Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 2 (P29709/02)  Year 2002  OctNov  (P29709/02)  Q#7
Hits: 223
Question
The equation of a curve is
i. Show that
ii. Find the coordinates of the points on the curve where the tangent is parallel to the xaxis.
Solution
i.
We are given;
We are required to find .
To find from an implicit equation, differentiate each term with respect to , using the chain rule to differentiate any function as .
We differentiate each term of the equation, on by one, with respect to x applying following rules.
Rule for differentiation of is:
If and are functions of , and if , then;
Rule for differentiation of is:
Now we can combines derivatives of all terms of the equation as;
ii.
We are given that tangent to the curve at a point is parallel to xaxis.
If two lines are parallel to each other, then their slopes and are equal;
Since slope of xaxis is ZERO, therefore, slope of tangent to the curve is also ZERO.
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, slope of the curve at the point where tangent meets the curve is equal to the slope of the tangent. Hence;
Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to is:
As demonstrated in in (i), gradient of the curve is given by;
As found above;
We can substitute this in given equation of the curve;
With xcoordinate of a point at hand, we can find the ycoordinate of the point by substituting value of xcoordinate of the point any of the two equations.
We substitute , one by one, in equation .








Therefore, at two points with coordinates and the slope of tangent to curve will be ZERO ie tangent will be parallel to xaxis.
Comments