Equations of Motion

The 4 equations of motion deal with an object which is travelling with constant acceleration (which can be 0 and therefore constant speed).

The equations are as follows:

(1) v = u + at
(2) s= t(v+u)/2
(3) v²=u²+2as
(4) s = ut + at²/2

Where
a = acceleration – this must be constant for equations to hold
u = initial velocity, ie at the start of the journey
v = final velocity, ie at the end of the journey
s = displacement which is a vector quantity for the distance of the object from its starting point
t = time taken for journey
Any of these values can be found using the equations if at least 3 of the other values are known.

Equation (1) comes from the definition of acceleration as acceleration is the rate of change of velocity and therefore for constant acceleration
a = (v-u)/t which you can rearrange to make
v = u + at

Equation 2 can be found by either considering a distance time graph or using the average speed. (v+u)/2 gives the average speed during the journey as a is constant and buy multiplying this by t we find the displacement. Or from the graph we can find it as the area under the graph is a trapezium of height t and sides u and v so using the area of a trapezium formulae we find 2.

Equation (4) can be formed by substituting 1 into 2 so
s = t(u+at+u)/2
s = t(2u + at)/2
s= 2ut/2 + at²/2
s = ut + at²/2

Equations 3 can also be found using 1 and 2 by rearranging 2 to get an expression for t we find
t=2s/(u+v)

If we substitute this into v = u+at we get
v=u+2as/(v+u)
And by bringing up the v+u we find
v²+vu = u²+vu + 2as
And because we have vu on both sides we can cancel these to find 3

v²=u²+2as

Transformations of Graphs

 

This looks at how a given graph will change when the the function is changed slightly eg how the graph y=x will change when it becomes y = 2x.

y = af(x)

The graph f(x) will "steeper" as the y value of each point is multiplied by a. It will appear like a "stretched" version of the graph y=f(x) 

 

 

y =f(x) + a

The graph f(x) will move up by the amount a as a is added to each y value. This means that the points of intersection of the graph and the y axis will increase by the amount a. The intersection of the graph and the x-axis will depend upon the function of the graph.

 

y = f(ax)

This will make the graph appear "narrower" beacuse y is taking the value of f(x) a time across. So if the value of f(x) is 3 when x =12 then the value of f(4x) = 3 when x=3 as 12/4=3.

y=f(x+a)

This will shift the graph to the left by a. This is because the value of f(x) at x+a is displayed at the point x so effectively the graph occurs a earlier and therefore shifts to the left.

Sec, Cosec, Cot

Sec, cosec and cot are all functions in trigonometry. They are simply equal to one over on of the other functions, ie cos, sin and tan.

so

Sec = 1/cos

Cosec = 1/sin

cot = 1/tan

You can remember which is paired with which using the third letter rule. This is that the third letter is the first letter of the corresponding function ie)

sec goes with cos
cosec goes with sin
cot goes with tan

Differentiate Inverse Sine (arcsin) - proof

How to differentiate cos^-1x

y=cos^-1x

Bring the cos across

cosy = x

Differentiate both sides, remember when differentiating y time by dy/dx

-sin(y) dy/dx = 1

dy/dx = -1/siny

However we want to get the differential in terms of x, to do this we can use the identity

sin²t+cos²t = 1

so

sint = √(1 - cos²t)

putting this into our expression for dy/dx we get

dy/dx = 1/√(1-cos²y)

but cosy = x so

dy/dx = 1/√(1-x²)

by David Woodford

Differentiate Inverse Sine

This tutorial explain how to differentiate inverse sine, this applies when using radians.

begin with

y = sin^-1 x
bring sin^-1 across to become sin
sin y = x
differentiate
cos y dy/dx = 1
note that the derivative of sint wrtt is cos t as explained in an earlier tutorial and by the chain rule when we differentiated sin y it became cosy time dy/dx as we are differnetiatiny a y and the derivative of y is dy/dx

then make dy/dx the subject

dy/dx = 1/cosy

We know the identity
sin²t + cos²t = 1
so we can wrtie
cos t =√(1 - sin²t)

we can now put this into the expression for dy/dx to get
dy/dx = 1/√(1 - sin²y)
but we know from the second line that sin y = x so

dy/dx = 1/√(1 - x²)

Factorising Quadratics

Factorising quadratics is basically putting them in brackets. In this section we will look at two different ways of factorising quadratics( for simple and complex ones) and when they should be used.

Note/ sometimes a quadratic cannot be factorized using whole numbers, this is when you must use the quadratic equation to find the values of x. See my earlier post and c++ program

Simple type

Use when there is no coefficent of x²

eg)x²+2x-8

start by opening 2 brackets with an x in each

(x )(x )

put the first sign in the first bracket. If the second sign is + put the same sign in both, if its - put the opposite sign in the second bracket

(x+ )(x- )

find the 2 numbers that will add(if both signs in brackets are +) or subtract(if the signs in the brackets are different) to make the middle number(2) and multiply to make the end number(8)

(x+4)(x-2)

and thats your quadratic factorized

Complex Type

Use when there is a coefficent of x²

eg) 8x²-14x-15

Before we can open brackets we need to split up the 14x

8x² ?x ?x-15

the rule for the signs is the same as in the simple case, put the first sign before the first x term. If the second sign is + put the same sign in both, if its - put the opposite sign before the second x term

8x²-?x+?x-15

we also use the same rule for the to coefficients of x, they must add or subtract to make the middle number(14) but they must times to make the end number times the first(15×8=120)

8x²-20x+6x-15

we then take out the common factor of the first 2 terms(4x)

4x(2x-5) + 6x -15

we use the bracket(2x-5) as the common factor for the second 2 terms and find what we need to multiply by(3)

4x(2x-5)+3(2x-5)

we then take the 2 numbers in front of the brackets(4x and 3) as our second bracket

(4x+3)(2x-5)

and there we have a fully factorized quadratic

Matrix Multiplication

This is a simple tutorial on how to multiply 2 matrices together. You can speed up doing this using my c++ matrix calculator (download matrix calculator) but I strongly recommend learning how to do them long hand first as you should only use formulae and program that you understand how they work. Multiplying matrices is a useful thing to know as it enables complex algebra to be simplified and i used extensively in 3d computer graphics.

What Matrices can i Multiply Together

You may fin it surprising that only certain size matrices can be multiplied together, this is because of the way the multiplication is done and how the size of the answer is determined.

The size of a matrix is given as rows by columns

eg) 2 3 1

3 6 1

is a  2×3 matrix. When multiply matrices the columns of the first matrix must be equal to the rows of the second. The size of the new matrix is given by the rows of the first by the columns of the second

eg) Fig1

How is the multiplication Done

The way in which matrix multiplication is done is surprisingly simple.  Go to the first point in the new matrix. Look at the row it is in and go to the start of that row in the first matrix and the start of the column it is in in the second matrix. Multiply these to numbers together, then move along one in the first matrix and down one in the second matrix and multiply these together. Keep going till you have reached the end of the row of the first matrix, now add all of these values up and that is the first value for your new matrix.

So you go to the row in the first, the column in the second, times each pair of values and add them up for each value.

In the above diagram (fig 1)

u=ag + bk

v=ah+bl

w=ai+bm

x=cg+dk

y=ch+dl

and so on….

That may sound a little confusing but you’ll soon get used to it and it is very useful. If you want to practise and check your answers please download my c++ matrix calculator, the command for multiply is mlt, though you will have to define you matrices using the dim command first. All the data it needs off you is asked for

If you have any comments please leave them below or email me at woodford_4@hotmail.co.uk