Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 2 (P29709/02)  Year 2019  FebMar  (P29709/22)  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 xcoordinates 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.










Comments