Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2019 | Oct-Nov | (P1-9709/13) | Q#11

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Question

The diagram shows part of the curve , and the lines x = 1 and x = 3. The point A  on the curve has coordinates (2, 3). The normal to the curve at A crosses the line x = 1 at B.

(i)       Show that the normal AB has equation .

(ii)    Find, showing all necessary working, the volume of revolution obtained when the shaded region  is rotated through about the x-axis.

Solution


i.
 

We are required to show that line AB, which is normal to the curve at point A (2, 3), has equation;

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).

Therefore, to find the equation of line AB we already have coordinates of a point on the line i.e., A(2,  3).

Now we need slope of the line AB.

We are given that line AB is normal to the curve at point A(2, 3).

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 the slope of the given curve at point A(2, 3) we can find slope of the normal to  the curve at this point.

Let’s first find slope of the curve.

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

We have equation of the curve given as;

Therefore;

Rule for differentiation of  is:

Rule for differentiation of  is:

Rule for differentiation of  is:

Since line AB is normal to the curve at point A(2, 3), we need slope of the curve at this 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;

Therefore;

Hence, slope of the curve at point A(2, 3) is 

We can see that;

Now, with slope of the line AB, , and coordinates of a point A(2, 3) on the line AB, we can  write equation of the line AB.

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


ii.
 

We are required to find the exact volume of revolution obtained when the shaded region is rotated  through 360o about the x-axis.

Considering the diagram below we can see that;

We have equation of the curve given as;

We have found equation of the line AB as;

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

It is evident from the above diagram that for shaded region under the line AB x-axis varies from x=1  (Point B) to x=2 (Point A) whereas for shaded region under the curve x-axis varies from x=2 (Point  A) to x=3.    

Therefore;

 

Rule for integration of  is:

 

Rule for integration of  is:

Rule for integration of  is:

Rule for integration of  is:

 

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