Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 1 (P19709/01)  Year 2020  FebMar  (P19709/12)  Q#10
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Question
The gradient of a curve at the point (x, y) is given by . The curve has a stationary point at (a, 14), where a is a positive constant.
a)Find the value of .
b)Determine the nature of the stationary point.
c)Find the equation of the curve.
Solution
a)
We are given that gradient of the curve at the point (x, y) is;
We are also given that point at (a, 14) is a stationary point.
A stationary point on the curve is the point where gradient of the curve is equal to zero;
Therefore, at point (a, 14);
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 xcoordinates of that point in the expression for gradient of the curve;
Hence, gradient at point (a, 14);
Coordinates of stationary point on the curve can be found from the derivative of equation of the curve by equating it with ZERO. This results in value of xcoordinate of the stationary point on the curve.
Hence;
Now we have two options.




We are given that is a positive constant.
Hence;
b)
Once we have the xcoordinate of the stationary point of a curve, we can determine its nature, whether minimum or maximum, by finding 2^{nd }derivative of the curve.
Second derivative is the derivative of the derivative. If we have derivative of the curve as , then expression for the second derivative of the curve is;
We are given that gradient of the curve at the point (x, y) is;
Therefore, for second derivative;
Rule for differentiation of is:
Rule for differentiation of is:
Rule for differentiation f is:
Once we have the coordinates of the stationary point of a curve, we can determine its nature, whether minimum or maximum, by finding 2^{nd }derivative of the curve.
We substitute of the stationary point in the expression of 2^{nd} derivative of the curve and evaluate it;
If or then stationary point (or its value) is minimum.
If or then stationary point (or its value) is maximum.
We are given that point (a, 14) is a stationary point on the curve and we have found in (a) that , therefore, coordinates of given stationary point on the curve are (6, 14).
To find the nature of stationary point we substitute xcoordinate of the point in expression of second derivative of the equation of the curve;
Since , the stationary point (6, 14) is a maximum.
c)
We are required to find the equation of the curve.
We can find equation of the curve from its derivative through integration;
We are given that gradient of the curve at the point (x, y) is;
Therefore;
Rule for integration of is:
Rule for integration of is:
Rule for integration of is:
If a point lies on the curve , we can find out value of . We substitute values of and in the equation obtained from integration of the derivative of the curve i.e. .
We are given that point (a, 14) lies on the curve and we have found in (a) that , therefore, coordinates of given point on the curve are (6, 14).
Hence;
Hence;
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