Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 2 (P29709/02)  Year 2017  MayJun  (P29709/23)  Q#8
Question
The diagram shows the curve with parametric equations
for . The endpoints 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;
Comments