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

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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.

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