Finding the Stationary Points

We know that stationary point occur when the gradient is 0 so when the derivative of the graph is 0, so in order to find the stationary points we but first differentiate the curve.

For example lets consider the graph . We cab differentiate this to find

We must then equate the derivative to 0 and solve the resulting equation. This is because we are trying to find the points where the gradient is zero and these point occur exactly at the solutions of the equation we have formed.

So in our example we form the equation

by equating our expression for , , to 0
Solving this equation we find that stationary points occur exactly when

Note that there can be more than solution to this equation, each of which is a valid stationary point.

Finally we should also find the y co-ordinate for the stationary point by putting this value of x into the initial equation. So for this example
So the only stationary point is at

Nature of Stationary Points

The nature of a stationary point simply means what the graph is doing around it and are characterised by the second derivative, (found by differentiating the derivative). There are three types of stationary point:

1. Maximum Points: These are stationary points where the graph is sloping down on either side of the stationary point (a sad face type of curve).
   Here
2. Minimum Points: These are stationary where the graph is sloping upwards on either side of the point (a happy face)

   Here

3. Point of Inflection: Here the direction of the slope of the graph is the same either side of the stationary point, it can be in either direction.

   At a point of inflection but isn’t enough to ensure that a point really is a point of inflection as it could still be a maximum or minimum point

Checking the nature of a Stationary Point when
In this case the easiest thing to do is look a small distance either side of the point and see whether the y value is greater than or less than that of the stationary point. You can then draw yourself a picture to see what it is. For example if they are both greater than the stationary point you know it is a minimum point, but if one is greater and one is less than it is a point of inflection

Warning: checking points either side does not guarantee the correct result as there may be another stationary point or a break in the graph between where you are checking and the stationary point so you should always check using the derivatives if possible

Also see

• Points of Inflection (0)
• Product Rule for Differentiation (0)
• Monotonic (Increasing and Decreasing) Functions (0)
• Fundamental Theorem of Calculus (0)
• Integrating Fractions – using the natrual logarithm – Example tan(x) (0)
• Integration by Parts (0)

Examples

Some simple examples which contain only one variable are:

Solving and Manipulating Inequalities

Inequalities can be solved by rearranging them and isolating the variable you want to find in a similar way to normal equation (see the post quadratic inequalities to see how to solve quadratics). However, rather than getting an exact value such as x=3 we get a range (open or closed) of values such as x<2 or -3<-1.

Much of the manipulation is the same though there are slight variations when dividing or multiplying by negative numbers or taking the reciprocal. The important thing to remember is that like normal equations we must do the same to both sides.

Addition and Subtraction

Addition and subtraction are exactly the same to equalities. We can add or subtract whatever we like as long as we do the same to both the sides. This enables us to take expressions “to the other side” by reversing their sign. For example all the following manipulations are valid.

Multiplication and Division

Again we can perform multiplication and division in a similar way to the way we perform it with equalities by doing the same to both sides. However, if we are multiplying or dividing by a negative number we must reverse the direction of the inequality since

This means we must be careful when diving by an unknown since by definition we don’t know whether or not it is positive or negative. If this has to be done you should consider both the cases it is positive and negative separately and if it is only positive or negative then the other inequality should lead to a contradiction which can easily be spotted such as x<0 and x>3.

Examples of valid manipulation are below:


for and/or for

Reciprocals

When taking the reciprocal or “one over” of an expression you must reverse the inequality so

Also see

• Simplifying and Solving equations with Algebraic Fractions (0)
• Simultaneous Equations (0)
• First Order Differential Equations (6)

Today the administrators that have taken over Woolworths said that all stores will close by January 5th.

Unemployment has risen to 1.86 million.

The pound hit a low of 1.0796 euros.

Opec is making an output cut of 2.2million barrels per day.

Yesterday the US interest rates to between 0 and 0.25%.

National express is to cut 750 jobs

But on the positive side inflation has fallen to 4.1% !!!

Also see

• Making Sense of the Economic Climate (0)

Also see

• Economic Update 17th December (0)

This is a brief explanation of how to create basic 3D graphics. There are many different packages that will allow you to program using pre written 3D graphics library’s, such as Java 3D but this is aimed at using and understanding the actual formula that can be used to draw the graphics yourself.

This assumes you are looking straight ahead from a single point and will map points in the 3D space in front of you onto a single plane at a fixed distance in front of you, probably 1 unit to keep it simple.

First of all consider the vertical direction, y axis. If we consider the right angled triangle formed by taking the line from you eye to the point, the line from your eye to beneath the point (ie the line where y =0) and the final verical equal to the y cordinate, we can use trigonometry to work out the height that need to be displayed.

From this we can work out the angle at your eye would be tan^-1(y/(x²+z²))
This is because the length of the “opposite” is equal to the y co-ordinate and the adjacent can be found using pythagoras from the x and z co-ordinates.

We now can use the angle we have found to work out where the point needs to be displayed in our plane. Again we need to form a triangle similar to the first one but where the top corner is the final position of the point when it is projected. Ie the opposite is the y co-ordinate we need.
therefore,
y = tan(t)((x/z)²+1)
y=y((x/z)²+1)
(x²+z²)

Also see

• Area of a Triangle (3)
• Proof by Contradiction (0)
• Auxiliary Angle Method for Solving Trigonometry Equations (0)
• Rational and Irrational Numbers (5)
• How to use Surds – Add, Multiply and Rationalise (1)
• Moments – Turning Forces (6)

This is how I do it

1. get the diablo spinning fast(it tends to slow down when its on the stick) and do a basic toss — though not much higher than your sticks (because you don’t want it bouncing as it lands)
2. Turn one of the sticks inwards(so its parallel to the string) and try to catch the diablo on it, so that it lands as it would have on the string
3. Point the knotted end of the stick upwards (because the diablo is spinning if you don’t do this it will go flying off, aim to angle it such that the force of gravity pulling it down the stick matches the spinning trying to pull it forward, its something you get used to)
4. When you ready (probably after a few second to start with) tilt the stcik back down again so that the diablo can roll off back on to the string — alternatively you can flick the stick up so that the diablo goes into the air and catch it again like normal

This trick will take a while to master, it took me sometime, but once its done it will look pretty good. At first you may only be able to keep it there for a few seconds but with practice you can extend this time, concentrate on keeping it under control so if it begins to wobble return to normal even if you’ve only just go it there.

If you have any suggestions, improvements or comments please leave them below or email me at woodford_4@hotmail.co.uk >> please visit my site www.breakingwave.co.nr

Also see

• 3 Clubs and Mills Mess Juggling Tutrials (0)
• Diablo Trick >> Forced Loop (0)
• Proof of Cosine Rule (3)
• Understand the Sine and Cosine Rules (11)
• Juggling Tutorials && Dwebs has moved (0)
• Exams (Fun fun fun) and Juggling (0)

Clubs. These are to go with my written tutorials on this blog as well as on my site (www.breakingwave.co.nr)

Any way here they are

[youtube=http://www.youtube.com/watch?v=Ws2LR5N0yz8]

[youtube=http://www.youtube.com/watch?v=C9jv9DORuc8]

Also see

• DIABLO TRICK >> GRINDS (1)
• Diablo Trick >> Forced Loop (0)
• How to use Surds – Add, Multiply and Rationalise (1)
• Derive Quadratic Formula (16)
• Matrix Multiplication (1)
• Factorizing Quadratics (1)

Download here(this link now works)
I have now complied my “level 1″ for Trevor. This means that there is no something you can play. Click the link above to download it(it is a link, even though it might not be underlined). The aim of the level is to reach the finish(the spirally thing) without dieing but by getting as many points as possible. You can get 10 points for each triangle you collect, i think the maximum is 120. However you can get injured if you get caught by the fish or land on the “spikes” each of the these take 10 points off your life.

other objects:

• springs: These allow you to jump higher (9 squares instead of 5 i think) meaning you can reach some of the platforms. Just walk on one and watch yourself go up
• hearts: these give you extra 10 health points
• shields: This gives you 10 second (50 frames) protection from things that hurt you

How to play

1. Click the link at the start of this page to download the game as zip file, save it to your hard drive and extract it
2. If you havnt got java installed already on your computer go to java.sun.com to download it. Note you may need the developer version to download (this is the java jdk)
3. Run the file play.bat that you downloaded

Diablo Trick:

Trapeze

In the trapeze you make the diablo swing up and around on of your sticks and land back on the string, keep it there for a short period of time and then swing it back down

1. Get the diablo spinning fast
2. Swing it across you your weak handed side
3. And then back across quickly to your strong handed side so it comes up and over you strong handed stick
4. heres the hard part, try and land the diablo on the strings between your 2 sticks, this is now your trapeze
5. when the diablo starts to slow down swing it back down to its normal position

At first you wont be able to keep it on the strings for very long but after a while you should be able to lengthen the amount of time.

Also see

• Trevor Changes (0)
• Trevor Pythagoras 08 Download (0)
• Play Trevor >> now avalible for download (0)
• DIABLO TRICK >> GRINDS (1)
• Diablo Trick >> Forced Loop (0)
• Juggling Tutorials && Dwebs has moved (0)

1. get the diablo spinning normally.
2. Turn your weak handed stick through about 70^o so that it is in the 2 o’clock position (ifyou imagine a flat clock with the sticks normally at 6) You need the string ends of the sticks to be roughly level
3. Raise the weak handed stick about 1 inch above the other
4. quickly pull it down and around the other stick so that the diablo is forced around in a circle between your arms
5. as you do this you can turn the stick so that they are back in the normal position when you’ve finished
6. carry on with the diablo as normal ready for your next trick

You shouldn’t find this trick too hard after a bit of practice. A good way of making it look more impressive is to learn to do it in both directions so that you can alternate them.

Good luck

Your Trick

Do you have a trick tutorial that you want everyone to be able to read, if so email it to me at woodford_4@hotmail.co.uk if you want ill include a link to your website and mention your name (just send these in the email) thanxs

Also see

• DIABLO TRICK >> GRINDS (1)
• 3 Clubs and Mills Mess Juggling Tutrials (0)
• Juggling Tutorials && Dwebs has moved (0)
• Exams (Fun fun fun) and Juggling (0)
• Trevor Pythagoras Level 1 && Diablo trick::Trampeze (0)
• Area of a Triangle (3)

This simple trick looks impressive especially when considering how easy it is to perform. The aim is the hold the sticks so the string is vertically and watch the diablo climb from the bottom one to the top

how to do it

1. start the diablo normally and getting it spinning as fast as you can (whilst keeping it under control) It is more important to keep the diablo straight than spinning fast
2. get the diablo toward you strong hand (about 1/3 of the way)
3. loop you strong hands stick around the diablo
4. immediately pull the string tight by pulling your strong hand down and your weak hand up.
5. watch the diablo climb up the string
6. as it begins to slow down quickly unloop the string and carry on as normal(not the diablo will probably be spinning quite slowly so try to get its speed up as quick as you can)

Dwebs has moved

Yay great news dwebs has now moved away from its bravehost domain, although its still hosted by them because they offer ftp support, and has gone to a .co.nr domain.

Visit my site now at

Also see

• DIABLO TRICK >> GRINDS (1)
• Diablo Trick >> Forced Loop (0)
• Exams (Fun fun fun) and Juggling (0)
• 3 Clubs and Mills Mess Juggling Tutrials (0)
• Trevor Pythagoras Level 1 && Diablo trick::Trampeze (0)
• Area of a Triangle (3)