# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2017 | May-Jun | (P2-9709/23) | Q#8

Question

The diagram shows the curve with parametric equations

for . The end-points of the curve are (1, 4) and (3, 3).

i.
Show that .

ii.       Find the coordinates of the minimum point, giving each coordinate correct to 3 significant  figures.

iii.       Find the exact gradient of the normal to the curve at the point for which x = 2.

Solution

i.

We are required to show that  for the parametric equations given below;

If a curve is given parametrically by equations for  and  in terms of a parameter , then;

First we find . We are given that;

Therefore;

Rule for differentiation of  is:

First we differentiate .

Rule for differentiation of  is:

Next we differentiate  utilizing Chain Rule.

If we define , then derivative of  is;

If we have  and then derivative of  is;

Let ; then

Rule for differentiation of  is:

Since ; then

Rule for differentiation of  is;

Finally, we differentiate  utilizing Chain Rule.

If we define , then derivative of is;

If we have  and then derivative of  is;

Let ; then

Rule for differentiation of  is:

Since ; then

Rule for differentiation of  is;

Hence;

Similarly, we find , when we are given that;

Rule for differentiation of  is:

First we differentiate .

Rule for differentiation of  is:

Next we differentiate .

We utilize Chain Rule to differentiate .

If we define , then derivative of  is;

If we have  and then derivative of  is;

Let , then;

Rule for differentiation of  is;

Since ;

Rule for differentiation of  is:

Hence;

Now we can find .

Since ;

ii.

We are required to find the coordinates of the minimum point of the curve;

A stationary point on the curve is the point where gradient of the curve is equal to zero;

Since point M is minimum point, therefore, it is stationary point of the curve and, hence, gradient of  the curve at point M must ZERO.

We can find expression for gradient of the curve at point M and equate it with ZERO to find the x- coordinate of point M.

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

As demonstrated in (i);

Now we need expression for gradient of the curve at point M.

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;

Since point M is a minimum point, the gradient of the curve at this point must be equal to ZERO.

Using calculator;

To find the coordinates of point M on the curve we substitute  in parametric equations of the curve.

Therefore, coordinates of point M.

iii.

We are required to find exact gradient of the normal to the curve at the point for which x = 2.

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 gradient of the curve at point where x=2, then we can find the gradient of  normal to 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;

As demonstrated in (i);

Therefore;

Hence, we need to substitute value of on the curve where x=2 in above expression.

We can find this from given equation of the curve;

Therefore, gradient of the curve at point where x=2  is;

Hence;

Now we can find gradient of the normal to the curve at this point;