Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 1 (P19709/01)  Year 2016  OctNov  (P19709/11)  Q#11
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Question
The point P(3,5) lies on the curve .
i. Find the xcoordinate of the point where the normal to the curve at P intersects the xaxis.
ii. Find the xcoordinate of each of the stationary points on the curve and determine the nature of each stationary point, justifying your answers.
Solution
i.
The point where the normal to the curve at P intersects the xaxis is actually xintercept of the normal.
The point at which curve (or line) intercepts xaxis, the value of . So we can find the value of coordinate by substituting in the equation of the curve (or line).
Therefore, to find the xcoordinate of xintercept of the normal to the curve at P, we need equation of the normal.
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 are given that desired normal is at point P(3,5) on the curve. This gives coordinates of a point on the normal to the curve.
Next we need slope of the normal 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 curve at point P(3,5), we can find the slope of normal to the curve at this point.
Let’s find slope of curve at point P(3,5).
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 are given equation of the curve as;
Therefore;
Rule for differentiation of is:
Rule for differentiation of is:
Therefore, slope of the curve at point P(3,5);
Therefore slope of the curve at point P(3,5) is . Hence, we can find slope of normal to the curve at this point.
With coordinates of a point P(3,5) on the normal and slope of the normal in hand , we can write equation of the normal.
PointSlope form of the equation of the line is;
For the given case;
Finally, to find the coordinates of xintercept of this normal to the curve, we substitute in its equation;
ii.
We are required to find the xcoordinate of stationary points on the curve and determine the nature of each stationary point.
A stationary point on the curve is the point where gradient of the curve is equal to zero;
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.
From (i) we have found that expression for the gradient of the curve is;
To find the xcoordinate(s) of the stationary point;
Now we have two options.




















Two possible values of imply that there are two stationary points on the curve one at each value of .
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.
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;
From (i) we have found that expression for the gradient of the curve is;
Therefore;
Rule for differentiation of is:
Rule for differentiation of is:
We have found above that xcoordinates of two possible stationary points on the curve are;


Therefore, substituting these values in the above obtained expression of ; 























Maximum 
Minimum 
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