Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 2 (P29709/02)  Year 2019  MayJun  (P29709/22)  Q#6
Question
The diagram shows the curve with parametric equations
,
for 0 ≤ t ≤ 2. At the point P on the curve, the ycoordinate is 1.
i. Show that the value of t at the point P satisfies the equation .
ii. Use the iterative formula with t_{1} =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 ycoordinate 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 xcoordinates 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 xcoordinates of that point in the expression for gradient of the curve;
We are given that ycoordinate of point P is 1 and we have demonstrated in (iii) that, at point P of the curve.
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