Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2018 | May-Jun | (P1-9709/12) | Q#11

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Question

The diagram shows part of the curve . The line y = 4 intersects the curve at  the points P and Q.

i. Show that the tangents to the curve at P and Q meet at a point on the line y = x.

ii. Find, showing all necessary working, the volume obtained when the shaded region  is rotated through 360o about the x-axis. Give your answer in terms of .

Solution


i.
 

We are required to show that the tangents to the curve at points P and Q meet at a  point on the line y = x.

Therefore, we need to show that coordinates of point of intersection of the tangents  to the curve at points P and Q satisfy the equation of the line y=x.

Hence, we first need to find the coordinates of point of intersection of the tangents to  the curve at points P and Q.

If two lines (or a line and a curve) intersect each other at a point then that point lies  on both lines i.e. coordinates of that point have same values on both lines (or on the  line and the curve). Therefore, we can equate coordinates of both lines i.e. equate  equations of both the lines (or the line and the curve).

Hence, we need to find equations of the tangents to the curve at points P and Q. 

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 equations of the tangents to the curve at points P and Q we need  to find coordinates of points P and Q.

It is evident from the diagram that points P & Q are the points of intersection of the  given line and the curve. 

If two lines (or a line and a curve) intersect each other at a point then that point lies  on both lines i.e. coordinates of that point have same values on both lines (or on the  line and the curve). Therefore, we can equate coordinates of both lines i.e. equate  equations of both the lines (or the line and the curve). 

Equation of the line is;

Equation of the curve is;

Equating both equations;

Now we have two options;

Two values of x indicate that there are two intersection points.

Corresponding values of y coordinate can be found by substituting values of x in any  of the two equation i.e either equation of the line or equation of the curve.

However, we notice that both points P & Q lie on the line , therefore, their y- coordinates are same i.e. 4. It is evident from the diagram that; 

Hence;

Next we need slopes of the tangents to the curve at points P(2, 4) and Q(6, 4).  

The slope of a curve at a particular point is equal to the slope of the tangent to the  curve at the same point;

Therefore, if we can find slope of the curve at points P(2, 4) and Q(6, 4), then, we  can find slopes of the tangents to the curve at points P(2, 4) and Q(6, 4).  

Hence, we find slope of the curve at points P(2, 4) and Q(6, 4)

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, first we find expression for gradient of the curve at a given point. 

For the given case;

Therefore;

Rule for differentiation is of  is:

Therefore;

Rule for differentiation is of  is:

Gradient of the curve at points P(2, 4) and Q(6, 4) can be found as follows.

P(2, 4)

Q(6, 4)



Hence;

Now we can write equations of the tangents to the curve at points P(2, 4) and Q(6,  4). 

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

P(2, 4)

Q(6, 4)

Now we can find the coordinates of the point of intersection of the tangent to the  curve at points P(2, 4) and Q(6, 4).

If two lines (or a line and a curve) intersect each other at a point then that point lies  on both lines i.e. coordinates of that point have same values on both lines (or on the  line and the curve). Therefore, we can equate coordinates of both lines i.e. equate  equations of both the lines (or the line and the curve).

Equation of the line is;

Equation of the curve is;

Equating both equations;

Single value of x indicates that there is only one intersection point.

With x-coordinate of point of intersection of two lines (or line and the curve) at hand,  we can find the y-coordinate of the point of intersection of two lines (or line and the  curve) by substituting value of x-coordinate of the point of intersection in any of the  two equations.

We choose;

Hence, coordinates of the point of intersection of the tangent to the curve at points  P(2, 4) and Q(6, 4) are (3, 3).

It is evident that these coordinates satisfy the equation of the line y=x.


ii.
 

Consider the diagram below.

It is evident that volume under the line from point A to point B when rotated  around x-axis will form a cylinder with radius AP and height PQ.

Expression to find distance between two given points  and is:

We have from (i);

We can also see that point A is right below point P therefore . Therefore,

Expression for the volume of the cylinder is;

Therefore;

It is evident from the diagram that;

We find out the volume formed by the rotation of area under curve around x-axis.

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 that;

Therefore;

Rule for integration of  is:



Rule for integration of  is:

Rule for integration of  is:

Rule for integration of  is:

Finally;

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