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.

Tangent on curve method – The tangent method is shown on the graph, when x=2, touching the very edge of the curve. Once the line has been drawn, I shall draw a triangle structure, and look at the change in y and the change in x. This is the formula used for this method:

Gradient = Vertical (change in y axis)

Horizontal (change in x axis)

Therefore the gradient of the curve equals –

This is most likely wrong and most certainly not accurate – it is only estimation. The second method, the increment method is far more accurate.

Increment method – This is where you plot two points, represented by the letters P and Q, on the curve, and draw a line to join them. Here I have taken the point x = 2. I first get the co-ordinates for both points, which are 2 and 22, and Q, 3 and 32. After this I used the formula:

y – x

2     2                 =          3^2 – 2^2               =                   9-4             =                  5

y – x                                3   -    2                                         1

1     1

These, however are quite far apart. Moving Q closer to P will help me obtain the actual gradient of the curve. This applies to any value of x. In this next exercise; I will keep moving these two points together to obtain a gradient.

xn

P (2, 2²) and Q (2.6, 2.6²)

(2.6² - 2²) / (2.6 – 2) =

2.76 / 0.6 = 4.6

P (2, 2²) and Q (2.4, 2.4²)

(2.4² - 2²) / (2.4-2) =

1.76/0.4 = 4.4

P (2, 2²) and Q (2.2, 2.2²)

2.2² - 2² / 2.2 – 2 =

4.84 – 4 / 0.2 = 4.2

P (2, 2²) and Q (2.01, 2.01²)

2.01² - 2² / 2.01 – 2 =

4.0401 – 4/ 0.01 = 4.01

P (2, 2²) and Q (2.00001, 2.00001²)

2.00001² - 2²/ 2.00001 – 2=

4.0000400001 - 4/0.00001= 4.00001

As P and Q get closer to each other, they get closer to 4. From these results, when x = 2, the gradient of the curve is equal to 4.

General Proof - With this method, we take two points, like the other two methods. The co-ordinates of P remain the same as the increment method, but this time, Q, it is (x+h) x+h) ²). Here, “h” tends to 0 and can be any value on the x axis. Here, a different formula is used –

Gradient of PQ = (x+h) ² - x²         = x² + 2hx + h² - x²    =     2hx + h² = 2x+h

x +h-x                          h                                h

As Q gets closer to P, the value of H gets closer to 0 to the gradient of the chord PQ also becomes closer to the actual gradient of PQ. Here, he tends to 0, so therefore

2x + h = 2x + 0 = 2x. Therefore the gradient is 2x. According to the previous values used in the increment method, x² is equal to 2x, so this proves that proving it binomially is equal to the increment method.

When next using other values of n, I shall present a table of results when next picking two points, unlike when previously illustrated in the increment method for y =x².

Increment Method

y=x³

The general pattern here between the two values of x and the gradient, the value of the gradient function here is 3x². Using this formula, I can predict that the next 5 values of x and the gradient will be:

x                   x2          3x2

And now to the general proof -

(x+h)³ - x³              =             x³ +3x²h + 3h²x+h³ -x³    =  h(3x² +3hx + h²)

x + h – x                                                 h                                 h

=   3x² + 3hx +h²

This equation cannot be simplified any further without making it into an algebraic fraction. However, here h tends to 0. All terms here contain an h value, apart from 3x². So therefore it simplifies to…

3x² + (3 x 0 x x) + 0² = 3x². Therefore, this is the gradient function. This also corresponds to the increment methods in the tables. Therefore, my prediction was correct.

y=x^4

The general pattern here between the two values of x and the gradient, the value of the gradient function here is 4x³. Using this formula, the next 5 values of x and the gradient are:

x           x³             4x³

General proof –

(x+h)^4 – x^4                   = x^4 + h^4 +4hx³ + 6x²h² + 4xh³ - x^4    =

x + h – x                                                      h

h (h³ + 4x³ +6x²h +4xh²)                    =    h³ + 4x³ +6x²h +4xh²

h

H tends to 0 again here, and every term contains an h except for (4x³). Therefore, this is the gradient function for x4, and verifies that my ...

#### 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.