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**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 = x^{2} + 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 = x^{2}
+ 3 = 0^{2} +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 = 3x^{2} + 4 (ii)
y = 4 – 3x^{2} (iii) y = 4x^{2} - 8x +1 (iv) y = x^{3} + 2x^{2} + 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 = x^{3}
shown below.

When y = x^{3}, _{}= 3x^{2} and equating 3x^{2} = 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 = x^{3} + 3x^{2} + 4

Answer: _{}= 3x^{2} + 6x Equating
_{} = 0 x(3x + 6) = 0 gives x = 0 or -2.

Substituting
in y = x^{3} + 3x^{2} + 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 = x^{3} + 3x^{2} + 4 gives y = 6

Substitute
x = -0.9 and x = -1.1 into the gradient function _{}= 3x^{2} + 6x

_{}= -2.97 at x = -0.9 and -2.97 at x = -1.1

The function y = x^{3} + 3x^{2}
+ 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 = 2x^{2} – 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 = 2x^{2} – 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 + x^{2} – 2x^{3}

Answer 2: _{}= 4 + 2x -6x^{2} 4
+ 2x – 6x^{2} = 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 -6x^{2}

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 = 4x^{3}
+ x^{2} – 2x + 5 has a maximum value when x = _{}, a minimum value when x = _{}, and a point of inflexion when x = _{}.

Answer 1. y = 4x^{3}
+ x^{2} – 2x + 5

y’ = 12x^{2}
+ 2x – 2

Equating
y’ to 0 gives 12x^{2} + 2x – 2 = 0 2(6x^{2}
+ 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 = x^{2} + 4x + 3 (ii) y = 24x – 2x^{3} (iii) y = 16 – 9x + 3x^{2} + x^{3}

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)