# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2016 | Oct-Nov | (P1-9709/13) | Q#11

Question

A curve has equation , where k is a non-zero constant.

i.       Find the x-coordinates of the stationary points in terms of k, and determine the nature of each  stationary point, justifying your answers.

ii.

The diagram shows part of the curve for the case when k = 1. Showing all necessary working, find  the volume obtained when the region between the curve, the x-axis, the y-axis and the line x=2,  shown shaded in the diagram, is rotated through 360about the x-axis.

Solution

i.

We are given that equation of the curve is;

First we are required to find the x-coordinates of the stationary points on the curve.

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 x-coordinate of the stationary point  on the curve.

Therefore, first we need derivative of the equation of the curve.

Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve  with respect to  is:

Therefore;

Rule for differentiation of  is:

Rule for differentiation of  is:

To find the x-coordinate 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  .

Next we are required to determine the nature of these stationary points.

Once we have the x-coordinate of the stationary point  of a curve, we can determine its  nature, whether minimum or maximum, by finding 2nd 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 already have found that;

Therefore;

Rule for differentiation of  is:

Rule for differentiation of  is:

Rule for differentiation of  is:

We substitute  of the stationary point in the expression of 2nd 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 k is a non-zero constant. Therefore, whether k is positive or negative, .

Therefore;

 Minimum Maximum

ii.

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

We have;

We are given that k=1;

Therefore;

Rule for integration of  is:

Rule for integration of  is:

Rule for integration of  is: