Using a series of lines joining two points [starting with P (5.0000, 5.0990) and

Question

Using a series of lines joining two points [starting with P (5.0000, 5.0990) and P1 (1.0000, 1.4142), shown on the accompanying diagram], find increasingly better approximations to the slope of the tangent to the curve y = Root of x^2 + 1 (shown) at point P. For the first calculation use points P and P1. For the second calculation use points P and P2, where P2 is 5 times closer to P than P1 was. For the third calculation move point P3, 5 times closer to P than P2 was, etc. Keep going for 5 iterations (i.e. points P1, P2, P3, P4, and P5), each point 5 times closer to P than the previous point. Keep 6 decimal places in each of your calculations and give an estimate of the accuracy of the slope of the line P5 to P from your results (i.e. how accurate do you think the approximation of line P5 to P is to the actual slope of the tangent to the curve at point P from looking at the differences in the approximate slopes you have calculated).

Explanation / Answer

Distance between P2 and P is x, then distance between P and P2 will be 5x

x + 5x = 5 - 1

6x = 4

x = 2/3

Hence the point P2 will be x-coordinate at (5-2/3) = 13/3

y(P2) = ( (13/3)^2 + 1)^(0.5) = 4.4472

slope of the line between P and P1 = (y(P) - y(P1)/(x(P) - x(P1))

=> (5.0990 - 1.4142)/(5-1)

=> 0.9212

slope of the line between P and P2 = (y(P) - y(P2)/(x(P) - x(P2))

=> (5.0990 - 4.4472)/(5-13/3)

=> 0.9777

For point P3(x) = 1 + 2/3 = 5/3

y(P3) = ( (5/3)^2 + 1)^(0.5) = 1.9436

slope of the line between P and P3 = (y(P) - y(P3)/(x(P) - x(P3))

=> (5.0990 - 1.9436)/(5-1.666667)

=> 0.9466

Hence the slope will be the mean of all this slope = (0.9466 + 0.9777 + 0.9212)/3 = 0.94850

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