# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2005 | May-Jun | (P2-9709/02) | Q#5

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Question

i.
By differentiating , show that if then .

ii. The parametric equations of a curve are

x = 1 + tanθ , y = secθ ,

for . Show that .

iii.Find the coordinates of the point on the curve at which the gradient of the curve is .

Solution

i.

We are given that; We are required to show that; Since provided that ;  Therefore; If and are functions of , and if , then; If , then;  Rule for differentiation of is: Rule for differentiation of is;    Since provided that and , therefore;  ii.

We are required to show that for the parametric equations given below;  If a curve is given parametrically by equations for and in terms of a parameter , then; First we find . We are given that; Therefore; As demonstrated in (i); Therefore; Similarly, we find , when we are given that;  Rule for differentiation of is:  Rule for differentiation of is: Rule for differentiation of is;   Now we can find .   Since provided that and , therefore;   iii.

We are required to find the coordinates of the point on the curve at which the gradient of the curve  is .

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; We have found in (i) that; Therefore;   Using calculator; We can find coordinates of the point where by substituting this value in parametric equations  of the curve.      Using calculator; Since ;  Using calculator;  Hence, coordinates of the point on the curve where gradient is are 