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

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