Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 2 (P29709/02)  Year 2003  MayJun  (P29709/02)  Q#7
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Question
The parametric equations of a curve are
i. Show that
ii. Find the equation of the tangent to the curve at the point where .
iii. For the part of the curve where , find the coordinates of the points where the tangent is parallel to the xaxis.
Solution
i.
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;
Rule for differentiation of is:
Rule for differentiation of is:
If we define , then derivative 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:
Rule for differentiation of is:
If we define , then derivative of is;
Rule for differentiation of is;
Now we can find .
Now we can apply double angle formulae to simplify this expression.
Therefore;
We have the trigonometric identity;
So ;
We know that provided that , therefore;
ii.
We are required to find the equation of tangent 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 tangent).















Hence, coordinates of the point on the curve (and tangent) are .
Next we need to find slope of tangent at in order to write its equation.
The slope of a curve at a particular point is equal to the slope of the tangent to the curve at the same point;
Therefore, if we can find slope of the curve C at point then we can find slope of the tangent 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;
except where or undefined
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;
iii.
We are required to find the coordinates of the points on the curve where the tangent is parallel to the xaxis, for the part of the curve where .
If two lines are parallel to each other, then their slopes and are equal;
We know that xaxis is horizontal and has ZERO slope, therefore;
The slope of a curve at a particular point is equal to the slope of the tangent to the curve at the same point;
Therefore, gradient of the curve at the point where tangent is parallel to xaxis must be ZERO as well.
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;
Taking reciprocal of both sides;
We know that at;
Properties of 

Domain 

Range 

Periodicity 



Odd/Even 

Translation/ Symmetry 


We utilize the periodicity/symmetry property of to find other solutions (roots) of :
Therefore;
For;








Only following solutions (roots) are within the given interval ;


Hence, the curve and tangent will have gradient equal to slope of xaxis when and .
We can find coordinates of the points on the curve when and by substituting these values in parametric equations, one by one.
When 













Hence, coordinates of point are 
When 













Hence, coordinates of point are 
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