Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 2 (P29709/02)  Year 2012  MayJun  (P29709/22)  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 (TwoPoint form of Equation of Line) or coordinates of one point on the line and slope of the line (PointSlope 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.
PointSlope form of the equation of the line is;
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