# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2012 | May-Jun | (P2-9709/22) | Q#5

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Question

The parametric equations of a curve are

i.       Find an expression for in terms of t.

i.       Find the equation of the normal to the curve at the point for which t = 0. Give your answer in  the form ax + by + c = 0, where a, b and c are integers.

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 natural exponential function , or ;

Rule for differentiation of  is:

Similarly, we find , when we are given that;

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 equation of normal to the curve at the point where .

To find the equation of the line either we need coordinates of the two points on the line (Two-Point  form of Equation of Line) or coordinates of one point on the line and slope of the line (Point-Slope  form of Equation of Line).

We need coordinates of the point on the curve (and tangent).

We are given equation of the curve parametrically;

We substitute desired value of parameter in both equations of  and to find the  coordinates of the point on the curve (and normal).

Hence, coordinates of the point on the curve (and normal) are .

Next we need to find slope of tangent at in order to write its equation.

If a line is normal to the curve , then product of their slopes  and at that point (where line  is normal to the curve) is;

Therefore, if we can find slope of the curve C at point then we can find slope of the normal to  the curve at this point.

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;

Hence;

With coordinates of a point on the tangent  and its slope in hand, we can write equation of the tangent.

Point-Slope form of the equation of the line is;