# I am going to investigate the gradients of different curves and try to work out a pattern that I could use to find the gradient of any curve.

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Introduction

Maths Coursework Introduction I am going to investigate the gradients of different curves and try to work out a pattern that I could use to find the gradient of any curve. I will draw graphs of a selection of curves, some by hand, some using Autograph and some using Excel. I will use three methods to investigate the graphs. Firstly, I will draw tangents to the curves at 4 or 5 points and measure the gradients. Secondly, I will draw chords between x = 1 and 4 or 5 points and measure the gradients. Thirdly, I will use algebra to work out a formula for the gradient and see how this matches the first two methods. At first I split up the coursework into 3 main families (for each family there are additional equations to investigate): Part One: Curves involving x2 1. y = x2 2. y = 2x2 3. y = 3x2 4. y = 4x2 5. y = x2 + 1 6. y = 7x2 + 6 Part Two: Curves involving x2 + x 1. y = x2 + x 2. y = x2 + 2x 3. y = 7x2 + 4x + 5 Part Three: Curves involving x3 + x 1. y = x3 2. y = 2x3 3. y = 4x3 + 2x - 5 Finally, I will summarise my results in a series of tables and work out an overall formula that I could use to predict the gradient of any curve. PART ONE: CURVES CONTAINING X2 (1) y = x2 I am investigating the changes in gradient for the curve y = x2. To plot the curve, I will use the table of values given below. x 0 1 2 3 4 5 6 y 0 1 4 9 16 25 36 I will be working out the gradients of tangents to the curve by using the equation: Gradient = difference in y values difference in x values To help in this, I will draw the tangents to the curve at x values 1 to 4. ...read more.

Middle

Gradient = difference in y values = 2.5 = 5 difference in x values 0.5 (3,12) Gradient = difference in y values = 3.5 = 7 difference in x values 0.5 (4,20) Gradient = difference in y values = 4.5 = 9 difference in x values 0.5 Test x-axis value 1 2 3 4 5 Gradient 3 5 7 9 11 I predict that the gradient for x = 5 will be 11. I used the formula 2x + 1 for the gradient. Gradient = 5 x 2 + 1 = 11. Test: (5,30) Gradient = 5.5 = 11 0.5 My prediction was correct. Chords for x2 + x I will investigate a series of chords on the y = x2 + x graph. Gradient of chord = difference in y coordinates difference in x coordinates Chord AB Gradient = 42 - 2 = 40 = 8 6 - 1 5 Chord AC Gradient = 30 - 2 = 28 = 7 5 - 1 4 Chord AD Gradient = 20 - 2 = 18 = 6 4 - 1 3 Chord AE Gradient = 12 - 2 = 10 = 5 3 - 1 2 Chord AF Gradient = 6 - 2 = 4 = 4 2 - 1 1 Table of results can be seen below. Chord start point 1 1 1 1 1 Chord finish point 6 5 4 3 2 Gradient of chord 8 7 6 5 4 From this result and the table, I have noticed that as the finish point of the chord approached its limit of x = 1, the gradients of the chords approach 3. This is the gradient of the tangent at point (1,2). Using this result and algebra, I can prove that the value of the gradient of the curve will be given by 2x + 1. y P2 P1 x Coordinates of P1 are (x, x2 + x); coordinates of P2 are (x + d, (x + d)2 + (x + d)) ...read more.

Conclusion

Evidence to support this comes from looking at the result for y = 7x2 +4x + 5. Part Three: x3 and x3 + x graphs Equation Gradient Value of a Value of b 3ax2 + b y = x3 3x2 1 0 3x2 y = 2x3 6x2 2 0 6x2 Test: y = 4x3 + 2x - 5 12x2 + 2 4 2 12x2 + 2 If the number in front of x3 is called 'a' and the number in front of x is called 'b', then I have found that the gradient of this type of curve is always given by 3ax2 + b. Evidence to support this comes from looking at the result for y = 4x3 + 2x - 5. I can summarise these results in a single table. Part of equation Part of gradient formula ax3 3ax2 bx2 2bx cx c d Not present Prediction x4 4x3 This means that I can find the gradient of any curve involving x3, x2, x and a number at any value of x. Sample equation Formula for gradient Gradient at x = 1 y = x3 - 2x2 + 4x - 1 3x2 - 4x + 4 3 - 4 + 4 = 3 y = 2x3 - x2 + 5x -5 6x2 - 2x + 5 6 - 2 + 5 = 9 y = 3x3 + 2x2 - 2x + 3 9x2 + 4x - 2 9 + 4 - 2 = 11 y = 4x3 - 2x2 + 6x - 1 12x2 - 4x + 6 12 - 4 + 6 = 14 y = 5x3 + x2 + 2x + 1 15x2 + 2x + 2 15 + 2 + 2 = 19 SUMMARY The formula for gradient involves taking the power of x and multiplying it by x to one less power. y = Dxb gradient = bDxb - 1 Aidan Casey 11CS 1 ...read more.

This student written piece of work is one of many that can be found in our GCSE Gradient Function section.

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