt $\mathrm{m\,s^{-1}}$

The string is inextensible, so the jolt is transmitted to every part at once.
Kinetic energy is lost in the jolt, so conservation of energy must not be used across it.
The acceleration must be recalculated with the new masses afterwards.

p. 102

Impacts and collisions

15 formulas · pages 106-122

Conservation of momentum

Impacts

m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2

Total momentum immediately before an impact equals total momentum immediately after it.

$m_1$: mass of the first body $\mathrm{kg}$
$m_2$: mass of the second body $\mathrm{kg}$
$u_1$: velocity of the first body before impact $\mathrm{m\,s^{-1}}$
$u_2$: velocity of the second body before impact $\mathrm{m\,s^{-1}}$
$v_1$: velocity of the first body after impact $\mathrm{m\,s^{-1}}$
$v_2$: velocity of the second body after impact $\mathrm{m\,s^{-1}}$

No external impulse acts during the impact.
All velocities measured in the same positive direction along the line of centres.

p. 107

Impulse

Impacts

J = Ft = mv - mu

The impulse delivered to a body equals its change in momentum.

$J$: impulse $\mathrm{N\,s}$
$F$: average force during contact $\mathrm{N}$
$t$: duration of contact $\mathrm{s}$
$m$: mass $\mathrm{kg}$
$u$: velocity before $\mathrm{m\,s^{-1}}$
$v$: velocity after $\mathrm{m\,s^{-1}}$

$F$ is the average force over the contact time.

p. 107

Newton's law of restitution

Newton's Law of Restitution

e = \frac{v_2 - v_1}{u_1 - u_2}

The speed of separation is $e$ times the speed of approach.

$e$: coefficient of restitution, $0 \le e \le 1$
$u_1, u_2$: velocities before impact $\mathrm{m\,s^{-1}}$
$v_1, v_2$: velocities after impact $\mathrm{m\,s^{-1}}$

Applies only to the component along the line of centres.
$e = 1$ is perfectly elastic; $e = 0$ means the bodies coalesce.

p. 111

Rebound height from a fixed floor

Newton's Law of Restitution

h' = e^2 h

A ball dropped from height $h$ rebounds to $e^2 h$ , because $e$ scales the speed and height goes as the square of speed.

$h$: drop height $\mathrm{m}$
$h'$: rebound height $\mathrm{m}$
$e$: coefficient of restitution

Dropped from rest onto a fixed horizontal surface.
$g$ cancels, so no value of $g$ is needed.

p. 112

Kinetic energy

Direct collisions

\text{KE} = \frac{1}{2}mv^2

Kinetic energy of a body; sum over both bodies to compare before and after.

$m$: mass $\mathrm{kg}$
$v$: speed $\mathrm{m\,s^{-1}}$

Scalar, so no direction is involved - use speed, not velocity.

p. 110

Component parallel to a smooth surface

Oblique collisions

v_{\parallel} = u_{\parallel}

A smooth surface exerts no force along itself, so the parallel component is unchanged.

$u_{\parallel}$: component along the surface before impact $\mathrm{m\,s^{-1}}$
$v_{\parallel}$: component along the surface after impact $\mathrm{m\,s^{-1}}$

Smooth surface only; a rough surface changes this component.

p. 114

Component perpendicular to a smooth surface

Oblique collisions

v_{\perp} = e\,u_{\perp}

The component into the surface is reversed and reduced by the factor $e$ .

$u_{\perp}$: component normal to the surface before impact $\mathrm{m\,s^{-1}}$
$v_{\perp}$: component normal to the surface after impact, opposite in direction $\mathrm{m\,s^{-1}}$
$e$: coefficient of restitution

The normal to the surface is the line of centres for a wall impact.

p. 114

Rebound angle at a smooth wall

Oblique collisions

\tan\beta = e\tan\alpha

Relates rebound angle to incidence angle, with both measured from the wall.

$\alpha$: angle between the incoming velocity and the wall $^\circ$
$\beta$: angle between the outgoing velocity and the wall $^\circ$
$e$: coefficient of restitution

Both angles measured from the WALL. Measuring from the normal inverts the relation.
For $e < 1$ this gives $\beta < \alpha$ - the rebound is always flatter.

p. 115

Time of flight after a bounce

Projectiles which bounce

T = \frac{2v_y'}{g}

Duration of the hop following a bounce, from leaving the ground to landing on the same level.

$T$: time of flight of the hop $\mathrm{s}$
$v_y'$: vertical component of velocity just after the bounce $\mathrm{m\,s^{-1}}$
$g$: acceleration due to gravity, $9.8$ unless the question states otherwise $\mathrm{m\,s^{-2}}$

Launch and landing on the same horizontal level.
$v_y' = e\,v_y$ where $v_y$ is the vertical speed just before the bounce.

p. 119

Restitution as a ratio of velocities

Impacts

\frac{\text{New}}{\text{Old}} = -e

The book's working form for a body bouncing off a fixed surface: the velocity after divided by the velocity before is $-e$ . The minus sign carries the reversal, so the directions look after themselves.

$\text{Old}$: velocity just before impact, signed $\mathrm{m\,s^{-1}}$
$\text{New}$: velocity just after impact, signed $\mathrm{m\,s^{-1}}$
$e$: coefficient of restitution, $0 \le e \le 1$

One body bouncing off a fixed wall, floor or barrier, not two moving bodies.
Both velocities measured in the same direction, so one of them is negative.

p. 107

Successive bounces on horizontal ground

Impacts

h_n = e^{2n}h

Height reached after the $n$ -th bounce of a ball dropped from height $h$ . Each bounce multiplies the speed by $e$ and therefore the height by $e^2$ .

$h$: the original drop height $\mathrm{m}$
$h_n$: height reached after the $n$ -th bounce $\mathrm{m}$
$n$: number of bounces
$e$: coefficient of restitution

Dropped from rest onto horizontal ground, with the same $e$ at every bounce.
$g$ cancels, so no value of $g$ is needed.

p. 109

Restitution for a bounce off a barrier at an angle

Oblique collisions

e = \frac{\tan B}{\tan A}

For a ball striking a smooth barrier at angle $A$ and leaving at angle $B$ , both measured from the barrier, the coefficient of restitution is the ratio of the tangents.

$A$: angle of approach, measured from the barrier
$B$: angle of rebound, measured from the barrier
$e$: coefficient of restitution

Both angles measured from the same reference - the barrier, not the normal.
Smooth barrier, so the component along it is unchanged.

p. 110

The four velocities in an oblique collision

Oblique collisions

a\vec{i} + c\vec{j} \to p\vec{i} + s\vec{j}, \qquad b\vec{i} + d\vec{j} \to q\vec{i} + t\vec{j}

The book's standard layout with $\vec{i}$ along the line of centres. Two of the four unknowns are free: $s = c$ and $t = d$ , because the $\vec{j}$ components are unchanged.

$a, b$: components along the line of centres before impact $\mathrm{m\,s^{-1}}$
$c, d$: components perpendicular to the line of centres before impact $\mathrm{m\,s^{-1}}$
$p, q$: components along the line of centres after impact $\mathrm{m\,s^{-1}}$
$s, t$: components perpendicular after impact, equal to $c$ and $d$ $\mathrm{m\,s^{-1}}$

Both spheres smooth, so no force acts along the $\vec{j}$ direction.
Momentum and restitution are both applied to the $\vec{i}$ components only.

p. 114

Speed arriving at a collision from a swing

Harder Examples

v = \sqrt{2gL(1 - \cos\theta)}

A sphere on a string of length $L$ , released from rest at angle $\theta$ to the vertical, arrives at the lowest point with this speed. It is the input to the collision, not part of it.

$v$: speed at the lowest point of the swing $\mathrm{m\,s^{-1}}$
$L$: length of the string $\mathrm{m}$
$\theta$: angle to the vertical at release
$g$: acceleration due to gravity $\mathrm{m\,s^{-2}}$

Energy is conserved on the swing, because the tension does no work.
Energy is NOT conserved through the collision that follows, so the two stages must stay separate.

p. 114

Kinetic energy lost in a direct collision

Harder Examples

\text{loss} = \frac{m_1m_2}{2(m_1 + m_2)}(u_1 - u_2)^2(1 - e^2)

The energy lost, in one expression. It shows at a glance that no energy is lost when $e = 1$ , and that the loss is greatest when $e = 0$ .

$m_1, m_2$: the two masses $\mathrm{kg}$
$u_1 - u_2$: speed of approach $\mathrm{m\,s^{-1}}$
$e$: coefficient of restitution

Direct collision between two free bodies.
Useful as a check, but the exam expects the two kinetic energies computed and subtracted.

p. 117

Motion in a Circle

16 formulas · pages 123-142

Radian measure

Centripetal accelerations

\theta = \frac{s}{r}

An angle in radians is the arc it subtends divided by the radius. One radian is the angle whose arc equals the radius.

$\theta$: angle subtended at the centre $\mathrm{rad}$
$s$: length of the arc $\mathrm{m}$
$r$: radius of the circle $\mathrm{m}$

A full revolution is $2\pi$ radians, so $180^\circ = \pi$ radians.
Radians are dimensionless - a length divided by a length.

p. 124

Angular speed from revolutions per minute

Centripetal accelerations

\omega = \frac{2\pi n}{60}

Converts a rate given in revolutions per minute into radians per second, which is what every other formula needs.

$\omega$: angular speed $\mathrm{rad\,s^{-1}}$
$n$: number of revolutions per minute

Each revolution is $2\pi$ radians; dividing by 60 converts minutes to seconds.

p. 125

Linear speed in a circle

Centripetal accelerations

v = \omega r

The speed of a point on a rotating body. At the same angular speed, points further from the centre move faster.

$v$: linear speed along the circle $\mathrm{m\,s^{-1}}$
$\omega$: angular speed $\mathrm{rad\,s^{-1}}$
$r$: radius $\mathrm{m}$

$\omega$ must be in $\mathrm{rad\,s^{-1}}$ and $r$ in metres.
Page 51 of Formulae and Tables.

p. 125

Position, velocity and acceleration vectors

Centripetal accelerations

\vec{s} = r\cos\omega t\,\vec{i} + r\sin\omega t\,\vec{j}, \qquad \vec{a} = -\omega^2\vec{s}

Differentiating the position vector twice gives an acceleration that is $-\omega^2$ times the position vector - that is, directed straight at the centre. This is the derivation of the centripetal result.

$\vec{s}$: position vector from the centre $\mathrm{m}$
$\vec{a}$: acceleration vector $\mathrm{m\,s^{-2}}$
$t$: time $\mathrm{s}$

Constant angular speed $\omega$ .
The intermediate result is $\vec{v} = -\omega r\sin\omega t\,\vec{i} + \omega r\cos\omega t\,\vec{j}$ , whose magnitude is $\omega r$ and which is perpendicular to $\vec{s}$ .
Examinable bookwork at Higher Level.

p. 126

Centripetal acceleration

Centripetal accelerations

a = \omega^2 r = \frac{v^2}{r}

Magnitude of the acceleration of a particle moving in a circle, always directed towards the centre.

$a$: centripetal acceleration $\mathrm{m\,s^{-2}}$
$\omega$: angular speed $\mathrm{rad\,s^{-1}}$
$v$: linear speed $\mathrm{m\,s^{-1}}$
$r$: radius $\mathrm{m}$

Present even when the speed is constant, because the direction of the velocity is changing.
Page 51 of Formulae and Tables.

p. 126

Centripetal force

Centripetal accelerations

F = m\omega^2 r = \frac{mv^2}{r}

What $F = ma$ becomes for circular motion. This is the resultant of the real forces, not an extra force to be drawn.

$F$: resultant force towards the centre $\mathrm{N}$
$m$: mass $\mathrm{kg}$
$\omega$: angular speed $\mathrm{rad\,s^{-1}}$
$r$: radius $\mathrm{m}$

Supplied by a real force - tension, reaction, friction or gravity.
There is no outward 'centrifugal force' in this model; never draw one.

p. 126

The two equations for a horizontal circle

Motion in a horizontal circle

F_{\text{up}} = F_{\text{down}}, \qquad F_{\text{centripetal}} = m\omega^2 r

Every horizontal-circle question is these two equations. Nothing accelerates vertically, and the horizontal resultant is centripetal.

$F_{\text{up}}, F_{\text{down}}$: total upward and downward forces $\mathrm{N}$
$F_{\text{centripetal}}$: resultant horizontal force, towards the centre $\mathrm{N}$

The speed is constant in a horizontal circle, so no energy equation is needed.
The radius is the horizontal distance from the axis, not the length of any string.

p. 127

The conical pendulum

Motion in a horizontal circle

T\cos\alpha = mg, \qquad T\sin\alpha = m\omega^2 r

A mass whirled on a string that sweeps out a cone. The vertical component of the tension carries the weight; the horizontal component provides the centripetal force.

$T$: tension in the string $\mathrm{N}$
$\alpha$: angle of the string to the vertical
$r$: radius of the horizontal circle, $= L\sin\alpha$ $\mathrm{m}$

$\alpha$ measured from the vertical. If the question measures from the horizontal, swap $\sin$ and $\cos$ .

p. 128

Angle of a conical pendulum

Motion in a horizontal circle

\tan\alpha = \frac{\omega^2 r}{g}

Dividing the conical pendulum's two equations eliminates the tension, and the mass cancels - the angle does not depend on how heavy the bob is.

$\alpha$: angle of the string to the vertical
$\omega$: angular speed $\mathrm{rad\,s^{-1}}$
$r$: radius of the circle $\mathrm{m}$

Faster rotation means a larger angle, so the cone opens out as $\omega$ increases.

p. 128

Slipping on a rotating turntable

Motion in a horizontal circle

\omega = \sqrt{\frac{\mu g}{r}}

Angular speed at which a body held on a rotating horizontal surface by friction alone is on the point of slipping outwards.

$\mu$: coefficient of friction
$r$: distance from the axis $\mathrm{m}$
$g$: acceleration due to gravity $\mathrm{m\,s^{-2}}$

Friction is the only horizontal force, so $\mu mg = m\omega^2 r$ and the mass cancels.
Slipping begins further out first, since $\omega$ falls as $r$ grows.

p. 131

Energy equation for a vertical circle

Motion in a vertical circle

mgh + \frac{1}{2}mv^2 = \text{constant}

In a vertical circle the speed varies, so energy conservation is needed to find $v$ at each point before any force equation can be written.

$h$: height above the chosen standard level $\mathrm{m}$
$v$: speed at that point $\mathrm{m\,s^{-1}}$

Valid because the tension acts along the radius, perpendicular to the motion, and so does no work.
Not valid if friction or air resistance acts.

p. 133

Radial equation in a vertical circle

Motion in a vertical circle

T - mg\cos\theta = \frac{mv^2}{r}

Resolving along the radius at a general point, with $\theta$ measured from the downward vertical. At the lowest point $\theta = 0$ and at the highest $\theta = 180^\circ$ .

$T$: tension in the string $\mathrm{N}$
$\theta$: angle from the downward vertical
$v$: speed at that instant $\mathrm{m\,s^{-1}}$

Resolve along and perpendicular to the radius, never horizontally and vertically.
At the lowest point this gives $T = mg + \frac{mv^2}{r}$ ; at the highest, $T = \frac{mv^2}{r} - mg$ .

p. 133

Tension at any point of a vertical circle

Motion in a vertical circle

T = \frac{mu^2}{l} - 2mg + 3mg\cos\theta

Combines the energy and radial equations for a particle struck horizontally with speed $u$ at the lowest point of a string of length $l$ .

$u$: speed at the lowest point $\mathrm{m\,s^{-1}}$
$l$: length of the string $\mathrm{m}$
$\theta$: angle from the downward vertical

$\theta = 0$ at the lowest point gives $T = \frac{mu^2}{l} + mg$ .
$\theta = 180^\circ$ at the highest point gives $T = \frac{mu^2}{l} - 5mg$ .

p. 133

Condition for a complete vertical circle

Motion in a vertical circle

u \ge \sqrt{5gl}

A particle on a string must be launched from the lowest point at least this fast to get all the way round. It comes from requiring $T \ge 0$ at the highest point.

$u$: speed at the lowest point $\mathrm{m\,s^{-1}}$
$l$: length of the string or radius of the circle $\mathrm{m}$

For a string, which can pull but not push. A rod or a groove can push, so there the condition is only $v \ge 0$ at the top.
The corresponding minimum speed at the highest point is $v = \sqrt{gl}$ .

p. 134

Hooke's Law

F = k(l - l_0)

An elastic string or spring stretched beyond its natural length pulls back with a force proportional to the extension.

$F$: restoring force, or tension $\mathrm{N}$
$k$: elastic constant of that string $\mathrm{N\,m^{-1}}$
$l$: actual stretched length $\mathrm{m}$
$l_0$: natural length $\mathrm{m}$

Zero force when $l \le l_0$ - a slack elastic string exerts nothing, and it can never push.
Sometimes written $F = -ks$ with $s$ the extension, the minus sign recording that the force opposes the stretch.
Page 57 of Formulae and Tables.

p. 137

Elastic string providing the centripetal force

Hooke's Law

k(r - l_0) = m\omega^2 r

A particle whirled in a horizontal circle on an elastic string. The stretched length is the radius, so $r$ appears on both sides and the equation must be solved for it.

$r$: radius of the circle, equal to the stretched length $\mathrm{m}$
$l_0$: natural length of the string $\mathrm{m}$
$\omega$: angular speed $\mathrm{rad\,s^{-1}}$

Requires $r > l_0$ , otherwise the string is slack and there is no tension to turn the particle.
Rearranges to $r = \frac{kl_0}{k - m\omega^2}$ , which needs $k > m\omega^2$ for a solution to exist.

p. 138

Difference Equations

16 formulas · pages 143-164

First difference

First and Second differences

\Delta T_n = T_{n+1} - T_n

Each term subtracted from the one after it. Repeating the operation on the result gives the second differences.

$T_n$: the $n$ -th term of the sequence
$\Delta T_n$: the $n$ -th first difference

Subtract in this order, or every difference comes out with the wrong sign.

p. 144

Which differences are constant

First and Second differences

\text{1st constant} \Rightarrow \text{linear}, \quad \text{2nd constant} \Rightarrow \text{quadratic}, \quad \text{3rd constant} \Rightarrow \text{cubic}

The level at which the differences settle to a constant gives the degree of the polynomial rule generating the sequence.

$n$: position in the sequence

Compute at least three differences at a level before declaring them constant.
If no level is ever constant, the rule is not polynomial.

p. 145

Leading coefficient from the second difference

First and Second differences

\text{second difference} = 2a

For a quadratic sequence $T_n = an^2 + bn + c$ , the constant second difference is exactly twice the leading coefficient.

$a$: coefficient of $n^2$

Quadratic sequences only - constant second differences.

p. 146

Arithmetic and geometric recurrences

Recurrence Relations

u_{n+1} = u_n + d \quad \text{(arithmetic)}, \qquad u_{n+1} = ru_n \quad \text{(geometric)}

The two simplest recurrence relations: one adds a fixed amount each step, the other multiplies by a fixed factor.

$d$: common difference
$r$: common ratio

Each needs one initial value to determine the sequence.

p. 148

The Fibonacci recurrence

Recurrence Relations

u_n = u_{n-1} + u_{n-2}

Each term is the sum of the two before it. Second-order, so two initial values are required.

$u_n$: the $n$ -th term

With $u_1 = u_2 = 1$ this gives $1, 1, 2, 3, 5, 8, 13, 21, \ldots$
Neither arithmetic nor geometric.

p. 148

General first-order difference equation

Difference Equations

T_{n+1} = aT_n + b

The standard first-order form. It is homogeneous when $b = 0$ and inhomogeneous otherwise.

$a$: multiplier applied to the previous term
$b$: constant added each step

First-order: only the immediately preceding term appears.

p. 150

General second-order homogeneous equation

Difference Equations

pu_{n+2} + qu_{n+1} + ru_n = 0

The standard second-order homogeneous form. The order is the gap between the highest and lowest subscripts.

$p, q, r$: constant coefficients

Homogeneous: nothing but terms of the sequence, and zero on the right.
$u_{n+2} - 9u_n = 0$ is still second-order despite the missing middle term.

p. 155

Solution of a first-order homogeneous equation

Solving First-Order Difference Equations

T_n = a^{n-1}T_1

Solution of $T_{n+1} = aT_n$ when the sequence is counted from $T_1$ . Counted from $u_0$ instead it reads $u_n = a^n u_0$ .

$a$: the multiplier
$T_1$: the first term

Watch the index: an off-by-one here shifts every term.
Verify by substituting $n = 1$ and checking you recover $T_1$ .

p. 150

Sum of a geometric series

Solving First-Order Difference Equations

S_n = \frac{a(1 - r^n)}{1 - r}

Needed to collapse the string of extra terms that the constant $b$ generates when iterating $T_{n+1} = aT_n + b$ .

$a$: first term of the series
$r$: common ratio
$n$: number of terms summed

$r \ne 1$ . As $n \to \infty$ with $|r| < 1$ this tends to $\frac{a}{1-r}$ .

p. 150

Shape of the first-order solution

Solving First-Order Difference Equations

u_n = Ca^n + L

Every solution of $u_{n+1} = au_n + b$ has this form: a geometric part that grows or decays, plus a constant.

$C$: constant fixed by the initial value
$L$: the limiting value, $\frac{b}{1-a}$

If $|a| < 1$ then $a^n \to 0$ and $u_n \to L$ .
If $|a| > 1$ the sequence has no limit.
$L$ can be found directly by setting $u_{n+1} = u_n = L$ in the original equation.

p. 151

Loan repayment difference equation

Interest Repayments

D_n = (1 + i)D_{n-1} - A

The outstanding debt grows by the interest rate first, and only then is the repayment subtracted.

$D_n$: debt owing after $n$ periods
$i$: interest rate per period, as a decimal
$A$: repayment made each period

$D_0$ is the amount borrowed.
The period of $i$ , of $n$ and of $A$ must all match - monthly with monthly, annual with annual.
The loan is cleared when $D_N = 0$ , and that equation determines $A$ .

p. 153

Monthly rate to annual rate

Interest Repayments

\text{APR} = (1 + \text{MPR})^{12} - 1

Interest compounds, so the annual rate is found by applying the monthly factor twelve times - not by multiplying the monthly rate by twelve.

MPR: monthly percentage rate, as a decimal
APR: annual percentage rate, as a decimal

An MPR of $0.4\%$ gives an APR of $(1.004)^{12} - 1 = 4.907\%$ , not $4.8\%$ .

p. 154

Characteristic quadratic equation

Second-order Homogeneous Difference Equations

px^2 + qx + r = 0

Formed from $pu_{n+2} + qu_{n+1} + ru_n = 0$ using exactly the same three coefficients. Its roots determine the solution.

$p, q, r$: the coefficients of the difference equation, in order
$x$: the unknown whose roots build the solution

Rearrange the difference equation so every term is on one side first, or the signs will be wrong.
A third-order equation gives a characteristic cubic in the same way.

p. 155

Difference Equation Theorem 1 - distinct roots

Second-order Homogeneous Difference Equations

u_n = l\alpha^n + m\beta^n

When the characteristic quadratic has two different roots $\alpha$ and $\beta$ , the solution is this combination of their powers.

$\alpha, \beta$: the two distinct roots
$l, m$: real constants, found from the two initial values

A root of $1$ contributes a constant; a negative root contributes an alternating term.
Two initial values give two simultaneous equations in $l$ and $m$ .

p. 155

Difference Equation Theorem 2 - double root

Second-order Homogeneous Difference Equations

u_n = l\alpha^n + mn\alpha^n

When the characteristic quadratic has a repeated root $\alpha$ , the second part of the solution carries an extra factor of $n$ .

$\alpha$: the repeated root
$l, m$: real constants, found from the two initial values

Without the factor $n$ the two terms would merge into one constant and could not meet two initial conditions.
Applies whenever the discriminant of the characteristic quadratic is zero.

p. 155

Solution of an inhomogeneous equation

Inhomogeneous Difference Equations

u_n = u_n^{\text{(particular)}} + u_n^{\text{(complementary)}}

The total solution is the sum of any one solution that produces the right-hand side and the general solution of the homogeneous version.

particular: a trial solution matching the right-hand side, e.g. $k\lambda^n$ for a right-hand side of $\lambda^n$
complementary: the general solution with the right-hand side set to zero

It works because the particular part yields the right-hand side and the complementary part yields zero.
Apply the initial values only to the combined solution, never to either piece on its own.
If the trial form already appears in the complementary solution, multiply it by $n$ first.

p. 158

Differentiation and Integration

18 formulas · pages 165-182

Log of a product and a quotient

Logarithms

\ln a + \ln b = \ln ab, \qquad \ln a - \ln b = \ln\frac{a}{b}

Adding logs multiplies the arguments; subtracting them divides.

$a, b$: positive numbers or expressions

Both arguments must be positive.
$\ln(a+b)$ is NOT $\ln a + \ln b$ .
Page 2 of Formulae and Tables.

p. 166

Log of a power

Logarithms

n\ln a = \ln a^n

A coefficient in front of a log becomes a power inside it. Use this first when combining several logs.

$n$: any real number, including fractions
$a$: a positive quantity

$\tfrac{1}{2}\ln 4 = \ln 4^{1/2} = \ln 2$ , so fractional coefficients become roots.

p. 166

Definition of the natural logarithm

Logarithms

\ln x = y \iff x = e^y

The step that turns a logarithmic equation into an answer. $\ln$ and $e^x$ are inverse functions.

$x$: a positive quantity
$y$: any real number

Equivalently $e^{\ln x} = x$ and $\ln(e^x) = x$ .
A bare constant in a log equation must first be rewritten as a log: $1 = \ln e$ .

p. 166

Power rule for integration

Indefinite integrals

\int x^n\,dx = \frac{x^{n+1}}{n+1} + c

Raise the index by one and divide by the new index. The workhorse of integration.

$n$: any number except $-1$
$c$: constant of integration

Fails at $n = -1$ , which gives a division by zero - that case is a logarithm.
Divide by the NEW index, not the original one.

p. 166

The reciprocal integral

Indefinite integrals

\int\frac{1}{x}\,dx = \ln x + c

The exception to the power rule, and the reason logarithms are needed in this chapter at all.

$x$: a positive quantity

Strictly $\ln\lvert x\rvert$ , though this course works with positive arguments.

p. 166

Linear expression inside a function

Indefinite integrals

\int\frac{1}{ax+b}\,dx = \frac{1}{a}\ln(ax+b) + c, \qquad \int e^{ax}\,dx = \frac{1}{a}e^{ax} + c

A linear expression inside brings a factor $\frac{1}{a}$ outside. Forgetting it is the commonest slip in this section.

$a, b$: constants, with $a \ne 0$

Only for a LINEAR expression inside - it does not work for $e^{x^2}$ .
Check by differentiating: the chain rule reproduces the $a$ that the $\frac{1}{a}$ cancels.

p. 166

The inverse tangent standard form

Indefinite integrals

\int\frac{1}{x^2 + a^2}\,dx = \frac{1}{a}\tan^{-1}\frac{x}{a} + c

A standard form worth recognising on sight; many substitution problems reduce to it.

$a$: a positive constant

The companion form is $\int\frac{1}{\sqrt{a^2 - x^2}}\,dx = \sin^{-1}\frac{x}{a} + c$ .
Page 26 of Formulae and Tables.

p. 167

Definite integral

Definite integrals

\int_a^b f(x)\,dx = F(b) - F(a)

Integrate, substitute the upper limit, subtract the value at the lower limit. The result is a number.

$a$: lower limit
$b$: upper limit
$F$: any antiderivative of $f$

No constant of integration - the two copies of $c$ cancel.
Reversing the limits changes the sign of the answer.

p. 168

Integration by parts

\int u\,dv = uv - \int v\,du

The product rule for differentiation, run backwards. It handles products that no direct rule can integrate.

$u$: the factor chosen by the InLATE rule
$dv$: the remainder of the integrand, including $dx$
$v$: the integral of $dv$

If the resulting integral is harder than the original, the choice of $u$ was the wrong way round.
Some integrals, such as $\int x^2e^x\,dx$ , need the formula applied twice.
Page 26 of Formulae and Tables.

p. 169

The InLATE rule

Integration by parts

\text{Inverse} \to \text{Logarithms} \to \text{Algebra} \to \text{Trigonometric} \to \text{Exponential}

Whichever factor appears earlier in this list becomes $u$ ; the other becomes $dv$ .

In: inverse functions such as $\tan^{-1}x$ , $\sin^{-1}x$
L: logarithms such as $\ln x$
A: algebra such as $x$ , $x^2$
T: trigonometric functions such as $\sin x$ , $\cos x$
E: exponentials such as $e^x$ , $e^{2x}$

In $\int xe^x\,dx$ , Algebra beats Exponential, so $u = x$ .
In $\int x\ln x\,dx$ , Logarithms beat Algebra, so $u = \ln x$ .

p. 169

The chain rule

The chain rule for composite functions

f(x) = u(v(x)) \Rightarrow f'(x) = \frac{du}{dv}\cdot\frac{dv}{dx}

To differentiate a function of a function, differentiate the outer and multiply by the derivative of the inner.

$u$: the outer function
$v$: the inner function

There is no corresponding chain rule for integration - substitution takes its place.
Page 25 of Formulae and Tables.

p. 171

Integration by substitution

The chain rule for composite functions

u = v(x), \qquad du = \frac{du}{dx}\,dx

Let $u$ be the inner function and swap $dx$ for $du$ , turning the integral into a standard form in $u$ alone.

$u$: the chosen inner function

Works when $\frac{du}{dx}$ already appears in the integrand, up to a constant factor.
Every $x$ must be eliminated before integrating.
For a definite integral, convert the limits to $u$ values or convert back to $x$ first.

p. 171

Velocity and acceleration as derivatives

Rates of change

v = \frac{ds}{dt}, \qquad a = \frac{dv}{dt} = \frac{d^2s}{dt^2}

Velocity is the rate of change of displacement; acceleration is the rate of change of velocity.

$s$: displacement $\mathrm{m}$
$v$: velocity $\mathrm{m\,s^{-1}}$
$a$: acceleration $\mathrm{m\,s^{-2}}$

$\frac{d^2s}{dt^2}$ is not the same as $\left(\frac{ds}{dt}\right)^2$ .

p. 173

Turning point condition

Rates of change

\frac{dh}{dt} = 0 \text{ at a maximum or minimum}

A quantity stops changing at a turning point. Solve this for $t$ , then substitute back to get the value.

$h$: the quantity being maximised or minimised
$t$: the variable it depends on

Solving gives the TIME; substituting back gives the VALUE. Both steps are needed.
A zero derivative locates a minimum just as readily as a maximum.

p. 173

Integration in kinematics

Using Integration

v = \int a\,dt, \qquad s = \int v\,dt

Integration runs the chain $a \to v \to s$ , the reverse of differentiation. It works even when the acceleration varies.

$a$: acceleration, possibly a function of $t$ $\mathrm{m\,s^{-2}}$
$v$: velocity $\mathrm{m\,s^{-1}}$
$s$: displacement $\mathrm{m}$

Each integration introduces a constant, fixed by an initial condition.
Unlike the equations of motion from Chapter 2, this does not require a constant acceleration.
A definite integral $\int_{t_1}^{t_2} v\,dt$ gives a distance without needing the constants.

p. 174

Differentiating a vector

Differentiating Vectors

\vec{s} = x(t)\vec{i} + y(t)\vec{j} \Rightarrow \vec{v} = \frac{dx}{dt}\vec{i} + \frac{dy}{dt}\vec{j}

Differentiate the $\vec{i}$ and $\vec{j}$ components separately; the results reassemble into the velocity vector.

$\vec{s}$: displacement vector $\mathrm{m}$
$\vec{v}$: velocity vector $\mathrm{m\,s^{-1}}$

Differentiating again gives the acceleration vector.
Speed is the magnitude $\sqrt{v_x^2 + v_y^2}$ - a number, not a vector.

p. 177

Work done by a variable force

W = \int_A^B F(x)\,dx

The definition of work done. Slice the journey, sum $F\,\Delta x$ over the pieces, and let them shrink to zero.

$W$: work done $\mathrm{J}$
$F(x)$: force as a function of position $\mathrm{N}$
$x$: displacement from the origin $\mathrm{m}$

The limits are positions, not times, so integrate with respect to $x$ .
$W = Fs$ from Chapter 5 is the special case of a constant force.

p. 179

Work done stretching an elastic string

Work done by a variable force

W = \frac{k}{2}\left(x_2^2 - x_1^2\right)

Obtained by integrating Hooke's law $F = kx$ from extension $x_1$ to extension $x_2$ . It is also the potential energy stored.

$k$: elastic constant of the string $\mathrm{N\,m^{-1}}$
$x_1$: initial extension beyond the natural length $\mathrm{m}$
$x_2$: final extension beyond the natural length $\mathrm{m}$

$x_1$ and $x_2$ are EXTENSIONS, not total lengths.
It is the difference of the squares, not the square of the difference.
A second equal stretch costs more work than the first, because the force grows with the extension.

p. 179

Differential equations

14 formulas · pages 183-196

Separation of variables

Differential equations

\frac{dy}{dx} = f(x)g(y) \Rightarrow \int\frac{dy}{g(y)} = \int f(x)\,dx

Move every $y$ term to the $dy$ side and every $x$ term to the $dx$ side, then integrate both sides. This is the whole method for the equations on this course.

$f(x)$: the part of the equation depending on $x$
$g(y)$: the part depending on $y$

Only works when the right-hand side factorises into an $x$ part times a $y$ part.
$\frac{dy}{dx}$ may be treated as a fraction for this purpose.

p. 184

Order of a differential equation

Differential equations

\frac{ds}{dt} = 2s \text{ is first-order}, \qquad \frac{d^2s}{dt^2} = 2\frac{ds}{dt} \text{ is second-order}

The order is the order of the highest derivative in the equation, not the highest power.

$s$: the dependent variable
$t$: the independent variable

$\left(\frac{ds}{dt}\right)^2$ is still a first derivative, so an equation containing it is first-order.

p. 184

The logarithmic integral

Differential equations

\int\frac{dy}{y} = \ln y

The integral that appears in nearly every separated equation, and the reason the solutions come out exponential.

$y$: a positive quantity

Undo it at the end with $\ln y = f(x) + c \Rightarrow y = e^{f(x)+c}$ .

p. 184

General versus particular solution

Differential equations

\ln y = f(x) + c \Rightarrow y = e^{f(x)+c} = Ae^{f(x)}

A general solution keeps the arbitrary constant; a particular solution uses the given values to fix it. Note $e^c$ can be renamed as a single constant $A$ .

$c$: the constant of integration
$A$: $e^c$ , a single positive constant

Only one constant is needed, and by convention it goes on the right-hand side.
$e^{f(x)+c}$ is NOT $e^{f(x)} + e^c$ .

p. 184

The two forms of acceleration

Solving real-life problems

a = \frac{dv}{dt} = v\frac{dv}{ds}

The second form comes from the chain rule, $\frac{dv}{dt} = \frac{dv}{ds}\cdot\frac{ds}{dt}$ . Which one you choose decides what the answer will be about.

$v$: velocity $\mathrm{m\,s^{-1}}$
$s$: displacement $\mathrm{m}$
$t$: time $\mathrm{s}$

$\frac{dv}{dt}$ leads to a relationship between $v$ and $t$ - use it when the question asks for a time.
$v\frac{dv}{ds}$ leads to a relationship between $v$ and $s$ - use it when the question asks for a distance.

p. 187

Velocity of a body falling against resistance

Solving real-life problems

v = \frac{g}{k}\left(1 - e^{-kt}\right)

A particle of mass $m$ falling from rest with air resistance $mkv$ . Obtained from $\frac{dv}{dt} = g - kv$ by separating the variables.

$v$: velocity at time $t$ $\mathrm{m\,s^{-1}}$
$k$: resistance constant $\mathrm{s^{-1}}$
$g$: acceleration due to gravity $\mathrm{m\,s^{-2}}$

Starts from rest, so $v = 0$ when $t = 0$ .
Downwards is taken as positive, so the resistance $mkv$ is negative.
The mass cancels out of the equation entirely.

p. 187

Terminal velocity

Solving real-life problems

v_{\text{terminal}} = \frac{g}{k}

As $t$ grows, $e^{-kt} \to 0$ , so the velocity approaches this limit and the body falls at a steady speed.

$g$: acceleration due to gravity $\mathrm{m\,s^{-2}}$
$k$: resistance constant $\mathrm{s^{-1}}$

Reached only in the limit - the body never quite attains it.
Equivalently, set $\frac{dv}{dt} = 0$ in $\frac{dv}{dt} = g - kv$ .

p. 187

Tractive force at constant power

Problems involving power

F = \frac{P}{v}

An engine working at a constant rate produces a force that falls away as the vehicle speeds up.

$P$: power output $\mathrm{W}$
$F$: tractive force $\mathrm{N}$
$v$: speed $\mathrm{m\,s^{-1}}$

Convert kilowatts to watts before using it.
Rearranged from $P = Fv$ .

p. 191

Equation of motion at constant power

Problems involving power

ma = \frac{P}{v} - R

$F = ma$ with the tractive force written as $\frac{P}{v}$ . Because $v$ appears on both sides, this is a differential equation, not an algebraic one.

$m$: mass of the vehicle $\mathrm{kg}$
$R$: total resistance to motion $\mathrm{N}$
$a$: acceleration, written as $\frac{dv}{dt}$ or $v\frac{dv}{ds}$ $\mathrm{m\,s^{-2}}$

Multiply through by $v$ early to clear the fraction.
On a hill, add $mg\sin\alpha$ to the resistance $R$ .

p. 191

Maximum speed at constant power

Problems involving power

\frac{P}{v_{\max}} = R

At maximum speed the acceleration is zero, so the tractive force exactly balances the resistance. No differential equation is needed for this part.

$v_{\max}$: maximum speed $\mathrm{m\,s^{-1}}$
$R$: total resistance $\mathrm{N}$

Set $a = 0$ in the equation of motion.

p. 191

Exponential growth

Problems involving populations, finance, cooling

\frac{dP}{dt} = kP \Rightarrow P = P_0e^{kt}

Models any quantity whose rate of increase is proportional to the amount already present - populations, investments, bacterial cultures.

$P$: the quantity at time $t$
$P_0$: the quantity at $t = 0$
$k$: growth constant

A second data point is needed to determine $k$ ; take logs of both sides to find it.
A negative $k$ gives exponential decay instead.

p. 192

Newton's Law of Cooling

Problems involving populations, finance, cooling

\frac{dT}{dt} = -k(T - A)

The rate of cooling is proportional to the difference between the body's temperature and its surroundings. The minus sign records that the temperature falls.

$T$: temperature of the body $^\circ\mathrm{C}$
$A$: ambient temperature of the surroundings $^\circ\mathrm{C}$
$k$: positive cooling constant $\mathrm{min^{-1}}$

It is the excess $T - A$ that decays, not $T$ itself.
Omitting the minus sign models a body heating up.

p. 193

Solution of the cooling equation

Problems involving populations, finance, cooling

T = (T_0 - A)e^{-kt} + A

Separating and integrating Newton's law. As $t$ grows the exponential dies away and the temperature settles at the ambient value $A$ .

$T_0$: initial temperature of the body $^\circ\mathrm{C}$
$A$: ambient temperature $^\circ\mathrm{C}$

The body never cools below the ambient temperature - a useful sanity check.
Find $k$ from a second temperature reading by taking logs.

p. 193

Difference or differential?

Difference: Differential and Difference Equations

u_{n+1} = au_n + b \quad \text{(discrete)}, \qquad \frac{dy}{dx} = f(x)g(y) \quad \text{(continuous)}

Difference equations model change that happens in steps; differential equations model change that happens at every instant.

$n$: step number, for the discrete case
$x$: a continuously varying quantity

Interest credited monthly is discrete - use a difference equation.
A body cooling continuously is smooth - use a differential equation.
Both are classified by order and both need initial conditions for a particular solution.

p. 193

Networks and Graphs

17 formulas · pages 197-235

A graph as a pair of sets

Graphs

G = (V, E), \quad |V| = n, \quad |E| = m

A graph is a set of nodes together with a set of edges joining pairs of them. Counting the two sets is the opening part of most questions.

$V$: the set of nodes (vertices)
$E$: the set of edges (arcs)
$n$: the number of nodes
$m$: the number of edges

Only a drawn dot is a node. Two edges crossing on the page do not create one.

p. 198

Subgraph test

Graphs

H \subseteq G \iff V(H) \subseteq V(G) \ \text{ and } \ E(H) \subseteq E(G)

A subgraph may delete nodes and edges from the parent graph, but may never add an edge the parent does not have.

$H$: the candidate subgraph
$G$: the parent graph

Check edge by edge; one edge not in $G$ is enough to fail the test.
A subgraph of a digraph is called a subdigraph.

p. 199

Length of a walk

Walks, Paths and Cycles

\text{length} = \text{number of edges} = (\text{number of letters}) - 1

The length of a walk counts edges traversed, so a walk written with $k$ letters has length $k - 1$ .

$\text{length}$: the number of edges in the walk

A repeated edge is counted every time it is used.
In a digraph the walk must follow the direction of every arrow.

p. 205

Path and cycle conditions

Walks, Paths and Cycles

\text{path} \iff \text{all nodes different}, \qquad \text{cycle} \iff \text{closed and all intermediate nodes different}

A path is a walk that never revisits a node; a cycle is a closed walk whose intermediate nodes are all distinct.

$\text{closed}$: the walk starts and finishes at the same node

It is the nodes, not the edges, that must all differ.
$ABEDCFA$ is a cycle; $ABCDEFBA$ is not, because $B$ repeats.

p. 206

Connectedness and components

Connected and Disconnected Graphs

G \text{ is connected} \iff c(G) = 1

A graph is connected exactly when it falls into a single component, that is, when a path joins every pair of nodes.

$c(G)$: the number of components of $G$

An isolated node of degree $0$ is a component in its own right.
To prove a graph disconnected, name one pair of nodes with no path between them.

p. 206

Degree of a node

Incident and Adjacent Nodes

\deg(v) = \text{number of edge-ends at } v, \quad \text{a loop contributes } 2

The degree, valency or order of a node counts the edges incident with it, with a loop counted twice because both of its ends arrive there.

$\deg(v)$: the degree of node $v$
$v$: a node of the graph

Degree, valency and order are three names for the same number.
A node of degree $0$ is isolated.

p. 209

Hand-shaking lemma

Incident and Adjacent Nodes

\sum_{v} \deg(v) = 2|E|

The degree-total of a graph is twice the number of edges, because every edge has two ends - just as every handshake involves two people.

$\deg(v)$: the degree of node $v$
$|E|$: the number of edges

Holds for every graph, loops and multiple edges included.
The degree-total is therefore always even; an odd total means you have miscounted.

p. 210

Number of odd-degree nodes

Incident and Adjacent Nodes

\#\{\, v : \deg(v) \text{ is odd} \,\} \text{ is even}

Because the degree-total is even, the nodes of odd degree must pair off - so there is always an even number of them.

$\deg(v)$: the degree of node $v$

Use it to reject an impossible degree list: $2, 1, 1, 1$ totals $5$ , so no such graph exists.

p. 210

Adjacency matrix entry

Adjacency Matrices

M_{ij} = \text{number of edges from } i \text{ to } j \qquad (\text{read Left to Top})

Each entry counts the edges joining the row node to the column node - Lawn Tennis, so the row is where you start and the column where you finish.

$M$: the adjacency matrix of the graph
$i$: the row node (where the journey starts)
$j$: the column node (where it finishes)

Multiple edges give an entry of $2$ , $3$ , and so on, not $1$ .
This book enters a loop as $1$ .
An undirected graph gives a symmetric matrix.

p. 212

Multiplying two matrices

Adjacency Matrices

\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} w & x \\ y & z \end{pmatrix} = \begin{pmatrix} aw + by & ax + bz \\ cw + dy & cx + dz \end{pmatrix}

Row of the first matrix into column of the second. The same rule extends to $3 \times 3$ matrices.

$a, b, c, d$: entries of the first matrix
$w, x, y, z$: entries of the second matrix

Matrix multiplication is not commutative: $AB \neq BA$ in general.
Never multiply entry by corresponding entry.

p. 213

Counting walks with matrix powers

Adjacency Matrices

(M^{n})_{ij} = \text{the number of walks of length } n \text{ from } i \text{ to } j

Raising the adjacency matrix to the power $n$ counts every walk of exactly $n$ edges between each pair of nodes.

$M$: the adjacency matrix
$n$: the length of the walks being counted
$i, j$: the start and finish nodes

It counts walks, not paths - nodes and edges may repeat.
A closed walk it counts is a cycle only if the intermediate nodes are all different.

p. 214

Digraph adjacency matrix

Adjacency Matrices for Digraphs

M \neq M^{T} \implies \text{the graph is directed}

In a digraph the entry in row $i$ , column $j$ counts only the arcs pointing from $i$ to $j$ , so the matrix is generally not symmetric - and that asymmetry identifies it as a digraph.

$M$: the adjacency matrix of the digraph
$M^{T}$: the transpose of $M$

A row of zeros is a node you can never leave; a column of zeros is one you can never reach.
$M^{n}$ still counts walks of length $n$ , following every arrow the right way.

p. 217

Edge count of a tree

Trees

|E| = n - 1

A tree is a connected graph with no cycles, and holding $n$ nodes together without a cycle takes exactly $n - 1$ edges.

$n$: the number of nodes in the tree
$|E|$: the number of edges

Applies to every tree, and so to every spanning tree of an $n$ -node graph.
Use it as the arithmetic check on any spanning tree you draw.

p. 219

Total weight of a spanning tree

Trees

W = \sum_{e \in T} w(e)

The total length of a spanning tree is the sum of the weights of the edges it uses.

$T$: the spanning tree
$w(e)$: the weight of edge $e$ $\mathrm{m}$ , $\mathrm{km}$ or euro
$W$: the total weight of the tree $\mathrm{m}$ , $\mathrm{km}$ or euro

A spanning tree must contain every node of the original graph.
Only edges already present in the graph may be used.

p. 219

Kruskal's algorithm

Minimum Spanning Tree: Kruskal's Algorithm

\text{take the next shortest edge unless it closes a cycle; stop at } n - 1 \text{ edges}

Work through the edge list in ascending order of weight, accepting an edge whenever it does not create a cycle and rejecting it when it does.

$n$: the number of nodes in the network

Order the whole edge list by weight before starting.
The partly built tree may be disconnected until the last edge is added.
Ties may be broken either way; the total weight is unaffected.

p. 223

Prim's algorithm

Prim's Algorithm

e_{k+1} = \min\{\, w(u, v) \ : \ u \in T, \ v \notin T \,\}

At every step add the shortest edge running from a node already in the tree to a node not yet in it, so a cycle can never form.

$T$: the tree built so far
$u$: a node already in the tree
$v$: a node not yet in the tree
$w(u,v)$: the weight of the edge joining them $\mathrm{m}$ , $\mathrm{km}$ or euro

Stop at $n - 1$ edges, as with Kruskal's.
Any starting vertex gives the same total weight.
In distance-matrix form, cross out the row of each joined node, number its column, then search all numbered columns.

p. 226

Cost of building the network

Prim's Algorithm

\text{cost} = W \times (\text{rate per unit length})

How a minimum spanning tree becomes money in a road, cable or cycle-lane question.

$W$: the total weight of the minimum spanning tree $\mathrm{m}$ or $\mathrm{km}$
$\text{rate}$: cost of building one unit of lengtheuro per $\mathrm{m}$ or per $\mathrm{km}$

Check the units match before multiplying: a rate per km against a total in metres is the usual slip.

p. 226

Optimal Paths

20 formulas · pages 236-294

Dijkstra's update rule

Dijkstra's Algorithm

\text{working}(v) = \min\{\, \text{working}(v), \ \text{final}(u) + w(u,v) \,\}

When a node $u$ is completed, every neighbour $v$ gets a new working value only if the route through $u$ beats the one it already has.

$u$: the node just made permanent
$v$: a node one step from $u$
$w(u,v)$: the weight of the arc joining them $\mathrm{km}$ , minutes or euro

A working value is never replaced by a larger number.
Weights must be positive.

p. 237

Final value of a node

Dijkstra's Algorithm

\text{final}(v) = \min\{\, \text{working values at } v \,\}

A node's final value is the smallest working value it ever receives, which is the shortest distance to it from the start.

$v$: the node being completed
$\text{final}(v)$: shortest distance from the start node to $v$ $\mathrm{km}$ , minutes or euro

The start node is completed first with a final value of $0$ .
Complete the uncompleted node with the smallest working value anywhere in the network.

p. 237

Correct subtraction method

Dijkstra's Algorithm

\text{final}(v) - w(u,v) = \text{final}(u) \implies u \text{ lies on the shortest path to } v

How the optimal route is read back from the destination once every node has been completed.

$u$: the previous node on the route
$v$: the node being traced back from
$w(u,v)$: the weight of the arc joining them $\mathrm{km}$ , minutes or euro

Work backwards from the destination, then write the route out forwards.
Check by adding the arcs of the route: the total must equal the destination's final value.

p. 239

Duration of a dummy activity

Critical Path Analysis

t_{\text{dummy}} = 0

A dummy is a dotted, directed arc that carries a dependency but no work, so it contributes nothing to any time.

$t_{\text{dummy}}$: the duration of a dummy activitydays or hours

Use one when two activities would otherwise join the same pair of nodes.
Use one when a later activity depends on some, but not all, of the activities meeting at a node.
A dummy is still traversed on the forward and backward passes; it just adds $0$ .

p. 252

Forward pass - early times

Calculating the Time to Complete a Project

\text{early}(j) = \max_{i \to j}\{\, \text{early}(i) + t_{ij} \,\}

Working from the source, an event's early time is the Enormousest of (previous early time plus activity duration) over every path arriving at it.

$\text{early}(j)$: earliest time all activities into event $j$ can be finisheddays or hours
$t_{ij}$: duration of the activity from event $i$ to event $j$ days or hours

The source node has early time $0$ .
Every path into the node must be considered, dummies included.

p. 257

Backward pass - late times

Calculating the Time to Complete a Project

\text{late}(i) = \min_{i \to j}\{\, \text{late}(j) - t_{ij} \,\}

Working back from the sink, an event's late time is the Least of (later late time minus activity duration) over every path leaving it.

$\text{late}(i)$: latest time event $i$ may happen without delaying the projectdays or hours
$t_{ij}$: duration of the activity from event $i$ to event $j$ days or hours

Complete the whole forward pass before starting the backward pass.
Early is the Enormousest, Late is the Least - the mnemonic exists because the two are easily swapped.

p. 257

Source and sink nodes

Calculating the Time to Complete a Project

\text{source}: \ \tfrac{0}{0}, \qquad \text{sink}: \ \tfrac{t}{t}

The source node always has early and late times of zero, and the sink always carries the project's minimum completion time twice.

$t$: the minimum time to complete the whole projectdays or hours

Fill the sink's late time in from its early time before starting the backward pass.

p. 256

Total float of an activity

Critical Activities and Critical Paths

\text{total float} = z - w - t

Latest finish minus earliest start minus duration: how long an activity's start may be delayed without delaying the project.

$w$: early time at the tail node of the activitydays or hours
$z$: late time at the head node of the activitydays or hours
$t$: duration of the activitydays or hours

Late at the head, early at the tail - taking them the other way round is the standard error.
A negative float means one of the early or late times is wrong.

p. 259

Critical activity and critical path

Critical Activities and Critical Paths

\text{critical activity} \iff \text{total float} = 0

An activity with no slack is critical; a path from source to sink joining only critical activities is a critical path.

$\text{total float}$: slack available to the activitydays or hours

The durations along a critical path total the project's minimum time.
There may be more than one critical path.

p. 259

Placing an activity on a Gantt chart

Gantt Charts

\text{solid bar}: [\, w, \ w + n \,], \qquad \text{dotted rectangle}: [\, w + n, \ z \,]

An activity is drawn solid from its early start for its own duration, with its total float shown as a dotted rectangle attached to the right.

$w$: early start time of the activitydays or hours
$n$: duration of the activitydays or hours
$z$: latest time the activity may finishdays or hours

The top row of the chart is reserved for a critical path, whose bars have no dotted extension.
The length of the dotted rectangle is exactly the total float, $z - w - n$ .

p. 263

Lower bound for the number of workers

Scheduling

\text{lower bound} = \left\lceil \dfrac{\text{total time of all activities}}{\text{critical time}} \right\rceil

The fewest workers that could conceivably finish the project in its minimum time - a floor, never a guarantee.

$\text{total time}$: sum of the durations of every activityhours or days
$\text{critical time}$: length of the critical pathhours or days

Always round up: $3.667$ workers means at least $4$ people.
Check the bound against the Gantt chart - the book's own example needs $5$ where the bound says $4$ .

p. 266

Bellman's Principle of Optimality

P = S \to \cdots \to X \to \cdots \to T \text{ optimal} \implies S \to \cdots \to X \text{ optimal}

Any part of an optimal path is itself optimal - the single fact that makes dynamic programming work.

$S$: the source vertex
$T$: the sink vertex
$X$: any state on the optimal path

'Optimal' means least for costs and times, greatest for profits.
Bellman's algorithm is not greedy: nothing is committed until the whole table is built.

p. 272

Optimal value of a state

Using Bellman's Principle of Optimality to solve Multi-Stage problems

V(X) = \operatorname{opt}_{X \to Y}\{\, w(X,Y) + V(Y) \,\}

The optimal value of a state is built from the optimal values one step nearer the sink, which is why the table is filled in backwards.

$V(X)$: optimal value from state $X$ to the sinkeuro, minutes or $\mathrm{km}$
$w(X,Y)$: weight of the action taking $X$ to $Y$ euro, minutes or $\mathrm{km}$
$\operatorname{opt}$: min for costs and times, max for profits

Stages are numbered backwards from the sink, so stage $1$ is the last stage.
Only the starred optimal value of a state is carried into the next stage.

p. 273

The Value column

Dynamic Programming for a multi-stage problem

\text{Value} = w(\text{action}) + \text{OVD}

Every row of a dynamic programming table adds the weight of the action to the optimal value already found for its destination.

$w(\text{action})$: weight of the arc taken at this stepeuro, minutes or $\mathrm{km}$
$\text{OVD}$: optimal value of the destination, from the previous stageeuro, minutes or $\mathrm{km}$

Columns are Stage, State, Action, Destination, Value.
Star the optimal value for each state, then follow the stars to read the route.

p. 273

Value line for a routing problem

Type 1: Routing Problems

\text{Value} = \text{profit} - \text{travel expense} + \text{OVD}

In a routing problem the traveller earns at each place and pays to move on, so the value nets the two against the optimal value of where they land.

$\text{profit}$: expected takings at this stateeuro
$\text{travel expense}$: cost of the journey choseneuro
$\text{OVD}$: optimal value of the destinationeuro

These are profits, so the starred value is the greatest.
Stage counts periods still to go; the last stage pays the fare home and the first stage the fare out.
The OVD of the final stage is $0$ .

p. 281

Value line for a stock control problem

Type 2: Stock Control Problems

\text{Value} = \text{storage} + \text{overheads} + \text{hired labour} + \text{OVD}

The running cost of a period: holding stock, the fixed cost of producing at all, any extra labour, plus the optimal cost of everything that follows.

$\text{storage}$: cost of holding the items in stock this periodeuro
$\text{overheads}$: fixed cost incurred in any period where production happenseuro
$\text{hired labour}$: extra cost when output exceeds the normal limiteuro
$\text{OVD}$: optimal value of the destinationeuro

These are costs, so the starred value is the least.
Storage is charged on the number held (the state), not the number made.
There are no overheads in a period where nothing is produced.

p. 282

Stock carried into the next period

Type 2: Stock Control Problems

\text{destination} = \text{state} + \text{action} - \text{demand}

What is left in stock after the period's orders have been met: what you had, plus what you made, less what was demanded.

$\text{state}$: items in stock at the start of the period
$\text{action}$: items made during the period
$\text{demand}$: orders that must be met in that period

Write the period's demand in brackets beside the Stage heading.
Rows breaching a stock maximum, or leaving too little for the next period, are written down and marked impossible.

p. 282

Value line for an allocation problem

Type 3: Allocation of Resources

\text{Value} = \text{profit} + \text{OVD}

Allocating units to one product returns a profit, and the units left over are worth their own optimal value.

$\text{profit}$: return from allocating this many units to this producteuro
$\text{OVD}$: optimal value of the units remainingeuro

These are profits, so the starred value is the greatest.
The order in which the products are taken does not matter; use the order in the question.
Stage = the product, state = units available, action = units allocated, destination = units remaining.

p. 283

Value line for a replacement problem

Type 4: Equipment replacement and maintenance

\text{Value} = \text{replacement cost} + \text{maintenance} - \text{resale value} + \text{OVD}

The cost of buying the item, running it for the chosen number of years, selling it on, and then facing the years that remain.

$\text{replacement cost}$: price of a new itemeuro
$\text{maintenance}$: total upkeep over the years kepteuro
$\text{resale value}$: what the item fetches at that ageeuro
$\text{OVD}$: optimal cost over the years still remainingeuro

These are costs, so the starred value is the least.
State = years left when you buy; destination = years left when you sell.
After the final year of the horizon the OVD is $0$ .

p. 285

Cumulative maintenance cost

Type 4: Equipment replacement and maintenance

\text{maintenance over } k \text{ years} = \sum_{i=1}^{k} m_i

Upkeep accumulates, so keeping the item a further year adds that year's cost to everything already spent.

$m_i$: maintenance cost in year $i$ of ownershipeuro
$k$: number of years the item is keptyears

Work the running totals out once, before starting the table; almost every row uses them.
Using only the final year's figure instead of the total is the standard error.

p. 285

Knowing the formula is not the same as knowing when to use it

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