Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 2 (P29709/02)  Year 2012  MayJun  (P29709/21)  Q#6
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Question
The parametric equations of a curve are
The point P on the curve has parameter p and it is given that the gradient of the curve at P is −1.
i. Show that .
ii. Use an iterative process based on the equation in part (i) to find the value of p correct to 3 decimal places. Use a starting value of 0.7 and show the result of each iteration to 5 decimal places.
Solution
i.
We are given that gradient of the curve at point P is 1.
We are given the parametric equations;
Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that particular point.
Gradient (slope) of the curve at a particular point can be found by substituting x coordinates of that point in the expression for gradient of the curve;
Therefore first we find from given parametric equations of the curve.
If a curve is given parametrically by equations for and in terms of a parameter , then;
First we find . We are given that;
Therefore;
Rule for differentiation of is:
Similarly, we find , when we are given that;
Therefore;
Rule for differentiation of is:
Now we can find .
We are given that gradient of the curve at point P is 1.
Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that particular point.
Gradient (slope) of the curve at a particular point can be found by substituting x coordinates of that point in the expression for gradient of the curve;
Therefore;
Since we are given that the point P on the curve has parameter p.
ii.
We are required to to find the value of p from following equation using iterative.
If the sequence given by the inductive definition , with some initial value , converges to a limit , then is the root of the equation .
Therefore, if , then is a root of .
We can write it as;
We use as initial value.



1 


2 


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7 


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It is evident that .
Hence, is a root of .
The root given correct to 3 decimal places is .
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