# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2012 | May-Jun | (P2-9709/21) | 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  3  4  5  6  7  8  It is evident that .

Hence, is a root of .

The root given correct to 3 decimal places is .