Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2017 | May-Jun | (P1-9709/11) | Q#10

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Question

The diagram shows part of the curve .

     i.       Find the equation of the normal to the curve at the point where x=1 in the form ,  where m and c are constants.

The shaded region is bounded by the curve, the coordinate axes and the line x=1.

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

Solution

     i.
 

We are required to find the equation of normal to the curve at point where x=1.

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 have neither coordinates of a point on the normal nor the slope of the normal.

First we need to find the coordinates of a point on the curve where x=1 and then slope of the normal  to write its equation.

For coordinates of a point, the point on the curve where x=1 will also lie on the normal to the curve at this point.

Therefore, y-coordinate of this point can be found by substituting x-coordinate of this point x=1 in the given equation of the curve.

Therefore, point with coordinates (1,2) lies on both the curve and the normal to the curve at this same point.

Now we find the slope of the normal to the curve at  point where x=1.

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;

So let’s find slope of the curve at point where x=1.

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

We are given equation of the curve as;

Therefore;

Rule for differentiation of  is:

We can rewrite the above expression for derivative as;

Hence;

This is expression for gradient of the curve. We need gradient of the curve at point where x=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;

Therefore;

Hence, gradient of the curve at point where x=1, is;

Therefore;

Now we can write equation of the normal.

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


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 are given equation of the curve as;

It also can be seen from the diagram that shaded region extends from x=0 to x=1 along x-axis.

Therefore;

Rule for integration of  is:

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