The Gradient Function.

By Laurence Gagen

The Gradient Function

I am trying to find a formula that will work out the gradient of any line (the gradient function)

I am going to start with the simplest cases, e.g. g=c², g=c, g=c3 etc. as they are probably going to be the easiest equations to solve as they are likely to be less complex and hopefully the formulas to the more complex equations will be easier to discover by looking at the previous formulas.

I am going to look at the line g=c² first.

g=c²

Please see graph on separate piece of paper

One of the most obvious things I notice is that as the co-ordinates increase so does the gradient. Not only can you see that from the results below, but also on the graph you can that the line gets steeper and steeper. This makes sense as the higher the number c is the larger the difference between c² and c.

Another thing that I have noticed is that the larger the co-ordinates the smaller the increase in gradient.

As the table above shows there are two methods that I am using for calculating the gradient of line. The first being drawing a tangent at the point, working out the distances on the tangent using the scale on the graph and then using this formula:

dg/dc

However there is another way called small increment method. This method gives a more accurate approximation on the gradient. What you do is zoom in on the graph and take part of the curve you take a co-ordinate e.g. (3,9) and (3.01,9.0601).

Now you connect the two points together with a straight line. And because the graph is to a much larger scale the line should follow almost the same path as the curve. Of course the more you zoom in the more accurate the gradient will be. You then use the same formula to work out the gradient.

dg/dc

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dg/dc

This should then give you an accurate gradient, it tends to be more accurate than the other method although if you were to draw the tangent and graph perfectly you should get the exact answer. One of the good things about this method is that it isn't necessary to sketch the magnified area instead there is another method. All you need to do is take a point e.g. (3,9) and then make up another c co-ordinate which is very close to the first one e.g. 3.01. And then using the lines equation e.g. g=c² you can work out the Y co-ordinate. In this case you would do 3.01*3.01, which would be the same as 3.01². This then gives you the value 9.0601. Now that you have those values you know that the line must cross the point (3,9) and the point (3.01,9.0601). Now you use the formula:

dg/dc

Which gives you 0.0601¸0.01= 6.01

I have discovered that the formula for the equation g=c² is 2c. As you can see from the results table the accuracy using the tangent method hasn't been perfect, but using the small increment method it is possible to get much more accurate set of results.

I am now going to look at the line y=c. I already know the gradient of the line, as you can tell from the equation the gradient is always going to be one as there are always going to be the same values as each other. Below is a sketch of how the line would look.

There is no need to use the small increment method here, as I know that the gradient is accurate as c-c is always going to equal 1. A formula to work out the gradient function for this equation is:

c/c

Which is simple a way of getting to the number 1 which is of course always going to be the answer if the c and g coordinates are the same. Another formula is of course just simple 1.

Now I am going to look at the line g=c³. I predict that this line will look similar to Y=c² but it will be steeper and indeed it is, by entering the two equations into my calculator I can see that the equation g=c³ is steeper than Y=c². Going by the formula for the previous equation I would have thought that the formula for this equation might be 3c.

Please see graph on separate piece of paper

The formula I predicted doesn't work for this equation.

I have worked out the gradient function for this line it is more complex than the previous ones, it is:

3*c2

Below are the calculations that I used to check that the formula does work:

3*12=3
3*22=12
3*32=27
3*42=36

As you can see my formula does work, they give almost exactly the same answers for the gradient as my small increment result which means it must be right and also the small increment method must be more accurate, so from now on I am going to just use that method instead of using tangents as well. This will mean I don't have to draw out the graphs any more.

I am now going to look at the gradient of the line g=c , I think this could have a very similar formula to the previous ones as looking back on the other equations the rule that I have used for g=c³ works for the two previous ones assuming I adapt it, for example in the equation g=c³ the formula was 2c, but I can also use my new formula to get the same results except of course instead of multiplying by three you would times by 2 as the equation is to the power of 2. So I would expect 4*c3 will be the formula that works for the equation g=c4.

g=c4

My prediction was right the formula for this line is 4*c3

I am going to put the formulas that I have discovered so far into a table so that they are hopefully easier to interpret.

Looking at my previous formulas I have come up with a formula that will work for any equation, although I need to check that it does work for negative, fractions etc. The formula is as follows:

nc (n-1)

This formula has been developed from the other formulas that I have discovered for the equations, as you can see from the table above it is fairly obvious what the formula is going to have to be. The n stands for the power of e.g. in the equation g=c-1, n would equal -1.

I am going to just check that my formula does work with the equation g=c5although I am already convinced it will work as it has worked for all the previous equations.

g=c5

5*14=3
5*24=80
5*34=405
5*44=1280

Indeed my rule does work, however now that the gradients are such high numbers it is more noticeable that the small increment method isn't perfect. I am now going to look at the line g=c-1and see if my rule does or doesn't work of that and if not what does, the reason it might not work is because all the other equation I have used have been positive numbers.

g=c-1

-1*1-2=-1
-1*2-2=-0.25
-1*3-2=-0.1111
-1*4-2=-0.0625

As you can see the formula does work however to make sure I am going to look at the line g=c-2.

g=c-2

-2*1-3=-2
-2*2-3=-0.25
-2*3-3=-0.074
-2*4-3=-0.03125

The formula does work for this equation, which means it must work for other equation that involve negative integers.

Just to be certain I am going to look at the line g=c-3 and do the same thing as I did with the equations g=c-1 and g=c-2.

g=c-3

-3*1-4=-3
-3*2-4=-0.1875
-3*3-4=-0.037
-3*4-4=-0.011

As you can see the formula does work, this means that the formula will obviously work for more complex equations e.g. g=2c-3+5c-2 + 902c-324. To work these out you would split up the equations as I explained earlier.

You can get equations like y=3X and y=128X14 you can work these out using the formula nx(n-1).To work these out, I will take for example y=128X14 you first work out y=X14 and then you times it by 128, which makes sense as it is 128 lots of X14.

To work out slightly more complex equations such as g=2c3+5c2, all you do is use the equation nx(n-1) but break up the equation into two so you have two equations, take for example g=2c3+5c2 you would break it up to g=2c3 and g=5c2 . Then you work out the equations as normal as shown below.

g=2c3

g=5c2

What you then do is combine the two together like this:

g=2c3+5c2

Using my graphical calculator I can check that my method is correct and indeed it is. Of course this method will work for any other similar types of equations.

I think that it is extremely likely that the equations will also work for fractions e.g. g=c0.2, g=c-0.5 etc. To check I am going to test both of the examples, notice that the second one is a fraction and a negative so might is less likely to work using the formula.

g=c0.2

g=c-0.5

As you can see the formula does work for both fractions and negative fractions.

What I have discovered is, that it is possible to find the gradient of a line at any point by using the equation and the formula nc (n-1), and that this formula works with positives, negatives and fractions.