Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2015 | Oct-Nov | (P1-9709/13) | Q#9

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Question

A curve passes through the point A(4,6) and is such that  . A point P is moving along  the curve in such a way that the x-coordinate of P is increasing at a constant rate of 3 units per  minute.

     i.       Find the rate at which the y-coordinate of P is increasing when P is at A.

   ii.       Find the equation of the curve.

  iii.       The tangent to the curve at A crosses the x-axis at B and the normal to the curve at A crosses         the x-axis at C. Find the area of triangle ABC.

Solution


i.
 

We are given that x-coordinate of P is increasing at a constant rate of 3 units per minute;

We are required to find the rate at which the y-coordinate of P is increasing when P is at A.

We know that;

Since we are interested in rate of change of y-coordinate of P when it is at point A; we need  at
point A. We are given that;

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;

Therefore  at point A(4,6);

Therefore;


ii.
 

We are required to find the equation of the curve from the given derivative. 

We can find equation of the curve from its derivative through integration;

We are given;

Therefore;

Rule for integration of  is:

Rule for integration of  is:

Rule for integration of  is:

If a point   lies on the curve , we can find out value of . We substitute values of  and    in the equation obtained from integration of the derivative of the curve i.e. .

We are given that curve passes through the point A(4,6). Therefore;

Hence, equation of the curve is;


iii.
 

It is evident that tangent to the curve at point A and normal to the curve at point A make a right  angle. Therefore, triangle ABC is a right triangle.

We are required to find the area of right triangle.

Expression for the area of the triangle is;

It is evident that we need AB and AC as base and height of right triangle ABC with angle BAC being  right angle. To find lengths of AB and AC we need coordinates of point B and C.

This brings us to the requirement of coordinates of both points B and C.

Let us first find coordinates of point B.

We are given that tangent to the curve at A crosses the x-axis at B, This makes point B the x- intercept of tangent to the curve at A.

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, we need equation of tangent to the curve at A to find the coordinates of its x intercept. 

Let’s find the equation of the tangent.

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 know that tangent to the curve is at point A(4,6). Therefore, we already have coordinates of a  point on tangent but we need to find the slope of the tangent.

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 have slope of the curve at point A(4,6) then we can find slope of the tangent 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 slope of the curve at point A(4,6) is;

Therefore;

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

With coordinates of point on the tangent to the curve A(4,6) and slope of the tangent , we  can write equation of the tangent to the curve.

Hence;

Now we can find the coordinates of the x-intercept of the tangent to the curve i.e. point B.

We substitute  in the equation of the tangent to he curve;

Therefore, coordinates of point B(1,0).

Let us first find coordinates of point C.

We are given that normal to the curve at A crosses the x-axis at C, This makes point C the x- intercept of normal to the curve at A.

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, we need equation of normal to the curve at A to find the coordinates of its x intercept. 

Let’s find the 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 know that normal to the curve is at point A(4,6). Therefore, we already have coordinates of a  point on normal but we need to find the slope of the normal.

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 slope of the curve at point A(4,6) then we can find slope of the normal 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 slope of the curve at point A(4,6) is;

Therefore;

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

With coordinates of point on the normal to the curve A(4,6) and slope of the normal  , we can write equation of the normal to the curve.

Hence;

Now we can find the coordinates of the x-intercept of the normal to the curve i.e. point C.

We substitute  in the equation of the tangent to he curve;

Therefore, coordinates of point C(16,0).

Now we can find AB and AC to calculate area of right triangle ABC.

First we find length of AB from A(4,6) and B(1,0).

Expression to find distance between two given points  and is:

Now we find length of AC from A(4,6) and C(16,0).

Expression to find distance between two given points  and is:

Finally, we can calculate the area of the triangle ABC;

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