Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 2 (P29709/02)  Year 2005  MayJun  (P29709/02)  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
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