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Introduction

Jamie Warner 11M

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.

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

n + lxm.

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 (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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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

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