# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2019 | Feb-Mar | (P1-9709/12) | Q#9

**Question**

The diagram shows part of the curve with equation . The shaded region is bounded by the curve, the x-axis and the line x = 3.

** i. **Find, showing all necessary working, the volume obtained when the shaded region is rotated through 360^{O }about the x-axis.

** ii. **P is the point on the curve with x-coordinate 3. Find the y-coordinate of the point where the normal to the curve at P crosses the y-axis.

**Solution**

i.

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

Therefore, for the given case;

We are given that;

Hence, for to ;

Rule for integration of is:

Rule for integration of is:

** ii.
**

We are required to find the y-coordinate of the point where the normal to the curve at P crosses the y-axis.

It is evident that we are required to find the y-coordinate of the y-intercept of the normal to the curve at point P.

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).

Therefore, we need equation of the normal to the curve at point P which has x-coordinate 3 i.e., .

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 coordinates of a point on the normal but we do not have slope of the normal.

First, we need to find the y-coordinate of the point on the curve P (point A lies on both curve and its normal) and then slope of the normal to write its equation.

For y-coordinate of point P, the point on the curve where 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 in the given equation of the curve.

Substitute in equation of the curve;

It is evident from the diagram that point P lies on the curve on positive sides of both x and y-axis. Hence, coordinates of point P, where normal meets the curve, are .

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

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 .

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:

Rule for differentiation of is:

Rule for differentiation of is:

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

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 ;

Therefore;

Now we can write equation of the normal.

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

This is the equation of the normal to the curve at point P.

Now, we substitute in this equation to find the y-coordinate of the point on y-axis where the normal intercepts it.

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