Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2016 | Oct-Nov | (P1-9709/12) | Q#7

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Question

The equation of a curve is .

    i.       Obtain an expression for

   ii.       Explain why the curve has no stationary points. 

At the point P on the curve, x = 2.

  iii.       Show that the normal to the curve at P passes through the origin.

  iv.       A point moves along the curve in such a way that its x-coordinate is decreasing at a constant rate of 0.06 units per second. Find the rate of change of the y-coordinate as the point  passes through P.

Solution


i.
 

We are given that;

Therefore;

Rule for differentiation of  is:

Rule for differentiation of  is:

Rule for differentiation of  is:


ii.
 

A stationary point  on the curve  is the point where gradient of the curve is equal to zero;

Therefore, if curve has any stationary point(s) then;

From (i) we have;

Hence;

This is not possible, therefore, . Hence, the curve has no stationary point(s).


iii.
 

W need to find the equation of the normal to the curve at point P and see whether it satisfies coordinates of origin or not.

Let’s find equation of normal to the curve at point P.

To find the equation of the line either we need coordinates of the two points on the line (Two-Point  form of Equation of Line) or coordinates of one point on the line and slope of the line (Point-Slope  form of Equation of Line).

We need to find the coordinates of a point on the normal to the curve in order to write its equation.

We are given that desired line is normal to the curve at point P. Therefore, point P lies both on the  curve and normal.

We are already given x-coordinate of the point P as .

If we are given one coordinate of a point on the curve (or line) then we can find the other coordinate  of that point on the curve (or line) by substituting known coordinate in the equation of  the curve (or line).

We are given that x-coordinate of the point P as . Therefore, substituting it in the equation of  the curve;

Hence coordinates of .

After finding coordinates of a point on the normal to the given curve, next we need slope of the  normal to the curve in order 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 have slope (gradient) of curve at point P, we can find the slope of normal to the  curve at point P.

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);

Therefore, gradient of the curve at point P can be found by substituting x-coordinate of the point P  in equation of .

Therefore, gradient of the curve at point P;

Hence;

Now with coordinates of point  on the normal and slope of the normal  in hand we can  write equation of the normal to the curve.

Point-Slope form of the equation of the line is;

Therefore;

If this normal to the curve at point P passes through origin O(0,0), then its equation must satisfy the  coordinates of origin. 

 

Substituting  and  in equation of the  normal;

Since equation of normal is satisfied with the coordinates of the origin, the normal to the curve  passes through the origin.


iv.
 

We are given that;

We are required to find the rate of change of y-coordinates as point passes through the point P;

We know that;

We have found in (iii) that;

Therefore;

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