# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2010 | Oct-Nov | (P2-9709/21) | Q#4

Question

The parametric equations of a curve are  for t > 2.

i.       Show that 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 0.

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:    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 0.

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.     Three possible values of imply that there are three 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=3, we substitute this value of t in following parametric  equation of the curve;   To find the x-coordinate of the point where t=3, we substitute this value of t in following parametric  equation of the curve;     Hence coordinates of the point on the curve where t=3 are (1,6).