• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13
  14. 14
    14
  15. 15
    15
  16. 16
    16
  17. 17
    17
  18. 18
    18
  19. 19
    19
  20. 20
    20
  21. 21
    21
  • Level: GCSE
  • Subject: Maths
  • Word count: 4626

I have been given the equation y = axn to investigate the gradient function for varied values of "a" and "n".

Extracts from this document...

Introduction

I have been given the equation y = axn to investigate the gradient function for varied values of "a" and "n" Firstly I will choose n = 1 for varied values of a When n = 1 then the graph will be a straight line The general equation of a straight line is y = ax I am going to keep n constant and use varied values for "a".The gradient for a straight line remains the same, therefore I am going to calculate the gradient by taking the difference in the y co-ordinates divided by the difference in x co-ordinates. In order to calculate the gradient for a straight line we will have to go through the process of calculating the gradient by taking the difference in y co-ordinates divided by the difference in the x co-ordinates. I would then choose n = 2 for varying values of a where a = 1,2,3,4 The equations would be ; y = x2 , y = 2x2 , y = 3x2 , y = 4x2 These equations would get us curved graphs and the gradient function at each point in a curve is different therefore I will draw tangents using a capillary tube and calculate the gradient by taking the change in the value of the y co ordinates divided by the change in t value of the x co ordinates. I would then choose n = 3 for varying values of "a". The equations would be ; y = x3, y = 2x3,y = 3x3. Again these equations would get me curved graphs and I will establish a relationship Between the curve and the gradient. In order to calculate the gradient in a curve , we know we have to draw the tangent for a given point x and then calculate the gradient. The gradient can be calculated by the change in the y co-ordinates divided by the change in the x co-ordinates. ...read more.

Middle

1.1 1.05 1.01 1.001 1.0001 1.00001 y=x2 1.21 1.1025 1.0201 1.002001 1.00020001 1.00002 Change in y 0.21 .1025 .0201 .002001 .00020001 .00002 Change in x .1 .05 .01 .001 .0001 .00001 Gradient function 2.1 2.05 2.01 2.001 2.0001 2 Gradient function at x = 3 x 3.1 3.05 3.01 3.001 3.0001 3.00001 y=x2 9.61 9.3025 9.0601 9.006001 9.00060001 9.00006 Change in y 0.61 .3025 .0601 .006001 .00060001 .00006 Change in x .1 .05 .01 .001 .0001 .00001 Gradient function 6.1 6.05 6.01 6.001 6.0001 6 As the procedure using the caluculator is very tedious and since I am well versed with Microsoft Excel and calculations can be done quite easily using its functions I would like to proceed using Microsoft Excel. Gradient function at x = 4 Using Ms Excel 2000 x 4.1 4.05 4.01 4.001 4.0001 4.00001 y=x� 16.81 16.4025 16.0801 16.008 16.0008 16.00008 change in y 0.81 0.4025 0.0801 0.008 0.0008 0.00008 change in x 0.1 0.05 0.01 0.001 0.0001 0.00001 gradient function 8.1 8.05 8.01 8 8 8 After finding out the gradient function using the increment method I have come to the conclusion that the relation I had found earlier from the graphs holds true Therefore the gradient function of the graph of y=x� is 2ax Looking at the curve of y=x�,I will now use the small increment method to get the gradient function at x=2,again using microsoft excel 2000 x 2.1 2.05 2.01 2.001 2.0001 2.00001 2.00001 2.000001 y=x� 9.261 8.615125 8.120601 8.012006 8.0012 8.00012 8.00012 8.000012 change in y 1.261 0.615125 0.120601 0.012006 0.0012 0.00012 0.00012 1.2E-05 change in x 0.1 0.05 0.01 0.001 0.0001 1E-05 1E-05 1E-06 Gradient function 12.61 12.3025 12.0601 12.006 12.0006 12.00006 12.00006 12.00001 Looking at the curve of y=x�,I will now use the small increment method to get the gradient function at x=3. X 3.1 3.05 3.01 3.001 3.0001 3.00001 3.000001 y=x� 29.791 28.37263 27.2709 27.02701 27.0027 27.00027 27.00003 change in y 2.791 1.372625 0.270901 0.027009 0.0027 0.00027 2.7E-05 change in x 0.1 ...read more.

Conclusion

and Q(x+?x,x+?y) be two near by points on the curve of y = 2x then y + ?y = 2x+?x ??y = 2x+?x-y =2x+?x-2x Therefore ?y=2x(2?x-1) This gives : ?y= 2x(2?x-1) ?x ?x Now dy= limit (?x tends to 0) [dy/dx] dx Therefore we have : dy=lim(?x tends to 0) [2x(2?x-1)] dx ?x Taking small values for ?x and using the calculator it appears that Lim(?x tends to 0) (2?x-1) ? 0.693 ?x Therefore d_ (3x) = 3x X lim(3?x-1) dx (?x tends to 0) ?x Taking small values for ?x and again using the calculator it appears that Lim (3?x-1) ?1.099 (?x tends to 0) ?x Therefore d_ (3x) ? 1.099 X 3 dx From the above calculations we can see that in the case of y=2x the gradient of the curve y=2x the gradient of the curve is less than y=2x whereas in case of y=3x the gradient of the curve is greater than y=3x.In fact there exists an exponential function y=ex such that the gradient function equals y=ex From my investigation above I cans see that 2<a<3.In fact e = 2.71828 correct to 5 decimal places.The value of e is irrational. The function y=ex is called the exponential function and dy = ex dx The graph of y = ex is shown below : In general : d(eax)_= aeax dx I started with drawing the graphs and then calculating the gradient by drawing the tangents using a capillary tube after which I had to work out relations for between the gradient function and the point at which we draw the tangent which was a very tedious and time consuming process. I then proceeded using the small increment method which was an effective method but again was a very tedious process. After constant reference to advance level math books I came into the topic of differentiation which was a little complicated but was more effective as I was able to find out the derivative which simplifies everything. I have thus come to the conclusion that calculus is the best way to investigate the gradient function of any graph. ...read more.

The above preview is unformatted text

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

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Gradient Function essays

  1. Peer reviewed

    The Gradient Function Coursework

    5 star(s)

    Here the gradient according to my formula is correct, but I will have to test it more than one time to be sure. The gradient calculated by my formula should be the exact number, and the formula calculated by using the chord method is only an approximation of the gradient at this point.

  2. Peer reviewed

    Gradient Function

    4 star(s)

    As with my own results, it was fairly good with several values near the actual value than others. Moreover, the results were quite consistent. The mean of the difference of my result to the real result was 1.2. Graph: y=x3 Theoretical formula for y=x3 P(x, x3)

  1. The Gradient Function

    This means that the gradient function for the curve y = x2, is 2x. I am now going to carry on with the investigation. I am now going to investigate the graph, of y = 2x2. Below is a table of values for this graph.

  2. Investigate gradients of functions by considering tangents and also by considering chords of the ...

    The results are shown below: x g (The Gradient) Tangent Method Increment Method 1 2 2.01 2 4 4.01 3 6 6.01 4 8 8.01 First, I found that the gradients increase as the co-ordinates increase. I not only found this by looking at the table of the results, but

  1. The Gradient Function Investigation

    This method is only as accurate as the number of chords measured and the gradients calculated will never be exactly the same as the actual gradient required. What is a Gradient Function? A gradient function is a rule, specific to a certain graph (e.g.

  2. Maths Coursework - The Open Box Problem

    is easier to interpret and I have also added the maximum value in a pink text box and the equation of the graph in a blue text box. This is the graph for the 40:20 rectangle. This is the graph for the 60:30 rectangle.

  1. Gradient function

    -6.36 -0.6 10.6 5.7 32.49 -7.49 -0.7 10.7 5.8 33.64 -8.64 -0.8 10.8 5.9 34.81 -9.81 -0.9 10.9 6 36 -11 -1 11 Power: 2 Coefficient: 1 Fixed Point: 5 My third fixed point is -3, 9 x y change in y change in x gradient -4 16 -7 1

  2. Aim: To find out where the tangent lines at the average of any two ...

    following up the same steps but with other two combination of roots.) 2. Roots -1.5 & 1.5 Average of the two roots = Plug the point, (0, -13.5) into the function Intersection = (3,0) 3. Roots -3 & 1.5 Average of the two roots = Plug the point, (-0.75, -7.59375)

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work