Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2016 | Oct-Nov | (P1-9709/11) | Q#11

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Question

The point P(3,5) lies on the curve  .

     i.       Find the x-coordinate of the point where the normal to the curve at P intersects the x-axis.

   ii.       Find the x-coordinate of each of the stationary points on the curve and determine the nature of  each stationary point, justifying your answers.

Solution

     i.
 

The point where the normal to the curve at P intersects the x-axis is actually x-intercept of the  normal.

The point  at which curve (or line) intercepts x-axis, the value of . So we can find the  value of  coordinate by substituting  in the equation of the curve (or line).

Therefore, to find the x-coordinate of x-intercept of  the normal to the curve at P, we need equation  of the normal.

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 are given that desired normal is at point P(3,5) on the curve. This gives coordinates of a point  on the normal to the curve.

Next we need slope of the normal to write its equation.

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 can find slope of curve at point P(3,5), we can find the slope of normal to the curve  at this point.

Let’s find slope of curve at point P(3,5).

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 equation of the curve as;

Therefore;

Rule for differentiation of  is:

Rule for differentiation of  is:

Therefore, slope of the curve at point P(3,5);

Therefore slope of the curve at point P(3,5) is . Hence, we can find slope of normal to the  curve at this point.

With coordinates of a point P(3,5) on the normal and slope of the normal in hand , we can  write equation of the normal.

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

For the given case;

Finally, to find the coordinates of x-intercept of this normal to the curve, we substitute  in its  equation;

   ii.
 

We are required to find the x-coordinate of stationary points on the curve and determine the nature  of each stationary point. 

A stationary point  on the curve  is the point where gradient of the curve is equal to zero; 

Coordinates of stationary point on the curve  can be found from the derivative of equation of the  curve by equating it with ZERO. This results in value of x-coordinate of the stationary point  on the curve.

From (i) we have found that expression for the gradient of the curve is;

To find the x-coordinate(s) of the stationary point;

Now we have two options.

Two possible values of  imply that there are two stationary points on the curve one at each value of  .

Once we have the coordinates of the stationary point  of a curve, we can determine its  nature, whether minimum or maximum, by finding 2nd derivative of the curve.

We substitute  of the stationary point in the expression of 2nd derivative of the curve and evaluate it;

If  or  then stationary point (or its value) is minimum.

If  or  then stationary point (or its value) is maximum.

Second derivative is the derivative of the derivative. If we have derivative of the curve   as  , then expression for the second derivative of the curve  is;

From (i) we have found that expression for the gradient of the curve is;

Therefore;

Rule for differentiation of  is:

Rule for differentiation of  is:

We have found above that x-coordinates of two possible stationary points on the curve are;

Therefore, substituting these values in the above obtained expression of ;

Maximum

Minimum

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