The Gradient Function

  Aim: To find the gradient function of curves of the form y=axn. To begin with, I should investigate how the gradient changes, in relation to the value of x. Following this, I plan to expand my investigation to see how the gradient changes, and as a result how a changes in relation to this.

Method:

 At the very start of the investigation, I shall investigate the gradient at the values of y=xn. To start with, I shall put the results in a table, but later on, as I attempt to find the gradient through advanced methods, a table may be unnecessary. As I plot the values of y=x2, this should allow me to plot a line of best fit and analyze, and otherwise evaluate, the relationship between the gradient and x in this equation.

I have begun with n=2. After analyzing this, I shall carry on using a constant value of “a” until further on in the investigation, and keep on increasing n by 1 each time. I shall plot on the graphs the relative x values and determine a gradient between n and the gradients. Perhaps further on in the investigation, I shall modify the value of a, and perhaps make n a fractional or negative power.

Method to find the gradient:

 These methods would perhaps be better if I demonstrated them using an example, so I will illustrate this using y=x2.

This is the graph of y=x2. I will find out the gradient of this curve, by using the three methods – drawing a tangent to the curve, using an increment method, and proving it via Binomial Expansion.

4.001

1.414301942

0.125172

2

1.189207

2.1

1.203801344

0.175152

2

1.189207

2.01

1.190690845

0.178385

2

1.189207

2.001

1.189355738

0.178718

1

1

1.1

1.024113689

0.241137

1

1

1.01

1.002490679

0.249068

1

1

1.001

1.000249906

0.249906

Like before, these data are very hard to work with and I cannot find a general pattern by looking at the gradients – they are not whole numbers. This will again, however, support the gradient function I will come up with using binomial expansion.

Therefore, this proves my prediction was correct, and works with the rule nx n-1. I will now test this formula with each relevant datum in the table.

These are exactly the same. Therefore this proves that all positive values of n in the equation nx n-1 apply to this formula. After I try to prove negative powers are correct, I shall thoroughly investigate, in depth, the overall gradient function for the curves y =ax n-1, where “a” is not constant.

But for now, however, I shall investigate negative values of n. The value of n I will choose is -1, or y= 1/x.

y = x-1

I predict that the gradient function for when n=-1 = -x -2, in correspondence to the formula nx n-1. I can tell whether these data do in fact agree with this gradient function, but I shall binomially expand this to verify this is the case:

    1           .      x            -   x+h       .     1                  =              x             -         (x + h)

 x + h               x                x+h             x                               x(x+h)                  x(x+h)

                     x    +    h     -   x                                                                  h

=      x – (x+h)   =     -h           =      -1         =   -1         = -1x-2.

         hx(x+h)       hx(x+h)           x(x+0)*        x2

*h tends to 0.

Voila, this gradient function is in accordance with nx n-1. By getting this far, I have nearly convinced myself that all values work, whether they are integers or not, or positive or not. However, thus far I have not perhaps produced enough evidence to support this theory, and therefore I shall attempt to prove one last n value in this section, where n=-2. Since I am nearly convinced that this formula is proven for any value, where “a” is a constant 1, I will just binomially prove y = x -2.

In accordance to the formula nx n-1, I predict that the gradient function for this = -2x-3.

y=x-2.

    1           .        x²         -   (x+h)²       .     1*                  =     x²             -      (x²+2hx+h²)**

 (x + h) ²            x²              (x+h)²             x²                x²(x²+2hx+h²)       x²(x²+2hx+h²)

                     x    +    h     -   x                                                                  h

=    x²   -   (x²+2hx+h²)    = -2hx – h²     =    -2x – h        =     -2 – h/x         = -2 – 0/x

        hx²(x²+2hx+h²)     hx²(x²+2hx+h²) x²(x²+2hx+h²)   x(x²+2hx+h²)   x(x²+((2)(0x))+0²)

=        -2              = -2x-3    -  here we assume x, is not 0.

         x-3

* IN all the past equations I have rationalized the numerator to make them easier to work with and thus solve. Otherwise, they are thus irrational and impossible to solve, with all the knowledge of this topic I possess.

** h tends to 0.

I now move onto the second and final section of this piece of coursework – where we now change a new variable in this investigation - the value of a. I will try to combine this with different value of n, to see how it affects the end result.

axn

 In this section, I will change the values of a by an increase of 1 in each example, and furthermore do the same for the value of n. While it is bad practice to change 2 variables at once as a rule, I have found the effect the changing n in the previous investigation, so is therefore valid. Firstly, I shall try to investigate 2x², and if successful, move onto greater values of a.

y=2x²

The gradient here is equal to 4x – I can see this by investigating the values of x, as compared to the gradient. I predict that the gradient for 5 will be 20. This can be shown here:

My hypothesis is correct. It will be also be useful to prove this binomially, however:

2(x+h) ² - 2x²    = 2(x²+h²+2hx) – 2x²   =   2x²+2h²+4hx -2x² = 2h²+4hx

  x + h – x                      h                                       h                         h      

2h + 4x = ((2)(0)) +4x  = 4x

Therefore the gradient function is correct. Now, I will change the value of n by 1 to see whether the same rule applies with different values of “a” and n.

y = 2x3

After looking through each value, I have observed a pattern. I have noticed that the gradient is always equal to 6x². I predict that when x = 5, the gradient will be equal to 150.

This has proved my hypothesis correct. Now, again I shall prove this binomially:

2(x+h) ³ - 2x³    = 2(x³ +3x²h + 3h²x+h³) – 2x³   =    2x³ + 6x²h + 6h²x +h³ - 2x³

  x + h – x                                h                                              h                        

= 6x²h + 6h²x +h³   = 6x² + 6hx + h² = 6x² + (6)(0x) + 0² = 6x²

             h        

I need not explain this any further. I will now investigate 2x4, and then try to find a general pattern for all the gradients of where a=2.    

After looking through each value, I have observed a pattern. I have noticed that the gradient is always equal to 8x³. I predict that when x = 5, the gradient will be equal to 1000.

My theory was correct. I shall now try and binomially prove this…

2(x+h)4 – 2x4                   = 2(x4 + h4 +4hx³ + 6x²h² + 4xh³) - 2x4    =  

 x + h – x                                                        h

2x4 + 2h4 +8hx³ + 12x²h² + 8xh³ - 2x4 =     2h4 +8hx³ + 12x²h² + 8xh³

                                 h                                                          h

=    2h³ + 8x³ + 12x²h + 8xh³

After doing this so many times, I can tell that every term containing an h will disappear. The one exception to this = 8x³. This is the answer I had previously predicted, and therefore my theory was correct.

I have observed a pattern using these results. I have noticed that when a =2, if you multiply this through by the power, and subtract 1 from the power, you get the gradient function.

In other words = 2(nxn-1). This may apply to every value of a, thus giving us the overall gradient function, but I must find far more evidence than one solitary a value to give it any meaning.

If this is the case, however, I think that when the co-efficient is 3, the whole gradient will be 3(nxn-1). I shall try and investigate this.

3x²

After looking through each value, I have observed a pattern. I have noticed that the gradient is always equal to 6x. I predict that when x = 5, the gradient will be equal to 30.

The gradient here is 30, so therefore I was correct. Now, to the binomial proof…

3(x+h) ² - 3x²         = 3(x² + 2hx + h²) - 3x²    = 3x² + 6hx + 3h² - 3x²=   6hx + 3h²

    x +h-x                             h                                          h                              h

=    6x + 3h = 6x

2x² + 3x5

2(x+h)² -2x²       +  3(x+h)5 – 3x5      =  

  (x+h) – x                  (x+h) – x

2(x²+ 2hx + h²) – 2x²  + 3(x5 + 5x4h + 10x³h² + 10x²h² + 5xh4 +h5) – 3x5

            h                                                            h

= 2x² + 4hx + 2h² - 2x²    +    3x^5 + 15x4(h) + 30x³h² + 30x²h² + 15xh4 +h5 – 3x5

                h                                                                h

=   (4x + 2h) + (15x4 + 30x³h + 30x²h + 15xh3 +h4) = 4x + 15x4

= x (4 + 15x^3)

This seems to follow the same pattern as the previous investigation; each term follows the correct pattern in accordance with naxn-1 (2x² and 3x5 have the respective gradient functions of 4x and 15x4).  I will need to try this, however, with 2 more cases, to find a respective pattern between all of them.

With the next method, I shall use solely the table/increment method to find respective gradients. It is perhaps not a preferred method to algebraic proof. However, I shall need to do this to see whether it is a better method.

3x3 + 2x3

Gradient + Gradient =

Gradient  3x3 + Gradient 2x3

From these results, I can figure out that the overall gradient function of each value added together is :

9x2 + 6x2 = x(9x + 6x)

This proves quite a clear pattern. Now it is clear that both functions correspond to the gradient function nax^n-1. However, the formula

naxn-1 + naxn-1 is not valid : this is because we are assuming that the a and x values are not equal to each other. Therefore, I shall use a different letter to use in these equations. I shall substitute “n” with “m” and “a” with “l”. Therefore the formula for adding powers together is :

nxn + lxm = naxn-1 + lmxm-1.

This is the overall formula for working out the basic functions of any curve of axn + lxm. If the powers are being subtracted from each other, the + is simply replaced by a -.

Now I have proved this, the coursework is complete.