Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2009 | June | Q#3

Question

The curve with equation passes through the point P, where .

a.   Find

i. ii. b.   Verify that the point P is a stationary point of the curve.

c.

i.       Find the value of at the point P.

ii.       Hence, or otherwise, determine whether P is a maximum point or a minimum point.

d.   Find an equation of the tangent to the curve at the point where .

Solution

a.

We are given; Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to is:  Rule for differentiation is of is:  Rule for differentiation is of is: Rule for differentiation is of is:    b.

A stationary point on the curve is the point where gradient of the curve is equal to zero; For the given curve, from (a) we have; We are also given that x-coordinate of point P is Hence;    It shows that point P is a stationary point.

c.

i.

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;   Rule for differentiation is of is:  Rule for differentiation is of is: Rule for differentiation is of is:     To find the value of at point P we substitute x-coordinate of point P in this equation.    ii.

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.

As seen in (c:i), at point P, hence, point P is a maximum point.

d.

We are required to find the equation of tangent to the curve.

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).

Let’s first find coordinates of point on the tangent line.

We already have x-coordinate of a point on the tangent to curve .

To find y-coordinate of the point on the curve, we substitute value of x-coordinate of the point on the curve in the equation of the curve.

We are given equation of the curve as; Substituting ;   Hence a point (1,13) lies on the tangent to curve.

Next we need slope of the tangent to the curve.

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 gradient (slope) of the curve at point where x=1, we can find the slope of  tangent to curve at this point.

For the given curve, from (a) we have; 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 also given that x-coordinate of point is .

Hence;    Hence; Now with coordinates of point on the tangent (1,13) and slope of the tangent , we can write  equation of the tangent.

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