Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2015 | Oct-Nov | (P1-9709/12) | Q#10

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Question

The diagram shows part of the curve . The point P(2,1) lies on the curve and the  normal to the curve at P intersects the x-axis at A and the y-axis at B.

     i.       Show that B is the mid-point of AP.

The shaded region is bounded by the curve, the y-axis and the line y = 1.

   ii.       Find, showing all necessary working, the exact volume obtained when the shaded region is
rotated through 360o about the y-axis.

Solution

     i.
 

We are required to show that B is the mid-point of AP which is normal to the curve at point P(2,1).

We need to find and compare the coordinates of point B and mid-point of AP.

First we find coordinates of point B.

It is evident that point B is the y-intercept of the line AP. Therefore to find the coordinates of y- intercept of AP, we need to find equation of the line AP.

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).

It is evident that we have coordinates of a point on AP as P(2,1). Therefore, we only need to find  slope of AP 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 point P(2,1) we can find slope of AP.

Let’s find slope of the curve at point P(2,1).

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 equation of the curve given as;

Therefore;

Rule for differentiation of  is:

Rule for differentiation of  is:

Rule for differentiation of  is:

Rule for differentiation of  is:

Now we can find slope of the curve at point P(2,1).

Now we can find slope of AP.

With coordinates of a point P(2,1) and slope  in hand, we can write equation of AP.

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

Now we can find the coordinates of point B which is y-intercept of AP. 

Substituting  in the  equation of AP;

Hence coordinates of point B are .

Now we need to find coordinates of mid-point of AP.

To find the mid-point of a line we must have the coordinates of the end-points of the line. 

Expressions for coordinates of mid-point of a line joining points  and;

x-coordinate of mid-point  of the line

y-coordinate of mid-point  of the line

Therefore, we need coordinates of both point A and P to find the coordinates of mid-point of AP.

We already have coordinates of point P(2,1).

Let’s find coordinates of point A.

It is evident from the diagram that point A is x-intercept of line AP.

The point  at which curve (or line) intercepts x-axis, the value of . So we can find the  value of  coordinate by substituting  in the equation of the curve (or line).

We have found above the equation of AP as;

Substituting  in the equation of AP;

Hence, coordinates of point A are .

Now, we have A(-2,0) and P(2,1) and can find coordinates of mid-point of AP;

x-coordinate of mid-point  of the line

y-coordinate of mid-point  of the line

Hence coordinates of mid-point of AP are .

We have already shown that coordinates of point B are .

Therefore, point B is the mid-point of AP.


ii.
 

Expression for the volume of the solid formed when the shaded region under the curve  is rotated
completely about the y-axis is;

We are given equation of the curve as;

It is  evident that to find the desired volume, first we need to write the equation in terms of y rather  than x.

Now we can write the volume equation as;

 

We are given that the desired shaded region extends from line  and y-intercept of the curve. Therefore, we need the coordinates of  y-intercept of the curve.

The point  at which curve (or line) intercepts y-axis, the value of . So we can find the value of  coordinate by substituting  in the equation of the curve (or line).

It is evident from the diagram that it is

Hence, the limits of the integration are applied as;

Rule for integration of  is:

Rule for integration of  is:

Rule for integration of  is:

Rule for integration of  is:

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