Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2018 | May-Jun | (P2-9709/21) | Q#5

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Question

The parametric equations of a curve are

for .

    i.       Find the gradient of the curve at the point for which  radian.

   ii.       Find the value of at the point on the curve where the tangent is parallel to the y-axis.

Solution

     i.
 

We are required to find gradient of the curve at point where radian.

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 of the curve;

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  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 ;

Rule for differentiation of  is:

Next we differentiate .

Rule for differentiation of  is;

Therefore;

Similarly, we find , when we are given that;

Rule for differentiation of  is;

Now we can find .

Hence, gradient of the curve at point where ;

 

   ii.
 

We are required to find the value of at the point on the curve where the tangent is parallel to the y-axis.

If two lines are parallel to each other, then their slopes  and  are equal;

Therefore, gradient of the tangent to the curve at the desired point is equal to gradient of the y-axis.

The slope of a curve at a particular point is equal to the slope of the tangent to the curve at the  same point;

Therefore, we are required to find the value of at the point on the curve where the gradient of  the curve is equal to gradient of the y-axis.

We know that gradient of y-axis is undefined.

Therefore, gradient of the curve at desired point must also be undefined. 

From (i), we have expression for gradient of the curve;

Now, at the desired point the gradient of the curve must be undefined, and it is only possible when  denominator of the above expression is equal to zero.

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