# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2007 | May-Jun | (P2-9709/02) | Q#3

Question

The parametric equations of a curve are

for t > 1.

i.       Express in terms of t.

ii.       Find the coordinates of the only point on the curve at which the gradient of the curve is equal to 1.

Solution

i.

We are required to find  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;

Rule for differentiation of  is:

Rule for differentiation of  is:

Rule for differentiation of  is:

Similarly, we find , when we are given that;

Rule for differentiation of  is:

First we solve the first term.

Rule for differentiation of  is:

Next we solve the second term;

If we define , then derivative of  is;

If we have  and then derivative of  is;

Let , then , therefore;

For ;

Since ;

Rule for differentiation of  is:

Rule for differentiation of  is:

Rule for differentiation of  is:

Hence;

Finally;

ii.

We are required to find the coordinates of the only point on the curve at which the gradient of the  curve is equal to 1.

Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to  is:

From (i) we have;

Hence;

Now we have two options.

Two possible values of imply that there are two points on the curve, one at each value of , where  gradient of the curve is 1.

We are given that , therefore, only possible value is;

To find the y-coordinate of the point where t=2, we substitute this value of t in following parametric  equation of the curve;

To find the x-coordinate of the point where t=2, we substitute this value of t in following parametric  equation of the curve;

Hence coordinates of the point on the curve where t=2 are (6,5).