Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2013 | Oct-Nov | (P2-9709/22) | Q#5

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Question

The parametric equations of a curve are

     i.       Find the exact value of the gradient of the curve at the point P where y = 6.

   ii.       Show that the tangent to the curve at P passes through the point .

Solution

     i.
 

We are need  for the parametric equations, to find gradient of the curve at point P, 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;

For ;

Similarly, we find , when we are given that;

Rule for differentiation of  is:

Rule for differentiation of  is:

Rule for differentiation of  is:

Now we can find .

We need to find gradient at point P where y=6.

We have;

We can substitute y=6 to find the value of  at point P.

Taking anti-logarithm of both sides;

 and are inverse functions. The composite function is an identity function, with  domain the positive real numbers. Therefore;

Therefore;

Now we can find gradient of the curve at point P where y=6   by substituting in  expression for the gradient of the curve ;

   ii.
 

If coordinates of a point satisfies the equation of a curve (or line) then that particular point  lies on that curve (or line).

Therefore, to show whether tangent to the given curve at point P passes through the point (1,0) we  first need to find equation of the tangent to the curve at point P.

Hence, we are required to find the equation of tangent to the curve at the point P.

To find the equation of the line either we need coordinates of the two points on the line (Two-Point  form of Equation of Line) or coordinates of one point on the line and slope of the line (Point-Slope  form of Equation of Line).

We need coordinates of the point P on the curve (and tangent). 

We are already given that at point P .

We only need to find value of x-coordinates of point P. 

We have found in (i) that at this point  .

We are given equations of the curve;

Therefore, we substitute  in;

Hence, coordinates of point P on the curve are

Next we need to find slope of tangent at this point in order to write its equation.

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, if we can find slope of the curve P at point then we can find slope of the  tangent 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;

We have found in (i) that gradient of the curve at point  is;

Hence;

With coordinates of a point on the tangent  and its slope in hand, we can write equation of the tangent.

Point-Slope form of the equation of the line is;

Now we can substitute coordinates of the point (1,0) in this equation of tangent to see whether it  passes through origin or not.

Substitute ;

Hence, equation of tangent is satisfied by coordinates of the point (1,0), therefore, point lies on the  tangent.

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