A Level Maths

y=af(x)

stretch graph vertically by scale factor a

y=f(x)+a

translation in the y-axis of a

y=f(ax)

stretch graph horizontally by scale factor 1/a

y=f(x+a)

translation in the x-axis of -a

y=-f(x)

reflect over x-axis

y=f(-x)

reflection in y axis

y=(ax+b)

move -b horizontally, the horizontal stretch of 1/a

Given a centre (a, b) and radius r, the equation of a circle is

(x-a)^2+(y-b)^2=r^2

Given x^2+3x+y^2-4x+10=0. How do you find the centre and radius of the circle

Complete the square of the x terms, complete the square of the y terms

The angle between the radius and the tangent is

90 degrees

Midpoint between A and B

(average of x, average of ys)

Gradient of line between A and B

Difference between y's over difference between xs

Distance between A and B

Use Pythagaros

Equation of line between A and B

m(x-x_a)=(y-y_b)

kinematic equation without displacement

v=u + at

kinematic equation without final velocity

s= ut+at^2

kinematic equation without acceleration

s=(u+v)/t

kinematic equation without time

v^2=u^2+2as

The area under speed time graph is

the displacement

The gradient of a speed time graph is

acceleration

differentiate x^n

nx^(n-1)

Differentiate x

1

Differentiate 3

0

To find the gradient of a tangent a point P on the curve

Sub the x coordinate of into the gradient function

dy/dx is

the gradient function

Given the gradient (m) of a tangent at point P, find the point P

Set the gradient function to equal m, rearrange to find x. Stick x into the original function to find the y coordinate

The definition of differentiation is

limit as h tends to 0 of (f(x+h)-f(x))/h

To find the tangent at point P of a curve

Use m(x-x_1)=y-y_1. To find m stick x_1 into the gradient function, to find y_1 stick x_1 into the original function

To find the normal at point P of a curve

Use m(x-x_1)=y-y_1. To find m stick x_1 into the gradient function and then use the negative reciprocal, to find y_1 stick x_1 into the original function

To solve sinx=0.5

Draw y=sinx. Draw y=0.5. Use the calculator to find the initial solution. Use the symmetry of the diagram to find the other solutions

Sin^2(x)+cos^2(x)=

1

tanx=

sinx/cosx

To solve sin2x

Draw 2 sinx for double the boundary, then solve as normal

The first thing to do with a projectiles question is

break it into a vertical problem and a horizontal problem

The horizontal acceleration is

0

The vertical acceleration is

-9.8

You can often find the horizontal displacement by

Working with the vertical displacement

The highest point of the particle is when

the vertical vecocity is 0

The speed of the particle is

the combination of the horizontal and vertical velocity

To work out the speed of the particle you use

Pythagaros

To work out the angle of the speed at a certain time you use

Elementary Trig

P(A ∩ B)=

P(A)P(B|A)=P(B)P(A|B)

P(A∪B)=

P(A)+P(B)-P(A∩B)

Independence means in terms of the intersection

P(A∩B)=P(A)P(B)

Independence means in terms of a given event

P(A|B)=P(A)

P(A')=

1-P(A)

P(A'∩B) or similar requires

A venn diagram

Arc length is

rθ

Sector area is

1/2 θr^2

Sin rule is

a/sinA=b/sinB

Cosine rule is

a²=b²+c²-2bc cosA

Segment area is

Sector - triangle

Triangle area os

1/2 abSinC

A fish or a ball or a book etc. is represented by

a dot

A fish or a ball or a book etc. is modelled by

a particle

We use particles in models because

all of the weight is concentrated at one point

Inextensible means

Acceleration is constant throughout the model

Newtons first law is

A particle at rest or moving at constant speed will remain at rest or at constant speed unless a different force is applies

Newtons third law is

When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body

Light means

No mass

Smooth means

No friction

Dave resolves forces by

Magnitude of the force x cosθ (Where θ is the angle between the original force and the direction you are resolving)

Secx

1/cosx

cosecx=

1/sinx

cotx=

cosx/sinx

1+tan^2(x)

sec^2(x)

1+cot^2(x)

cosec^2(x)

To solve (x-2)(x+3)>0

Draw the curve and define the points above the x axis

To solve (x-2)(x+3)<0

Draw the curve and define the points below the x axis

Why would you not draw a curve when solving a quadratic inequality

There is no fathomable reason

2 roots means

b^2-4ac>0

equal roots means

b^2-4ac=0

no real roots means

b^2-4ac<0

When a particle is in limiting equilibrium

Friction=µR

If a particle is moving friction acts

in the direction opposite to movement

If a particle is static friction acts

in the direction opposite to the potential movement

P=f(x)/(x-a)(x-b)

P=f(x)/(x-a)(x-b)^2

P=A/(x-a)+ B/(x-b)+C/(x-b)^2

P=f(x)/g(x), if f(x) is of the same of higher order then

Use algebraic division first

Log(xy)=

Log(x)+log(y)

Log(x/y)=

log(x) - log(y)

log(x^n)=

n log x

log1=

0

log_a(a^x)=

x

a^(log_a(y))

y

The definition of e is

if y=a^x=dy/dx, then a=e

The inverse of e^x is

ln(x)

if f(x) is a polynomial and f(a)=0 then

(x-a) is a factor of f(x)

Given that f(x) is a polynomial, and then when f(x) is divided by (x-a) there is a remainder B, this implies

f(a)=b

To find a stationary point

the first derivative =0

A stationary point is a maximum if

the second derivative <0

A stationary point is a minimum if

the second derivative >0

A function is increasing if

the first derivative ≥0

A function is decreasing if

the first derivative ≤0

A function is strictly increasing if

the first derivative >0

A point of inflection is

the second derivative =0 and a higher derivative exists

Concave is when

the second derivative <0

Convex is when

the second derivative >0

Concave looks like

a cave

Convex looks like

Not a cave