# The Gradient Function

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Introduction

Jamie Warner 11M

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.

Middle

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

x | X 0.25 | Gradient |

1 | 1 | 0.25 |

2 | 1.414214 | 0.148651 |

3 | 1.732051 | 0.109673 |

4 | 2 | 0.088388 |

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.

x | X 0.25 | Gradient |

1 | 1 | 0.25 |

2 | 1.414214 | 0.148651 |

3 | 1.732051 | 0.109673 |

4 | 2 | 0.088388 |

Gradient | 1/4x¾ |

0.25 | 0.25 |

0.148651 | 0.148651 |

0.109673 | 0.109673 |

0.088388 | 0.088388 |

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

x | y | Second point x | second point y | gradient |

3 | 0.333333 | 3.1 | 0.322580645 | 0.104058 |

3 | 0.333333 | 3.01 | 0.332225914 | 0.110374 |

3 | 0.333333 | 3.001 | 0.333222259 | 0.111037 |

4 | 0.25 | 4.1 | 0.243902439 | 0.059488 |

4 | 0.25 | 4.01 | 0.249376559 | 0.062189 |

4 | 0.25 | 4.001 | 0.249937516 | 0.062469 |

2 | 0.5 | 2.1 | 0.476190476 | 0.226757 |

2 | 0.5 | 2.01 | 0.497512438 | 0.247519 |

2 | 0.5 | 2.001 | 0.499750125 | 0.24975 |

1 | 1 | 1.1 | 0.909090909 | 0.826446 |

1 | 1 | 1.01 | 0.99009901 | 0.980296 |

1 | 1 | 1.001 | 0.999000999 | 0.998003 |

x | x-1 | Gradient |

1 | 1 | -1 |

2 | 0.5 | -0.25 |

3 | 0.333333 | -0.111111111 |

4 | 0.25 | -0.0625 |

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

x | y | second value x | second value y | gradient |

4 | 32 | 4.1 | 33.62 | 16.4 |

4 | 32 | 4.01 | 32.1602 | 16.04 |

4 | 32 | 4.001 | 32.016002 | 16.004 |

3 | 18 | 3.1 | 19.22 | 12.4 |

3 | 18 | 3.01 | 18.1202 | 12.04 |

3 | 18 | 3.001 | 18.012002 | 12.004 |

2 | 8 | 2.1 | 8.82 | 8.4 |

2 | 8 | 2.01 | 8.0802 | 8.04 |

2 | 8 | 2.001 | 8.008002 | 8.004 |

1 | 2 | 1.1 | 2.42 | 4.4 |

1 | 2 | 1.01 | 2.0402 | 4.04 |

1 | 2 | 1.001 | 2.004002 | 4.004 |

x | x2 | 2x2 | gradient |

1 | 1 | 2 | 4 |

2 | 4 | 8 | 8 |

3 | 9 | 18 | 12 |

4 | 16 | 32 | 16 |

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:

5 | 25 | 50 | 20 |

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

x | y | second value x | second value y | gradient |

4 | 128 | 4.1 | 137.842 | 100.86 |

4 | 128 | 4.01 | 128.962402 | 96.4806 |

4 | 128 | 4.001 | 128.096024 | 96.04801 |

3 | 54 | 3.1 | 59.582 | 57.66 |

3 | 54 | 3.01 | 54.541802 | 54.3606 |

3 | 54 | 3.001 | 54.054018 | 54.03601 |

2 | 16 | 2.1 | 18.522 | 26.46 |

2 | 16 | 2.01 | 16.241202 | 24.2406 |

2 | 16 | 2.001 | 16.024012 | 24.02401 |

1 | 2 | 1.1 | 2.662 | 7.26 |

1 | 2 | 1.01 | 2.060602 | 6.1206 |

1 | 2 | 1.001 | 2.006006002 | 6.012006 |

x | x3 | 2x3 | gradient |

1 | 1 | 2 | 6 |

2 | 8 | 16 | 24 |

3 | 27 | 54 | 54 |

4 | 64 | 128 | 96 |

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.

5 | 125 | 250 | 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.

x | y | second value x | second value y | gradient |

4 | 512 | 4.1 | 565.1522 | 551.368 |

4 | 512 | 4.01 | 517.139232 | 515.8496 |

4 | 512 | 4.001 | 512.512192 | 512.3841 |

3 | 162 | 3.1 | 184.7042 | 238.328 |

3 | 162 | 3.01 | 164.170824 | 218.1672 |

3 | 162 | 3.001 | 162.216108 | 216.2161 |

2 | 32 | 2.1 | 38.8962 | 74.088 |

2 | 32 | 2.01 | 32.64481602 | 64.96481 |

2 | 32 | 2.001 | 32.06404802 | 64.09605 |

1 | 2 | 1.1 | 2.9282 | 10.648 |

1 | 2 | 1.01 | 2.08120802 | 8.242408 |

1 | 2 | 1.001 | 2.008012008 | 8.024024 |

x | x4 | 2x4 | gradient |

1 | 1 | 2 | 8 |

2 | 16 | 32 | 64 |

3 | 81 | 162 | 216 |

4 | 256 | 512 | 512 |

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.

5 | 625 | 1250 | 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

x | y | second value x | second value y | gradient |

4 | 48 | 4.1 | 50.43 | 24.6 |

4 | 48 | 4.01 | 48.2403 | 24.06 |

4 | 48 | 4.001 | 48.024003 | 24.006 |

3 | 27 | 3.1 | 28.83 | 18.6 |

3 | 27 | 3.01 | 27.1803 | 18.06 |

3 | 27 | 3.001 | 27.018003 | 18.006 |

2 | 12 | 2.1 | 13.23 | 12.6 |

2 | 12 | 2.01 | 12.1203 | 12.06 |

2 | 12 | 2.001 | 12.012003 | 12.006 |

1 | 3 | 1.1 | 3.63 | 6.6 |

1 | 3 | 1.01 | 3.0603 | 6.06 |

1 | 3 | 1.001 | 3.006003 | 6.006 |

x | x2 | 3x2 | gradient |

1 | 1 | 3 | 6 |

2 | 4 | 12 | 12 |

3 | 9 | 27 | 18 |

4 | 16 | 48 | 24 |

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.

5 | 25 | 75 | 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

axn value | gradient function |

x2 | 2x |

2x2 | 4x |

3x2 | 6x |

Conclusion

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

x | y | second value x | second value y | gradient |

4 | 192 | 4.1 | 206.763 | 151.29 |

4 | 192 | 4.01 | 193.443603 | 144.7209 |

4 | 192 | 4.001 | 192.144036 | 144.072 |

3 | 81 | 3.1 | 89.373 | 86.49 |

3 | 81 | 3.01 | 81.812703 | 81.5409 |

3 | 81 | 3.001 | 81.081027 | 81.05401 |

2 | 24 | 2.1 | 27.783 | 39.69 |

2 | 24 | 2.01 | 24.361803 | 36.3609 |

2 | 24 | 2.001 | 24.036018 | 36.03601 |

1 | 3 | 1.1 | 3.993 | 10.89 |

1 | 3 | 1.01 | 3.090903 | 9.1809 |

1 | 3 | 1.001 | 3.009009003 | 9.018009 |

x | y | second value x | second value y | gradient |

4 | 128 | 4.1 | 137.842 | 100.86 |

4 | 128 | 4.01 | 128.962402 | 96.4806 |

4 | 128 | 4.001 | 128.096024 | 96.04801 |

3 | 54 | 3.1 | 59.582 | 57.66 |

3 | 54 | 3.01 | 54.541802 | 54.3606 |

3 | 54 | 3.001 | 54.054018 | 54.03601 |

2 | 16 | 2.1 | 18.522 | 26.46 |

2 | 16 | 2.01 | 16.241202 | 24.2406 |

2 | 16 | 2.001 | 16.024012 | 24.02401 |

1 | 2 | 1.1 | 2.662 | 7.26 |

1 | 2 | 1.01 | 2.060602 | 6.1206 |

1 | 2 | 1.001 | 2.006006002 | 6.012006 |

Gradient + Gradient =

Gradient 3x3 + Gradient 2x3

Gradient (3x3) | Gradient (2x3) | a + b |

151.29 | 100.86 | 252.15 |

144.7209 | 96.4806 | 241.2015 |

144.072 | 96.04801 | 240.12001 |

86.49 | 57.66 | 144.15 |

81.5409 | 54.3606 | 135.9015 |

81.05401 | 54.03601 | 135.09002 |

39.69 | 26.46 | 66.15 |

36.3609 | 24.2406 | 60.6015 |

36.03601 | 24.02401 | 60.06002 |

10.89 | 7.26 | 18.15 |

9.1809 | 6.1206 | 15.3015 |

9.018009 | 6.012006 | 15.030015 |

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.

This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics section.

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## Here's what a teacher thought of this essay

Generally an excellent piece of work with some sophisticated results. A good conversational style describes clearly why he is following particular lines of investigation.

Marked by teacher Mick Macve 18/03/2012