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  • Level: GCSE
  • Subject: Maths
  • Word count: 3595

The Gradient Function Investigation

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Introduction

The Gradient Function Introduction Curves on a graph can be of varying steepnesses. This steepness also varies from point to point on many graphs. The steepness of a curve at a point is called its gradient. There are several methods for calculating the gradient at a certain point on a curve including the 'Tangent Method' and the 'Small Increments Method'. The Tangent Method Calculating the gradient of a straight line is simple. The formula is: Gradient = Change In Y Change In X This formula is demonstrated on Graph A. A curve proves more of a problem as the gradient is constantly changing. To calculate the gradient at a certain point, we must somehow be able to create a straight line from which to calculate this gradient. This can be achieved by drawing a tangent to the curve at the point in question. A tangent is a straight line which touches the curve at one point and one point only. Calculating the gradient of this tangent will give the gradient of the curve at this point. This is illustrated on Graphs B1, C1 and D1. This method is not very accurate though and large discrepancies in gradient can occur. This is because this method involves manually drawing a tangent which will often be drawn incorrectly. Errors in the measurement of changes in Y and X can also occur, even on a very large scale graph. The Small Increments Method This method is more accurate than the 'Tangent Method' as it does not involve manual drawing and measurement of changes. This method works on the basis of drawing a chord (a line joining two points on the curve) from the point where the gradient will be taken to another point close by. As the function of the graph (i.e. y = x�) can be used to calculate the precise X and Y co-ordinates of the second point on the curve, the exact changes in X and Y can be calculated. ...read more.

Middle

...etc. In each case the second bracket contains n terms. I can summarise this with the formula: an - bn = (a - b)(a(n-1) + a(n-2)b + a(n-3)b� +....+ b(n-1)) [With n terms in the second bracket] If I apply this rule to y = xn I can prove that my rule works for all values of n as a positive integer. Gradient = (x + h) - xn (x + h) - x = [(x + h) - x][(x + h) + (x + h) x +...+ x ] h = h[(x + h) + (x + h) x +...+ x ] (cancel x) h = [(x + h)(n-1) + (x + h)(n-2)x +...+ x(n-1)] (cancel h) If I now expand all the brackets and replace h with 0, the only terms I am left with are x(n-1). There will be n of these terms as there are always n terms in the second bracket. Gradient Function = x(n-1) + x(n-1) + x(n-1) +...+ x(n-1) [With n terms of x(n-1)] =nx(n-1) I have therefore proved that for all curves of the form y = xn, the formula nx(n-1) gives the gradient function of the curve for any value of n as a positive integer. Finding a Rule on Other Curves I now wish to see if I can find a rule for curves of the form y = Axn. I will again work through two examples of curves of this nature to attempt to find a rule which finds the gradient function of any of these curves. 1. y = 2x� Gradient = 2(x + h)� - 2x� (x + h) - x = 2(x� + 2xh + h�) - 2x� (expand brackets) h = 2x� + 4xh + 2h� - 2x� (expand brackets) h = 4xh + 2h� (cancel 2x�) h = 4x + 2h (cancel h) as h tends to 0 GF tends to 4x 2. ...read more.

Conclusion

Testing y = x-�; Hypothetical Gradient at Point x =0.5 is -16. Graph H shows that my rule also works for the graph y = x-� as my hypothetical gradient at the point x = 0.5 was the same as that which I calculated using the 'Small Increments Method'. Proving I will now try to prove my rule for graphs of the form y = x-n. My rule should lead to a formula of form GF = -nx-(n-1) y = x-n can be written as y = 1/xn Gradient = [1/(x + h) ] - [1/x ] (x + h) - x 1/{[(x + h) - x][(x + h) + (x + h) x +...+ x ]} h 1/{[h][(x + h) + (x + h) x +...+ x ]} h 1/[(x + h)(n-1) + (x + h)(n-2)x +...+ x(n-1)] (with n terms in the bracket) 1/n[(x + h)(n-1)] -nx-n-1 Fractional Powers I will now use my rule to hypothesise the gradient function for a graph of the form y = xp/q (where p and q are positive integers). I will then test it by using the 'Small Increments Method' to see if my rule works. I will then try to prove it using the 'Small Increments of Size "h" Method'. Testing y = x ; Hypothetical Gradient at Point x = 0.5 is 2. Graph I shows that my rule also works for the graph y = x1/2 as my hypothetical gradient at the point x = was the same as that which I calculated using the 'Small Increments Method'. Proving I will now try to prove my rule for graphs of the form y = xp/q. My rule should lead to a formula of form GF = (p/q)x[(p/q)-1]. Gradient = (x + h) - xp/q (x + h) - x = [(x + h) - x][(x + h) + (x + h) x +...+ x ] h = [h][(x + h) + (x + h) x +...+ x ] h = [(x + h)(p/q-1) + (x + h)(p/q -2)x +...+ x(p/q-1)] (with (p/q) terms in the bracket) = (p/q)x(p/q) ...read more.

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