# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2019 | May-Jun | (P2-9709/22) | Q#6

Question

The diagram shows the curve with parametric equations

,

for 0 t 2. At the point P on the curve, the y-coordinate is 1.

i.       Show that the value of t at the point P satisfies the equation .

ii.       Use the iterative formula  with t1 =0.7 to find the value of t at P  correct to 3 significant figures. Give the result of each iteration to 5 significant figures.

iii.       Find the gradient of the curve at P, giving the answer correct to 2 significant   figures.

Solution

i.

We are given that;

We are required to find the equation for variable t at the point P.

We are given that y-coordinate at point P is 1.

Therefore, we can find the value of t by substituting y=1 in following equation of the  curve;

ii.

Iteration method can be used to find the root of the given equation using sequence  defined by;

If the sequence given by the inductive definition , with some initial value  , converges to a limit , then  is the root of the equation .

Therefore, if , then  is a root of .

We use as initial value.

 1 2 3 4 5 6 7 8

It is evident that .

Hence, is a root of .

The root given correct to 3 significant figures is 0.715.

iii.

We are required to find gradient of the curve at the 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;

Therefore, first we need to find expression for gradient of the curve.

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

We are given that;

We are required to find .

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

First we find . We are given that;

Therefore;

We use Product Rule to differentiate .

If  and  are functions of , and if , then;

If , then;

Let  and ;

Rule for differentiation of  is:

Rule for differentiation natural exponential function , or ;

Hence;

Similarly, we find , when we are given that;

Rule for differentiation of  is:

Rule for differentiation of  is:

Rule for differentiation natural exponential function , or ;

Therefore;

Now we can find .

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 that y-coordinate of point P is 1 and we have demonstrated in (iii) that,  at point P of the curve.