Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2010 | Oct-Nov | (P2-9709/22) | Q#8

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Question

The diagram shows the curve , for . The point  lies on the curve.


i.      
Show that the normal to the curve at Q passes through the point .

   ii.       Find .

  iii.       Hence evaluate

Solution

     i.
 

If a point P(x,y) lies on a line (or the curve) then coordinates of P(x,y) must satisfy equation the line  (or the curve).

To show that the normal to the curve at Q passes through the point we need to show that  coordinates of the point satisfy equation of the normal to the curve at Q.

Therefore, we need to find equation of the normal to the curve at .

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-Sl ope   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 normal to the curve is at point . Therefore, point Q lies both on  the curve and normal.

Therefore, we only need slope of the normal to the curve 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 of the curve at a given point we can find slope of the normal to the  curve at that point.

Now we find slope (gradient) of the curve at point .

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;

If  and  are functions of , and if , then;

If , then;

Let and , then;

Rule for differentiation of  is:

Rule for differentiation of  is;

We are looking for gradient of the curve at point .

Now we can find slope of normal to the curve at point .

With coordinates of a point and slope at hand we can write equation of the  normal.

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

Now that we have equation of normal, we can see whether coordinates of point satisfy this  equation or not.

Hence, the normal to the curve at point passes through the point .


ii.
 

We are required to find;

Rule for differentiation of  is:

We differentiate both the terms one-by-one.

First we differentiate first term.

Rule for differentiation of  is;

Next we differentiate second term.

If  and  are functions of , and if , then;

If , then;

Let and , then;

Rule for differentiation of  is:

Rule for differentiation of  is;

Hence;

  iii.
 

We are required to evaluate;

From (ii) we know that;

Since integral and differentiation are inverse operations;

Therefore;

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