For more information visit www.caresa.com.au

 

Geometrical Applications of Differentiation (10.1 – 10.8)

 

10.1          Significance of the sign of the derivative.

10.2          Stationary points on curves.

10.3          The second derivative. The notation f ”(x), , y”.

10.4          Geometrical significance of the second derivative.

10.5          The sketching of simple curves.

10.6          Problems on maxima and minima.

10.7          Tangents and normals to curves.

10.8          The primitive function and its geometrical interpretation.”  (syllabus)

 

For a more detailed description of the requirements for this topic, see the mathematics syllabus on the Board of Studies website click here

 

Scroll down to page 2.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

TURNING POINTS

 

The derived function, , is also known as the gradient function because it gives the gradient.

When the gradient is zero it means that a tangent to the curve at that point has a gradient of zero. It also indicates that the curve changes direction at that point and the point is a maximum or minimum value. Finding the points where  is equal to zero will determine the turning points on a graph.

To find the turning points on a graph:

(i) Find .

(ii) Equate to zero and solve. This gives the x co-ordinates of the turning points.

(iii) Substitute the x-values from (ii) into the original equation to find the y co-ordinates.
 
 

 

 

 

 

 

 

 

 

 

 

 

 


Example: Find the turning point on the curve y = x2 + 3

 

Answer:            (i) Find .                 = 2x

                        (ii) Equate to zero and solve.           2x = 0              x = 0

(iii) Substitute the x-values from (ii) into the original equation to find the y co-ordinates.

y = x2 + 3 = 02 +3 = 3

Turning point is (0, 3)

 

 

Maximum or Minimum?

 

To determine whether the point is a maximum or minimum, substitute values a little below and a little above the x value of the turning point into the gradient function. It helps to roughly sketch the gradients at this point to picture what the curve looks like.

In the above example the gradient is 0 when x is 0.

Let x = -0.1      gradient = 2x = 2(-0.1) = -0.2 Note that the gradient is negative \.

Let x = +0.1     gradient = 2x = 2(0.1)  = 0.2                Note that the gradient is positive /.

Drawing the gradient before x=0, at x = 0 and after x = 0 gives \_/ which can be seen as a minimum value.

 

Exercise: Find the turning points for the following functions. For each turning point, state whether it is a maximum or minimum value.

(i) y = 3x2 + 4               (ii) y = 4 – 3x2     (iii) y = 4x2 - 8x +1    (iv) y = x3 + 2x2 + x + 5

 

Answers: (i) (0, 4) min.  (ii) (0, 4) max.   (iii) (1, -3) min. 

 

            (iv) (-1, 5) max (, ) min.

 

 

 

POINTS OF INFLEXION

 

Examine the graph of y = x3 shown below.

                       

When y = x3, = 3x2 and equating 3x2 = 0 gives x = 0

This would suggest that the graph would have a turning point at (0, 0)

Looking at the point (0, 0) reveals no turning point but the concavity of the graph changes, i.e. it changes from bending upwards to bending downwards.

Such a point is called a point of inflexion.

Points of inflexion can be found by equating the second derivative to zero but further checking has to be done.

 

 

Example: Find any turning points or points of inflexion for the function y = x3 + 3x2 + 4

 

Answer: = 3x2 + 6x            Equating  = 0          x(3x + 6) = 0 gives x = 0 or -2.

 

            Substituting in y = x3 + 3x2 + 4 when x = 0, y = 4, when x = -2, y = 8

            Find the second derivative: = 6x + 6

            Equating 6x + 6 to zero gives x = -1.

            Substituting x = -1 into y = x3 + 3x2 + 4 gives y = 6

            Substitute x = -0.9 and x = -1.1 into the gradient function = 3x2 + 6x

            = -2.97 at x = -0.9 and -2.97 at x = -1.1

 

The function y = x3 + 3x2 + 4 has turning points at (0, 4) and (-2, 8) and a point of inflexion at (-1, 6)

 

 

THE SECOND DERIVATIVE

 

While the first derivative can be used to determine the turning points on a graph, the second derivative can be used to determine whether these turning points are maximum or minimum values.

If the second derivative has a positive value at the turning point then the graph is concave up i.e. it has a minimum value.

If the second derivative has a negative value at the turning point then the graph is concave down i.e. it has a maximum value.

 

Example 1:       Find the turning point on the graph y = 2x2 – 8x + 5 and determine whether it is a maximum or minimum value.

Answer:            = 4x – 8     putting 4x – 8 = 0 gives x = 2

                        Substituting x = 2 in y = 2x2 – 8x + 5 gives y = 8 - 16 + 5 = -3

                        Turning point is (2, -3)

                        The second derivative = 4 which is positive.

Hence the graph is concave up at the point (2, -3) i.e. it has a minimum value.

 

Example 2:       Find any turning points or points of inflexion for the graph

y = 8 + 4x + x2 – 2x3

Answer 2:         = 4 + 2x -6x2         4 + 2x – 6x2 = 0           (4+ 6x)(1-x) = 0

                        x = 1 or

                        When x = 1, y = 11.     When x = , y =

                       

                        = 2-12x and when = 0          x = , y =

                       

                        Substitute x =  and x = into = 4 + 2x -6x2

           

                        x = , = ,       x = , =

                        The gradient has the same sign on both sides of (,)

                        (,) is a point of inflexion.

                        Substituting the x values of the turning points in 2 – 12x (i.e. ) gives:

2-12 = -10 when x = 1 Since -10 is negative then the curve is concave down

i.e. there is a maximum value at (1, 11)

2 -  -8 = 10 when x =   Since 10 is positive then the curve is concave up.

i.e. there is a minimum value at (,)

 

 

TURNING POINTS & INFLEXION

 

Exercise 1.       Show that the equation y = 4x3 + x2 – 2x + 5 has a maximum value when     x = , a minimum value when x = , and a point of inflexion when x = .

 

Answer 1.        y = 4x3 + x2 – 2x + 5   

y’ = 12x2 + 2x – 2

                        Equating y’ to 0 gives 12x2 + 2x – 2 = 0           2(6x2 + x – 1) = 0

                        2(3x - 1) (2x + 1) = 0   x =  or

                        y’’ = 24x + 2

                        When x = , y’’ =  + 2 = 10 which is positive.

                        the curve is concave up so there is a minimum when x = .

                        When x = , y’’ =  + 2 = -10 which is negative.

                        the curve is concave down so there is a maximum when x =

                        Equating y’’ to 0 gives 24x + 2 = 0       x =

            Check the gradient slightly below and slightly above x = , say x = and x = 0.

                        When x =  gradient  = 12+ 2() – 2 =  -  - 2 = -2

                        When x = 0, gradient = 0 + 0 – 2 = -2

            Since the gradient has the same sign on both sides of  x =  then there is a point of inflexion when x = .

Exercise 2.       Find the y-intercept, any turning points and points of inflexion for the following functions and use this             information to sketch the functions.

 

            (i) y = x2 + 4x + 3                     (ii) y = 24x – 2x3          (iii) y = 16 – 9x + 3x2 + x3       

 

Answers 2.       (i) y-intercept = 3, minimum at (-2, -1)

                        (ii) y-intercept = 0, min. at (-2, -32), max. at (2, 32) , inflexion at (0, 0)

                        (iii) y-intercept = 16, min. at (1, 11), max. at (-3, 43), inflexion at (-1, 27)