# The Gradient Function

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Introduction

The Gradient Function Introduction The gradient of any line is the steepness at which it slopes; on straight lines it can be worked out by drawing a right angled triangle using the line itself as the hypotenuse to find out the ?y, and ?x. The gradient of a line can then be worked out by dividing ?y by ?x. The following graphic shows an example: However, with a curved graph, the gradient is different at different points. To work out the gradient at a point of a curved graph, a tangent would have to be drawn, and the gradient of it measured. The longer the tangent is, the more accurate the result if done by eye. The following graphic is an example: Because this method is inherently inaccurate, to improve the accuracy we could either use a computer program to draw an accurate tangent, or use the small increment method. The small increment method is where the gradient of a chord, from the point, to another point of the line a short distance away, is worked out to find the gradient between the two points. So, for example, if it was a curve of y=x2, the gradient at x=3 would be measured using a chord from [3, 9] to [3.0001, 9.00060001] and so the gradient would be 0.00060001 divided by 0.0001 which is 6.0001 or 6 as an integer. The Small Increment Method I given a curve, y=x2, and a point, x=2, I can calculate an approximate gradient by using a chord with a second point a small distance away from x=2. ...read more.

Middle

I will use this data to work out the nth term of the gradient in the graph of y=x4. 4 32 108 256 500 +28 +76 +148 +244 +48 +72 + 96 +24 +24 As the differences are equal at the third level, the pattern for the nth term is an3 + b, if the value of a is the difference divided by 3, then the result for n=1 would be 8 which is twice the real value, so the value must be the difference over 6, this applies for all the values I have. This means that the nth term is 4n3; because it already fits, the value of b must be zero. To test this pattern, I can use it to predict the gradient at x=6. Using the nth term pattern it should be 4 x 63 which is 864. Using the small increment method, the answer is (6.0014 -64) divided by 0.001 which rounds to 864. Therefore the nth term is correct. Patterns for graphs of form y=xn In the graphs of this form I have worked out patterns for, there seems to be a global pattern in the nth term of each sequence: This means that for graphs of form y=xn the gradient at any point could be worked out using G=nxn-1. I will test this on the graph y=x5: PREDICTIONS X=1 5 x 14 = 5 X=2 5 x 24 = 80 X=3 5 x 34 = 405 X=4 5 x 34 = 1280 X=5 5 x 34 = 3125 Graphs of form y=ax2 I will extend my work to see what effect the value of a has on my results so far, initially I kept a =1. ...read more.

Conclusion

Using calculus to calculate gradients If I was to simplify (x+f)n it would be the following: xn + nxn-1f + ?xn-2f2... The value of ? is not relevant, but would in fact be a triangular number, because the line in the calculation that relates to this is as follows: G = (x+f)n - xn f This means that if it is simplified it would be: G = xn + nxn-1f + ?xn-2f2... - xn f But as it is subtracted by xn and divided by f it cancels out to: G = nxn-1 + ?xn-2f ... Therefore as f was zero (which it should be) the value of the ? makes no difference as all that is left is: G = nxn-1 Which is in fact the formulae I originally came up with. I have now demonstrated that calculus can be used to derive the gradient function for all graphs of the type y=axn. Conclusion To conclude, I have worked out that a gradient at any point on a graph of form y=axn can be worked out using the following formulae: G = anxn-1 I have tried four different methods to work this out: * Drawing tangents by hand * Using the small increment method * Using a computer program * Calculus From using these, I know: * That hand drawn tangents are inaccurate and so no use for working out a pattern * That the small increment method is not entirely accurate * That computer programs can generate perfectly accurate gradients * That calculus can work out the formulae in a single equation. I also know that in graphs of form y=axn + bxm can be treated as two gradients from graphs of form y=axn added together. ...read more.

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