Maths IB standard

1
Maths IB standard

• Calculus
• Yay!

2
Starter activity

• How do we know what the gradient of a line is?
• Y mx c
• If we are given two points how do we find the
  gradient?

3
Gradient/ Slope

• The gradient or slope of a line describes the
  steepness. To find the gradient we use the phrase
  rise
• run
• Remember you rise up in the morning before you go
  for a run
• Click here for a summary of everything

4
Gradient/ Slope

• To find the gradient/slope between two points
  you have to find the rise (difference between the
  ys) over the run (difference between the xs)
• Look at this applet

• MathsNet A Level Pure C1 Module

Click here for examples of finding the gradient
between two points
5
Formula for the gradient/slope

• Can you think of a formula to find the gradient
  between two points (x1,y1) and (x2,y2)?
• M y2-y1
• x2-x1
• This is the differences of the ys over the
  differences of the xs

6
Example 1

• Find the gradient between the following points
• 1) (4,2) (6,3)
• 2) (-1,3) (5,4)
• 3) (-4,5) (1,2)
• 4) (2,-3), (5,6)
• 5) (-3,4) (7,-6)

7
More practice

• Find the gradient between these points
• 1) (-2,-4) (10,2)
• 2) (-12,3) ( -2,8)
• 3) (-2,7) (4,5)
• 4) (2,3) (5,7)
• 5) (3b,-2b) (7b,2b)

8
Example 2

• Work out the gradients of the following
  equations
• 1) y -2x5
• 2) y -x7
• 3) x2y -4 0
• 4) -3x6y7 0
• 5) 9x6y2 0

9
Looking at the gradient on a curve

• Let's think the slope of the secant line that
  goes through two points
• (a,f(a)) and (ah,f(ah)) on y f(x).
• The slope is (f(ah)-f(a))/h.
• When we make h smaller, what does
  (f(ah)-f(a))/h approach ? We call this taking a
  limit when we investigate h becoming smaller.

10
Gradients of y x2

• Complete this table by looking at this applet

Xa x2 Gradient 2x
1
2
3
4
5
6
7
11
Gradients of y x3

• Complete this table by using the applet

x x3 gradient
1
2
3
4
5
6
7
12
Gradients of y x4

• Complete this table by using the applet

x x4 Gradient x
1
2
3
4
5
6
7
13
So far

• y x2 has a gradient function of
• Y x3 the gradient function is
• Y x4 the gradient function is

14
The magic phrase

• Put the power in front and then drop the power by
  one!!!!!!
• We call the gradient function the derivative of
  the function i.e. dy DONT WRITE IT LIKE
• dx THIS dy/dx

15
The Derivative

• So if y x n
• Then dy
• dx n x n-1

16
F?(x) or y?

• This is also called the derivative noun
• Another way to express dy
• dx
  the gradient function is to say f?(x) or y ?
• This is called f dash x or y dash
• The verb is differentiate
• These are all names for the GRADIENT FUNCTION

17
The derivative of a number

• What is the gradient of the line y 10?
• What is the gradient of the line y 5?
• What does this line look like?
• So what is the derivative of y x5 10?

18
Derivative

• So if y x n where n is a number
• Then dy
• dx n x n-1
• Example 1
• Find derivative of
• A) y 4x2
• dy
• dx 8x
• b) y 5x3
• dy
• dx 15x2

19
More examples

• Find the derivative of the following

Y f(x) x4 x5 x3 x2 x 7x X-1 X-2
dy f(x) dx
20
Answers

• Find the gradient function (dy when y is
  
  
• dx)
• A) 2x2 6x 3
• B) ½ x2 12x
• C) 4x2 6
• D) 8x27x12
• E) 54x-5x2

21
Some rules to follow

• When y kxn and k is a
  constant
• Then dy
• dx nkx
  n-1
• Example 1
• If y 4x3 then dy
• dx 12x2
• When y u v
• Then dy du dv
• dx dx dx
• Example 2
• If y 3x3 5x2 then dy/dx 9x2 10x
• Ditto for subtraction

22
Some rules to follow

• When y u v
• Then dy du dv
• dx dx dx

23
Example 2

• If y 3x3 5x4 then
• dy
• dx 9x2 20x3
• Ditto for addition

24
Example 3

• (i) What about the derivative of y 1
• 
  x3
• (ii) Find the derivative of Y 20 1/3x410x
• dy - 4/3 x3 10
• dx
• Rules 1) derivative of a number (constant) is
  zero
• 2) derivative of 10x is 10 why?

25
Example 4

• Find the derivative of
• Y 3x - 11/5 x2 x4
• The derivative is 3 2/5 x 4x3

26
Example 5

• Find the gradient of the curve with
• equation y (2x-1)(3x2) at the point x 3
• Here we multiply the brackets out first by using
  the moon!
• Y 6x2x-2
• dy dx 12x1
• Now the gradient at x3 is 12(3)1 37

27
Starter

• Quick questions
• Write down the magic phrase to find the
  derivative of a function.
• Write down at least three different types of
  notation used for the derivative of a function.
• What is another name for derivative?

28
Quick questions- find the derivative
29
(No Transcript)
30
In the next few lessons we will

• A) Find out how to find the second derivative.
• B) Use first derivative as a rate of change for
  real life situations
• C) Review Finding gradients at particular points
  on a curve
• D) Find the equation of a tangent and normal to a
  curve

31
Applications of the first derivative

• Some movies but got to get correct format

32
(No Transcript)
33
An online quiz

• Rates of change
• Distance, velocity and acceleration

34
Second derivative

• The second derivative means just differentiate dy
• dx again!
• The second derivative looks like this
• d2y
• dx2 or y or f (x)
• Download livemath plugin now to view inter active
  pages

35
An example of second derivative

• Find the second derivative

36
Quiz

• A MC quiz on differentiating polynomials

37
Gradient as a rate of change

• Remember that the first derivative is the
  gradient function or the rate of change of one
  variable y with respect to x.
• What if we were given an equation of displacement
  s with respect to time?
• For example s t23t
• If I differentiate this I get a rate of change of
  distance with respect to time- this is called
• Velocity! So v ds
• dt

38
The derivative as a rate of change

• The derivative tells us the gradient function of
  any curve. If we have a distance time graph what
  does the gradient tell us?
• The gradient tells us the rate of change of the
  distance with respect to time which is also
  called speed!
• Click here for the surfer
• Click here for AS guru Game

39
Example 1

• The displacement of a body at a time t seconds is
  given in metres by s t23t
• Find a) the velocity of the body at time t
• b) the initial velocity of the body
• Solution
• A)The velocity ds
• dt 2t3
• B) The initial velocity is when t0
• so v 2(0)3 3 m/s
• So the initial velocity is 3 m/s

40
The rate of change- gradient

• Example The volume of an expanding spherical
  balloon is related to its radius, r cm by the
  formula V 4/3 ?r3. Find the rate of change of
  volume with respect to the radius at the instant
  when the radius is 5 cm.
• Ex 7G

41
Example 1

• Find the gradient of the curve
• f(x) x(x1) at the point P(0,0).

42
Example 2

• Find the gradient of the curve y 3x2 x 3 at
  the point (1,1)

43
Example 3

• Find the gradient of the curve
• y ½ x2 3/2 x at the point (1,2)

44
Drag and drop

• Put these tangents onto the curve

45
Equations of tangents and normals

• Some definitions first

46
Finding the equation of a Tangent

• What does the curve y (x-3)(x2) look like?
• At the point x 2 we can draw a tangent to the
  curve.
• How do we find the equation of this line?
• What information do we know already?

47
Tangents

• If we know the gradient, m at any point on a
  curve and we have the coordinates of that point
  (x ,y) we can easily find the equation of the
  tangent line at that point.
• What is the equation of a straight line?

48
The steps

• Example 1
• Find the equation of the tangent of the curve
• y (x-3)(x2) at the point x 1 y - 6

49
The steps

• Complete these sentences
• 1) Find the gradient of the curve at the point x
  1 by.
• 2) The equation of a straight line is
• 3) Now substituting x 1 and y -6 and m
  to obtain the equation of the tangent.

50
An example

• Find the equation of the tangent on the curve
• Y (x3)(x-1) at the point (1,5).
• How do I find the equation of a line given a
  gradient and point?
• Y mx c
• How do I find the gradient?
• Y x23x-x-3
• Y x22x-3 dy
• dx 2x2
• Gradient at x 1 dy
• dx 2(1) 2 4
• Y-5 4(x-1)
• Y 4x1 is the equation of the tangent at the
  point x 1

51
Your example

• Find the equation of the tangent to the curve
  with equation y (2x-1)(x1) at the point
• x 3 y 20
• Y 2x2x-1
• So dy
• dx 4x1
• So the gradient is dy
• dx 4(3) 1
  13
• So using y mxc
• Y- 20 13(x-3)
• Simplifying this gives y 13x - 19

52
Example 3

• Find the equation of the tangent at the
• point x -2 on the curve y 3x3 4x2-7
• Step 1 dy
• dx 9x2 8x
• Step 2 x -2 into dy
• dx
  9(-2)28(-2) 20
• Step 3 m20 x -2 find the y
  coordinate
• y 3(-2)34(-2)2-7
• y -2416-7 -15
• One point (-2,-15) and gradient m
  20
• y mxc
• Y 20x25

53
Follow these steps to find the equation of a
tangent

• Step 1 Find dy/dx
• Step 2 Sub the x value into dy/dx to find m, the
  gradient
• Step 3 Use y mxb or y-y1 m(x-x1)to find the
  equation
• A game for tangents As guru

54
Example

• Here are some guided examples.
• So we use y mxc to find the equation of the
  tangent

55
Tangents

• You will be given a card.
• Order these steps to find the tangent

56
What is a normal?

• A normal is perpendicular to the tangent

57
Equations of normals

• To find the equation of a normal we follow the
  steps but we use -1/m instead of m.
• Find the equation of the normal on the curve
• Y (x3)(x-1) at the point (1,5).
• Check the steps here to find the normal
• Ex 7H, 7L

58
Quiz

• A quiz on differentiating polynomials

59
A beautiful application problem

• Jordan wants to design a drink container for
  orange juice in the shape of a cylinder with NO
  lid. The volume of the orange juice is 1000 cc.
  Show that the surface area of this container with
  NO lid is given by
• Show that when the rate of the change of the area
  with respect to r is zero then r3 500/?

60
Solution
61
(No Transcript)
62
Card A

• When x 1
• dy/dx 5

63
Card B

• 2 5(1) C

64
Card C

• A curve has equation
• Y 3x2-x

65
Card D

• Second find the gradient at x 1

66
Card E

• Simplify

67
Card F

• Use y mx C

68
Card G

• When x 1
• y 2

69
Card H

• First find a point on the curve

70
Card I

• Y 5x-3

71
Card J

• The tangent at x 1 is to be found

72
Card K

• The derivative is dy/dx 6x-1

73
A lot of activities

• Click here for mathsnet.net

74
Derivatives

• What does a derivative look like?

75
C) Gradient as a rate of change

• Remember that the first derivative is the
  gradient function or the rate of change of one
  variable y with respect to x.
• What if we were given an equation of displacement
  s with respect to time?
• For example s t23t
• If I differentiate this I get a rate of change of
  distance with respect to time- this is called
• Velocity! So v ds
• dt

76
Example 1

• The displacement of a body at a time t seconds is
  given in metres by s t23t
• Find a) the velocity of the body at time t
• b) the initial velocity of the body
• Solution
• A)The velocity ds
• dt 2t3
• B) The initial velocity is when t0
• so v 2(0)3 3 m/s
• So the initial velocity is 3 m/s

77
What is dv dt?

• If we differentiate displacement w.r.t. to time
  we get velocity
• What do we get if we differentiate velocity?
• What is the rate of change of velocity?
• This is called acceleration!
• So dv
• dt acceleration,a.

78
Example 2

• The displacement s metres of body at time, t is
  given by s 2t3 t22. Find
• A) the velocity after 1 second
• B) the acceleration after 1 second
• C) the time at which the acceleration is zero.

79
Homework Quiz

• A quiz on differentiating polynomials
• Here you must print this out with your test
  results

80
D) Second derivative

• The second derivative means just differentiate dy
• dx again!
• The second derivative looks like this
• d2y
• dx2 or y or f (x)
• Download livemath plugin now to view inter active
  pages

81
More practice with the second derivative

• Find the second derivative of the following
  functions.
• A) f(x) x4-3x31
• B) y 8x2-7x
• C) f(x) ½ x2
• D) y (x32x)2
• E) f(x) 7x3-5x2
• F) y x1
• x

82
More practice with the second derivative

• Find the second derivative of the following
  functions.
• A) f(x) x24x-3
• B) y x3-3x
• C) f(x) x2 5
• D) y 2x3-2x
• E) f(x) 3x3

83
Finding the second derivative
84
Second derivative

• Look at this applet to understand the second
  derivative

85
Stationary points

• A stationary point is when the gradient is zero
  or when the curve is flat.
• There are three types of stationary points

86
Stationary points

• A turning point is when the gradient changes in
  sign. Look at this diagram

87
Maximum and minimums

• For a maximum you can see that we walk uphill ,
  flat and then downhill. This means the gradient
  is positive, zero and then negative. A maximum
  moves to a negative gradient
• For a minimum we walk downhill, flat and then
  uphill. This means the gradient is negative, zero
  and then positive. For a minimum it moves to a
  positive gradient.

88
Stationary points

• An good explanation

A little game with a stationary point
89
Second derivative

• Look at this applet to understand the second
  derivative and stationary points
• Here are some more practice questions

90
How do we find out where the stationary points
are?

• We firstly find the first derivative dy/dx and
  see when this equals zero!
• Example 1
• Find the where the stationary point is for the
  curve
• y x2
• dy
• dx 2x
• When does dy
• dx 0?
• 2x 0 so x 0 and y 0
• Example 2 Find where the stationary point is for
  the curve y x2 5x6
• Find the first derivative and let this equal
  zero.
• So dy/dx 2x 5 this equals zero when x 2.5
• Y -1/4

91
Follow these steps to test for stationary points-
maximum or minimum

• Step 1
• Find the values of x for which dy
• 
  dx 0
• Step 2 Find d2y
• dx2
• Step 3 Substitute the x values you got for
• dy/dx 0 into the second derivative to test
  what kind of stationary point it is (the nature).
• Now substitute x value into function to find
  corresponding y value

92
Second derivative

• The second derivative tells us whether a
  stationary point is a maximum, minimum or
  inflexion point

A sad negative maximum
A happy positive minimum
t
t
93
Complete this table when dy
dx 0
d2y/dx2 gt 0 second derivative is positive d2y/dx2 lt 0 second derivative is negative
Minimum Maximum
Happy face positive Sad face negative
94
An example

• Find the coordinates of the stationary points on
  the curve y x3-6x2-15x1, using the second
  derivative determine their nature.
• Two stat points are min (5,-99) and max (-1,9)

95
Another example

• Find the coordinates of the stationary points on
  the curve y x4-4x3 and determine their nature.
• Min (3,-27)

96
Drag and drop

• Put these tangents onto the curve

97
Maxima and minima problems

• Look at this link for lots of examples
• Look at this applet to see maximization

98
Maxima and minima

• We can use calculus to solve many practical
  problems such as finding the maximum area or
  volume.

99
Example 2

• A closed right cylinder of base radius r cm and
  height h cm has a volume of 54?. Show that S the
  total surface area of the cylinder is given by
• S 108 ? 2 ? r2
• r
• Find the maximum surface area

100
(No Transcript)
101
Hints!

• Steps
• 1) Find the volume of the cylinder and let this
  equal 54?. Make h the subject of this formula.
• 2) Find an expression for the total surface area.
• 3) Substitute what you got in 1) into the total
  surface area expression from 2) to give you S in
  terms of r.
• 4) Find the derivative of this expression to find
  the stationary values and determine their nature
  by using the second derivative.
• 5) Find the corresponding height.

102
More examples

• The profit, y, generated from the sale of x
  items of a certain luxury product is given by the
  formula
• y 600x15x2-x3. Calculate the value of x
  which gives a maximum profit, and determine that
  maximum profit.
• Solution
• X 20 profit 10,000

103
Quiz

• A MC quiz on differentiating polynomials