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
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4. Gradient/ Slope
To find the gradient/slope between two points you have to find the rise (difference between the y’s) over the run (difference between the x’s)
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 y’s over the differences of the x’s

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 = -2x+5
2) y = -x+7
3) –x+2y -4 = 0
4) -3x+6y+7 = 0
5) 9x+6y+2 = 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 (a+h,f(a+h)) on y = f(x).
The slope is (f(a+h)-f(a))/h.
When we make h smaller, what does (f(a+h)-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 X=a
Gradient 2x 1 2 3 4 5 6 7

11. Gradients of y= x3 Complete this table by using the applet x x31 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. Put the power in front and then drop the power by one!!!!!!
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 DON’T 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
dx= 15x2

19. More examples Find the derivative of the following: Y = f(x) 7x
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) 8x2+7x+12
E) 5+4x-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 – 1/3x4+10x
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 - 1+1/5 x2 +x4
The derivative is 3 + 2/5 x + 4x3

26. Example 5 Find the gradient of the curve with
equation y = (2x-1)(3x+2) at the point x= 3
Here we multiply the brackets out first by using the moon!
Y= 6x2+x-2
dy dx = 12x+1
Now the gradient at x=3 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

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

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)
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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 = t2+3t
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!
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39. Example 1
The displacement of a body at a time t seconds is given in metres by s = t2+3t
Find a) the velocity of the body at time t
b) the initial velocity of the body
Solution
A)The velocity = ds
dt = 2t+3
B) The initial velocity is when t=0
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(x+1) 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)(x+2) 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)(x+2) 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= (x+3)(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 = x2+3x-x-3
Y= x2+2x-3 dy
dx = 2x+2
Gradient at x = 1 dy
dx = 2(1) + 2= 4
Y-5 = 4(x-1)
Y = 4x+1 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)(x+1) at the point
x = 3 y =20
Y= 2x2+x-1
So dy
dx = 4x+1
So the gradient is dy
dx = 4(3) +1 = 13
So using y = mx+c
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 dy
dx = 9x2 +8x
Step x = -2 into dy
dx = 9(-2)2+8(-2) = 20
Step m=20 x = -2 find the y coordinate
y = 3(-2)3+4(-2)2-7
y = = -15
One point (-2,-15) and gradient m = 20
y = mx+c
Y = 20x+25

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 = mx+b 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 = mx+c 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= (x+3)(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

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. Second find the gradient at x = 1
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. First find a point on the curve
Card H
First find a point on the curve

70. Card I
Y = 5x-3

71. The tangent at x = 1 is to be found
Card J
The tangent at x = 1 is to be found

72. The derivative is: dy/dx = 6x-1
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 = t2+3t
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 = t2+3t
Find a) the velocity of the body at time t
b) the initial velocity of the body
Solution
A)The velocity = ds
dt = 2t+3
B) The initial velocity is when t=0
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 –t2+2. 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)
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81. More practice with the second derivative
Find the second derivative of the following functions.
A) f(x) = x4-3x3+1
B) y = 8x2-7x
C) f(x) = ½ x2
D) y = (x3+2x)2
E) f(x) = 7x3-5x2
F) y = x+1 x

82. More practice with the second derivative
Find the second derivative of the following functions.
A) f(x) = x2+4x-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 –5x+6
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
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
second derivative is positive
d2y/dx2 < 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= x3-6x2-15x+1, 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

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 Solution X = 20 profit = $10,000
The profit, $y, generated from the sale of x items of a certain luxury product is given by the formula
y = 600x+15x2-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