Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2002 | Oct-Nov | (P2-9709/02) | Q#7
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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 x-axis.
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 x-axis.
If two lines are parallel to each other, then their slopes and
are equal;
Since slope of x-axis 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 x-coordinate of a point at hand, we can find the y-coordinate of the point by substituting value of x-coordinate of the point any of the two equations.
We substitute , one by one, in equation
.
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Therefore, at two points with coordinates and
the slope of tangent to curve will be ZERO ie tangent will be parallel to x-axis.
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