Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2019 | Feb-Mar | (P2-9709/22) | Q#7

Hits: 47

Question

The parametric equations of a curve are

for . At the point P on the curve, the gradient of the curve is 2.

     i.       Show that the value of the parameter at P satisfies the equation

.

   ii.       By first expressing in the form , where R > 0 and   , find the coordinates of P. Give each coordinate correct to 3 significant 

figures.

Solution

     i.
 

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;

We are given the parametric equations as 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  using 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,  , therefore;

Rule for differentiation of  is:

Therefore;

Similarly, we find , when we are given that;

Rule for differentiation of  is:

Rule for differentiation of  is:

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,  , then;

Rule for differentiation of  is:

Now we can find .

We are given that gradient of the curve at point P is 2, hence;

   ii.
 

We are required to express  in the form .

We are given the expression;

We are required to write it in the form;

If  and are positive, then;

can be written in the form

can be written in the form

where,

 and , , with

Considering the given equation, we have following case at hand;

can be written in the for

Comparing it with given equation Therefore

Therefore;

Finally, we can find , utilizing the equation;

Using calculator we can find that;

Therefore;

We are required to find the coordinates of point P.

We can find coordinates of point P by substituting value of  at
point P in parametric equations of the curve.

As demonstrated in (i), value of at point P is given by;

We have shown above that;

Hence;

Therefore, we need to solve;

Let , then:

Using calculator we can find that;

We utilize the symmetry property of   to find another solution (root) of :

Properties of

Domain

Range

Odd/Even

Periodicity

Translation/

Symmetry

Hence;

Therefore;

We have found two solutions of equation .

Since given interval is , for  interval for is found as follows.

Multiplying entire inequality with 2;

Subtracting  from inequality;

 

To find all other solutions of the equation  we utilize the periodicity property of :

For the given case,

For

For

Now;

For 

We have found following solution of  for the interval;

Since , therefore;

Now we can find coordinates of point P by substituting in parametric  equations of the curve.

Please follow and like us:

Comments