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All  Differentiation
All  Differentiation
All  Differentiation
All  Differentiation
All  Differentiation
All  Differentiation
All  Differentiation
All  Differentiation
All  Differentiation
All  Differentiation
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All  Differentiation
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All  Differentiation
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All  Differentiation
All  Differentiation
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All  Differentiation
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All  Differentiation
All  Differentiation
All  Differentiation
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All  Differentiation
All  Differentiation
All  Differentiation
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All  Differentiation
All  Differentiation
All  Differentiation
All  Differentiation
All  Differentiation
All  Differentiation
All  Differentiation
All  Differentiation
All  Differentiation
All  Differentiation
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All Differentiation

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  • 1. Higher Unit 1 Differentiation
  • 2. Introduction to 1. differentiation
  • 3. Introduction to Differential Calculus Do not go from speed x to zero instantaneously. Up to this moment, your experience of speed and the relationship with distance and time has been restricted to questions such as those represented by this travel graph. However, there are two problems: 1. We can only find the AVERAGE speed between 2 time points, we cannot find the actual speed when t = a given time. 2. The graph doesn’t realistically depict changes in speed
  • 4. Therefore we will look at methods for finding the instantaneous Speed or rate of change for a given graph or function. This is known as Differential Calculus and was developed by Sir Isaac Newton and others and is used in many areas of maths. “If I have seen further it is by standing on the shoulders of giants.” 50 40 Sir Isaac Newton(1642-1727) 30 20 10 0 1 2 3
  • 5. To Speed or Not To Speed? A couple of motorists have been photographed by a speed camera and charged with speeding. The speed limit is under 30m/s in the area they were driving. They claim they were under the speed limit. Use the table below to determine if they broke the law or not. The camera flashed at 3 seconds. Time 0s 1s 2s 3s 4s Distance 0m 5m 20m 45m 80m
  • 6. Finding The Average Speed. Time 0s 1s 2s 3s 4s Distance 0m 5m 20m 45m 80m The formula for finding the speed is: Distance Speed  Time Calculate the average speed between 0 and 3 seconds. 45  0 So far so Speed  = 15m/s good for our 30 dastardly duo.
  • 7. So far we have worked out the average speed between 0 and 3 seconds as shown below: f=at2 Now repeat for 1 to 3 seconds. 50 And 2 to 3 seconds 40 45m 30 20 10 0 1 2 3 3s
  • 8. The Instantaneous Speed. We use these results to complete the table below: Time Interval 0-3 s 1-3 s 2-3 s Average Speed 25m/s 15m/s 20m/s It is clear that the closer the times are taken together the more accurately the speed of the car can be measured at 3 seconds. How can we get more accurate readings ? Use d = 5t 2 you snickering fools !!
  • 9. Using the formula d = 5 t 2 we find the table below: Time 2.6s 2.7s 2.8s 2.9s 3.0s Distance. 33.8m 36.45m 42.04m 45m 39.2m Time 2.6-3.0 2.7-3.0 2.8-3.0 2.9-3.0 Interval Average 28.5m/s 29m/s 29.6m/s 28m/s Speed. Guilty or not guilty ? The verdict is yours !!!
  • 10. The Tangent To The Curve. To calculate the average speed we have been calculating the gradients of lines as shown below: 50 40 30 20 10 0 1 2 3
  • 11. To get the instantaneous speed of the car at three seconds we require to calculate the gradient of the line which is tangential to the curve at t = 3seconds.That is to say the line contacts the curve at almost a single point at t= 3. The line is shown below: The tangent to the curve 50 at t = 3 . 40 30 20 10 0 1 2 3
  • 12. The Final Verdict. To calculate the value of the gradient of the tangent you require to consider a time as close to 3 seconds as possible. You might consider t= 3 and t= 2.99s or t= 3 and t=2.999s and so on… Investigate and find out if the car travelled at 30m/s. Learn your calculus numb skulls or you’ll never catch us !!
  • 13. The Derived Function 2.
  • 14. Required skills Before we start ………. You will need to remember work with Indices, as well as what you have learned about Straight Lines from Unit 1.1. Lets recall the rules on indices …..
  • 15. Rules of indices Rule Examples 1 a0 = 1 12.3140 = 1 1 1  3 a= x= 5 -m -5 nn 3 x am 2 m 6 = x x 3 2 3 am  a n n y 55 y6 3 1 = 6a a 2a × 3a 3/2 1/2 2 2 am × an = am+n 2 6a = x2 x -3= x 23 a a =a m n m-n x5 = q2 ᄡ q2 ᄡ q2 (q ) = 23 (am)n = amn 6 =q
  • 16. What is Differentiation? Differentiation is the process of deriving f ′(x) from f(x). We will look at this process in a second. f ′ (x) is called the derived function or derivative of f(x). The derived function represents: • the rate of change of the function • the gradient of the tangent to the graph of the function.
  • 17. Tangents toofcurves or slope of a The derivative function is a measure the gradient function at any given point. This requires us to consider the gradient of a line. We can do this if we think about how we measure the gradient from Unit 1.1 B(x2,y2) The gradient of AB = mAB y2 – y1 Change in y  Change in x A(x1,y1) x2 – x1 y2  y1 C  x 2  x1 y2  y1 m , Gradient Formula x 2  x1
  • 18. Tangents to curves We will look at a function and think about the gradient of the function at any given point. The function itself is not important, the process we go through to get the gradient is. We want to find the gradient of a curve. A is the point (x, f(x)) and B is a point on the function a short distance h from A. This gives B the coordinates (x+h, f(x+h)) The line AB is shown on the diagram. We want to find the gradient of the curve at A. If we find the gradient of the line AB and move B towards A we should get the gradient at A.
  • 19. Gradient of a function y = f(x) y A (( x  h), f ( x  h)) f ( x  h) ( x  h) y2  y1 f ( x  h) M AB  ( x  h) x2  x1 (( x), f ( x)) f ( x  h) ( x) B f ( x) ( x  h) x ( x  ( x  h) f x  h f ( x) hf)( x) f ( x) ( x) x f ( x) f ( x) f ( x) f ( x  h) f ( x ) ( x) ( x  h) f ( x  h)  f ( x ) ( x) M AB  ( x  h) ( x)  ( x) ( x) ( x) ( x) ( x) ( x) ( x)
  • 20. h What happens if we make smaller? f ( x  h)  f ( x )  M AB y = f(x) ( x  h)  x y A (( x  h), f ( x  h)) When x  h  x Then gradient of (( x), f ( x)) AB is the B gradient of the tangent at B x  x xhxhxhxhxhhh h xhxhx hx xhh  xx  hx x xh x
  • 21. Shrinking the tangent line is the same as letting h get very small y = f(x) y The gradient doesn't change as the line gets smaller and so the gradient at B must be the B same as the gradient of the line x
  • 22. What are we saying? We know f ( x  h)  f ( x )  M AB ( x  h)  x But f ( x  h)  f ( x ) M  lim B h 0 ( x  h)  x This is the basic, first principle definition of the derived function. Usually written f / ( x )
  • 23. Putting it together…. f ( x) f ( x  h) f ᄁ x)  lim ( Differentiate the function f(x) = x2 h hᆴ 0 from first principles. ( x  h) 2  x 2 f ᄁ x)  lim ( h hᆴ 0 The derivative is the same as the gradient of the tangent to the curve so we can go straight to the gradient formula we x 2  2 xh  h 2  x 2 f ᄁ x)  lim saw in the previous slides. ( h hᆴ 0 2 xh  h 2 f ᄁ x)  lim ( h hᆴ 0 h(2 x  h) f ᄁ x)  lim ( h hᆴ 0 h gets so f ᄁ x)  lim(2 x  h) small its ( hᆴ 0 effectively The limit as h → 0 is f ᄁ x)  2 x ( zero. written as lim hᆴ 0
  • 24. What does the answer mean? For each and every point on the curve f(x)=x2 , the gradient of the Value of x Gradient of the tangent tangent to the curve is given by -3 -6 the formula -2 -4 f ′(x)=2x y y y -1 -2 10 10 10 0 0 8 8 8 m= 1 2 6 6 6 -6 4 2 4 =4 4 4 m A table of results 2might make this = m 3 6 m 2 -4 2 =2 = clearer -2 4 8 m=0 m x –4 –2 2 4 x x –4 –2 2 4 –4 –2 2 4 –2 –2 –2 Is there an easier way to do –4 –4 –4 this? –6 –6 –6 –8 –8 –8 – 10 – 10 – 10
  • 25. Rules for differentiation There are four rules for differentiating – remember these and you can differentiate anything … Rule Examples f(x) = xn  f ′(x) = nxn-1 f(x) = x6  f ′ (x) = 6 x 6-1 = 6 x 5 f(x)= 4x2  f ′ (x) = 4 x 2 x 2-1  f ′ (x) = cnx f(x) = cx n n-1 = 8 x 1 or 8 x f(x) = c  f ′ (x) = 0 f(x) = 65  f ′ (x) = 0 f(x) = g(x) + h(x) f(x)= x6 + 4x2 + 65  f ′ (x) = 6 x 5 + 8x  f ′ (x) = g′ (x) + h′ (x)
  • 26. Derivatives of f ( x)  ax n 3.
  • 27. Copy the following: The Derivative ' The process of deriving f ( x) from f(x) is called differentiation. f ' ( x) represents two things: • The rate of change of the function Power must • The gradient of the tangent to the function be rational Basic Rule If f(x) = axn then f '(x) = naxn-1 multiply by reduce the the power power by 1
  • 28. Example 1 Find the derivative of: 9 4x 9 x (b) (a) Think “Flower Power”:  9 ᄡ 4x (9 1) ( 9 1)  9x 10  9x  36x 8 Power on top 2 Root at bottom 5 3 (d) x 7 (c) x 7 3  2x x 5 2 Must be if form axn (55) 3 5x 5 3 ( 7  2 )   ᄡ 2x 7 22 2 5 2 9 x  7x 3 2 5
  • 29. Heinemann , p.95, EX 6F, Q1-10
  • 30. Example 2 1 Find the gradient of the tangent to the curve f ( x)  5 at x = 2 x 1 f ( x)  5 Solution: x f ( x)  x 5 1. Prepare for differentiation (ie must be if form axn) f ' ( x)  5 ᄡ 1x ( 51) 2. “Multiply by power then f ' ( x)  5 x 6 reduce power by 1” 5 f ( x)  6 ' x 3. Tidy up 5 f (2)  ' (2)6 4. Substitute in given value for x 5 f (2)   ' 64
  • 31. Example 3 The volume in a container can be calculated using V (t )  3 2 t Calculate the rate of change of the volume after 8 seconds. V (t )  t 3 2 Solution: 2 1. Prepare for differentiation V (t )  t 3 (ie must be if form axn) 2 ( 33) 23 V (t )  t ' 2. “Multiply by power then 3 1 reduce power by 1” V (t )  t ' 2 3 3 21 3. Tidy up V (t )  ᄡ 3 ' 3 t 4. Substitute in given value for t 21 V (8)  ᄡ 3 ' 3 8
  • 32. Example 3 The volume in a container can be calculated using V (t )  3 2 t Calculate the rate of change of the volume after 8 seconds. 21 V (8)  ᄡ 3 ' Solution: 3 8 1. Prepare for differentiation (ie must be if form axn) 21 V (8)  ᄡ ' 32 2. “Multiply by power then reduce power by 1” 21 V (8)  ' 63 3. Tidy up 4. Substitute in given value for t
  • 33. Heinemann , p.92, EX 6E
  • 34. Derivatives of complex 4. expressions f ( x )  g ( x ) ᄡ h( x ) f ( x )  g ( x )  h( x ) g ( x) f ( x)  h( x )
  • 35. The Derivative of Multiple Terms in x So far we have differentiated functions with only one term in x. Our basic rule still applies when we have multiple terms in x. We simply have to ensure that each individual term has been prepared for differentiation. Then differentiate each term separately. f ( x )  g ( x )  h( x ) This means that if f ' ( x)  g ' ( x)  h' ( x) then
  • 36. Example 4 3 f ( x)  x  3x  9 4 2 1 Find the derivative of 2 2x f ( x)  1 x 4  3 x 2  3 x 1  9 2 2 Solution: f ' ( x)  4 ᄡ 1 x (41)  2 ᄡ 3x (21)  1ᄡ 3 x ( 11)  0 1. Prepare for differentiation 2 2 (ie must be if form axn) f ' ( x )  2 x 3  6 x  3 x 2 2 2. “Multiply by power then 3 reduce power by 1” for each term f ' ( x)  2 x 3  6 x  2x2 3. Tidy up
  • 37. Heinemann , p.94, EX 6F, Q19-27
  • 38. Example 5 Find the derivative of f(x) = (2x – 3) (3x + 6) Solution: f ( x)  2 x(3 x  6)  3(3 x  6) 1. Prepare for differentiation f ( x)  6 x 2  12 x  9 x  18 (ie expand brackets and simplify) f ( x)  6 x 2  3 x  18 2. “Multiply by power then f ' ( x)  12 x  3 reduce power by 1” 3. Tidy up
  • 39. Heinemann , p.95, EX 6G, Q 1-6
  • 40. Example 6 x3  4 x 2  x  3 NAB f ( x)  Find the derivative of x2 Solution: x3 4 x 2 x 3 1. Re-write expression with each f ( x)  2  2  2  2 term in the numerator over the x x x x denominator f ( x)  x  4  x 1  3 x 2 2. Prepare for differentiation (ie use laws of indices) 3. “Multiply by power then f ' ( x)  1  x 2  6 x 3 reduce power by 1” 16 f ( x)  1  2  3 ' 4. Tidy up x x
  • 41. Heinemann , p.95, EX 6G, Q 17-24
  • 42. Leibniz Newton Leibniz Notation 5. dy  ax  bx  c 2 dx
  • 43. Gottfried Wilhelm Leibnitz (1646 - 1716) On the accession in 1714 of his master, George I., to the throne of England, Leibnitz was thrown aside as a useless tool; he was forbidden to come to England; and the last two years of his life were spent in neglect and dishonour. He was overfond of money and personal distinctions; was unscrupulous, as perhaps might be expected of a professional diplomatist of that time; but all who once came under the charm of his personal presence remained sincerely attached to him. He also held eminent positions in diplomacy, philosophy and literature. The last years of his life - from 1709 to 1716 - were embittered by the long controversy with John Keill, Newton, and others, as to whether he had discovered the differential calculus independently of Newton's previous investigations, or whether he had derived the fundamental idea from Newton, and merely invented another notation for it. The controversy occupies a place in the scientific history of the early years of the eighteenth century quite disproportionate to its true importance, but it materially affected the history of mathematics in western Europe . From `A Short Account of the History of Mathematics' (4th edition, 1908) by W. W. Rouse Ball.
  • 44. Place in History 1660 Charles ii restores English monarchy 1665/6 Plague & Great fire of London Newton’s correspondence refers to “fluxions” 1666 and “the sum of infinitesimals” 1675 Leibniz makes the calculus public (published 1684) 1689 William of Orange takes English Crown Massacre in Glencoe 1692 Newton publishes work on the calculus 1693 1694 Bank of England Formed 1695 Bank of Scotland Formed 1707 Act of Union 1712 First steam engine George I begins Hanovarian dynasty leading 1714 to Robert Walpole becoming first Prime Minister.
  • 45. Leibniz Notation Liebniz Notation is an alternative way of expressing derivatives to f'(x) , g'(x) , etc. Regardless of the notation, the meaning of the result is the same: • the rate of change of the function • the gradient of the tangent at a given point If y is expressed in terms of x then the derivative is written as /dx . dy “dee y by dee x” so /dx = 6x - 7 . dy eg y = 3x - 7x 2 If function is given as y=, or x=, t= etc. we use Leibniz notation.
  • 46. Example 1 If Q = 9R2 - 15 find /dR ! dQ R3 Solution: Q  9 R 2  15 R 3 1. Prepare for differentiation dQ 2. “Multiply by power then  2 ᄡ 9 R (21)  (3) ᄡ 15 R ( 31) reduce power by 1” dR dQ  18 R  45 R 4 3. Tidy up dR dQ 45  18R  4 dR R
  • 47. Example 2 A curve has equation y = 5x3 - 4x2 + 7 . Find the gradient of the tangent at x = -2 Solution: y  5x  4 x  7 3 2 1. Prepare for differentiation (NB. y =, so Leibniz notation) dy 2. “Multiply by power then  15 x 2  8 x dx reduce power by 1” dy  15(2) 2  8(2) 3. Substitute in given x-value dx(2) dy  60  16 dx(2) 4. Communicate that you have Gradient at dy  76 answered question. x = -2 is 76 dx (2)
  • 48. Extra bit for “FIZZY SYSTS” or even Physicists. Newton’s 2ndLaw of Motion s = ut + 1/2at2 where s = distance & t = time. /dt gives us a “diff in dist”  “diff in time” Finding ds ie speed or velocity so /dt = u + at ds but /dt = v so we get v = u + at ds and this is Newton’s 1st Law of Motion
  • 49. Heinemann , p.100, EX 6I, Q 2-5
  • 50. The Equation of the tangent 6. dy y  y1  m( x  x1 )  f ( x) ' dx
  • 51. The Tangent to the Graph y = mx + c y = f(x) A (a,b) tangent So far we have differentiated functions, stating that the derived function provides us with the gradient of the tangent to the graph. We have seen that whilst the derived function is the same throughout the curve the actual gradient is dependant on the x-coordinate.
  • 52. The Tangent to the Graph y = mx + c y = f(x) A (a,b) tangent Now we want to find the equation of the tangent. The tangent is a straight line. We know from previous work on the straight line that y  y1  m( x  x1 ) . How can we find m?
  • 53. The Tangent to the Graph y = mx + c y = f(x) A (a,b) tangent At any point gradient of curve = gradient of tangent As the gradient of a straight line is the same throughout the line if we can find the gradient of the curve at A this is the gradient we can use for the equation of the tangent. So m  f (a ) '
  • 54. Finding the Equation of the Tangent Copy the following: To find the equation of the tangent we: 1. Must have both coordinates of the point on the curve 2. Must have the gradient at that point. y  y1  m( x  x1 ) 3. Substitute these details into:
  • 55. Example 1 NAB Find the equation of the tangent to the curve y  5 x  1 4 at x = 1 y  5x4  1 Solution: y  5(1) 4  1 =6 1. If not given, find y-coordinate dy  20 x3 2. Find dy / dx dx dy 3. Find gradient at x-coordinate  20(1)3 = 20 dx(1) Equation of tangent at (1,6): y  6  20( x  1) 4. Substitute values into y  y1  m( x  x1 ) y  6  20 x  20 y  20 x  14
  • 56. Example 2 Show that there is only one tangent to the curve y  5 x  6 x 2 with gradient 36 Solution: 1. Note we are being asked to prove that dy/dx = 36 has only one solution. dy  10 x  6 2. Find dy/dx dx 36  10 x  6 3. Make an equation with given 10 x  30 gradient and solve x3 As dy/dx = 36 has only one 4. Make statement solution there is only one tangent with that gradient.
  • 57. Example 3 Find the point of contact of the curve y  x  3 x  8 2 at which the gradient is 9. y  x 2  3x  8 Solution: dy 1. Find dy / dx  2x  3 dx 2. Make dy / dx equal to given 2x  3  9 gradient and solve. 2x  6 x3 3. Substitute into original equation to y-coord when x = 3: find y-coordinate. y  (3) 2  3(3)  8 (DO NOT USE dy / dx.) y  9  9  8  10 4. Make statement Point of contact (3 , 10)
  • 58. Heinemann , p.101, EX 6J, Q1, 3, 5, 6 & 7
  • 59. Increasing / Decreasing 7. Functions f ( x) > 0 f ( x) < 0 ' '
  • 60. Is the function increasing or decreasing? y = mx + c y = f(x) Whilst f(x) is increasing gradient is POSITIVE. A (a,b) tangent Our next task is to establish whether or not a function is increasing or decreasing. What can we say about the gradient of the tangent as the y- coordinates increase i.e. the value of the function is increasing?
  • 61. Is the function increasing or decreasing? y = f(x) Whilst f(x) is decreasing gradient is NEGATIVE. What can we say about the gradient of the tangent as the y- coordinates decrease i.e. the value of the function is decreasing?
  • 62. Is the function increasing or decreasing? y = f(x) Whilst f(x) is not changing gradient is ZERO What happens when the graph is changing direction? The tangent is horizontal and so the gradient is ZERO. These are called stationary points.
  • 63. Increasing and Decreasing Functions Copy the following: y = f(x) f ' ( x) > 0 f ( x) < 0 ' f ( x)  0 ' f ( x) > 0 ' Function increasing f ( x) < 0 ' Function decreasing f ( x)  0 ' Stationary Point
  • 64. Example 1 State whether the function y  4 x  3 x  3 is increasing or 3 2 decreasing at x = -2 y  4 x  3x  3 3 2 Solution: dy  12 x 2  6 x 1. Find dy / dx dx dy  12 ᄡ (2) 2  6 ᄡ (2) 2. Find gradient at x-coordinate dx(2) dy  48  12 = 36 dx(2) 3. Make statement As dy/dx is positive, function is increasing at x = -2
  • 65. Example 2 Find the intervals in which the function y  4 x  6 x  2 3 2 is increasing and decreasing. y  4 x3  6 x 2  2 Solution: dy 1. Find dy/dx  12 x 2  12 x dx 2. Set dy/dx = 0 and solve to find 12 x 2  12 x  0 x-coords of Stationary Points 12 x( x  1)  0 Factorise 12 x  0 x 1  0 or x0 x  1 or 3. Make a table to show what -1 0 X gradient is before and after SP’s dy + - + 0 0 dx
  • 66. Example 2 Find the interval in which the function y  4 x  6 x  2 3 2 is increasing and decreasing. Solution: x < 1 f(x) increasing: 4. Make statement f(x) decreasing: 1 < x < 0 x>0 f(x) increasing: -1 0 X dy + - + 0 0 dx
  • 67. The Proof y  4 x3  6 x 2  2 x < 1 f(x) increasing: f(x) decreasing: 1 < x < 0 x>0 f(x) increasing:
  • 68. Example 3 x3 is never decreasing. Show that the function y   3 x 2  9 x  4 3 x3 y   3x 2  9 x  4 Solution: 3 y  1 x3  3x 2  9 x  4 1. Prepare for differentiation 3 dy  x2  6x  9 2. Find dy / dx dx dy  ( x  3) 2 3. Factorise (all terms involving x dx must have even powers). Since (x - 3)2 will always 4. Make statement produce a positive gradient function is never decreasing.
  • 69. The Proof x3 y   3x 2  9 x  4 3
  • 70. Heinemann , p.104, EX 6L, Q1 to 6
  • 71. Stationary Points 8. f ( x)  0 '
  • 72. What if the function is neither increasing nor decreasing? y = f(x) Whilst f(x) is not changing gradient is ZERO Recall from the last lesson that when the gradient is changing there will be points where the tangent is horizontal and so the gradient=ZERO. At these points f ( x )  0 ' These are called stationary points.
  • 73. The Nature of a Stationary Point Copy the following: The nature of a stationary point is determined by the gradient either side of it. There are four classifications: Maximum Turning Point Minimum Turning Point
  • 74. The Nature of a Stationary Point Copy the following: The nature of a stationary point is determined by the gradient either side of it. There are four classifications: Rising Point of Inflection Falling Point of Inflection
  • 75. Example 1 NAB Determine the coordinates and nature of the stationary points on the curve y  5 x ( x  4) 3 y  5 x3 ( x  4) Solution: y  5 x 4  20 x3 1. Prepare for differentiation dy  20 x 3  60 x 2 dx 2. Find dy / dx At SP’s dy/dx = 0 3. Set dy/dx equal to zero, factorise 0  20 x3  60 x 2 if possible, and solve for x 0  20 x 2 ( x  3) 20 x 2  0 or ( x  3)  0 In exams must x = 0 or x=3 make this statement
  • 76. Example 1 Determine the coordinates and nature of the stationary points on the curve y  5 x ( x  4) 3 x=0 or x=3 Solution: For x = 3: 4. Find y-coordinates by subbing these For x = 0: values into original equation y  5(0)3 ((0)  4) y  5(3) ((3)  4) 3 (NOT dy/dx) y  5 ᄡ 27 ᄡ 1 y0 y  135 SP’s are (0,0) and (3,-135) 5. State coords of Stationary Points
  • 77. Example 1 Determine the coordinates and nature of the stationary points on the curve y  5 x ( x  4) 3 Solution: SP’s are (0,0) and (3,-135) dy 6. Draw a NATURE TABLE for each  20 x 3  60 x 2 (0,0) dx SP to determine its nature.   0 0 0 X dy  20(1)3  60(1) 2  80 dy d (1) -ve 0 -ve dx dy  20(1)3  60(1) 2  40 Slope d (1) 7. Make statement (0,0) is a falling point of inflexion.
  • 78. Example 1 Determine the coordinates and nature of the stationary points on the curve y  5 x ( x  4) 3 Solution: SP’s are (0,0) and (3,-135) dy 6. Draw a NATURE TABLE for each  20 x 3  60 x 2 (3,-135) dx SP to determine its nature.   3 3 3 X dy  20(1)3  60(1) 2  40 dy d (1) -ve 0 +ve dx dy  20(4)3  60(4) 2  320 Slope d (4) 7. Make statement (3,-135) is a minimum turning point.
  • 79. The Proof y  5 x3 ( x  4)
  • 80. Heinemann , p.106, EX 6M
  • 81. Curve Sketching 9.
  • 82. What do we need to know to sketch the curve? Stationary Points y = f(x) y-intercept x-intercept Behaviour as x gets small Behaviour as x gets big If asked to sketch this curve what information do you think you would need?
  • 83. Curve Sketching Copy the following: If asked to sketch a curve we need to establish the following: 1. y-intercept (x = 0) 2. x-intercept(s) (y = 0) 3. Stationary Points (dy/dx = 0 ) xᆴ ᆬ 4. Behaviour as x gets very big and very small x ᆴ ᆬ
  • 84. Example 1 Sketch the graph of y  8 x  3 x 3 4 y  8 x3  3x 4 Solution: y  8(0)3  3(0) 4 = 0 1. Find y-intercept (x = 0) y-intercept is (0,0) 0  8 x3  3x 4 2. Find x-intercept (y = 0) 0  x3 (8  3 x) or (8  3 x)  0 x3  0 3x  8 8 x x=0 3 x-intercepts (0,0) & ( 2 2,0) 3
  • 85. Example 1 Sketch the graph of y  8 x  3 x 3 4 y  8 x3  3x 4 dy Solution:  24 x 2  12 x 3 dx 3. Find Stationary Points and their At SP’s dy/dx = 0 nature (dy/dx = 0 ) 0  24 x 2  12 x 3 0  12 x 2 (2  x) 12 x 2  0 or (2  x)  0 x=0 x =2 y  8(0)3  3(0) 4 y  8(2)3  3(2) 4 y = 16 y=0
  • 86. Example 1 dy Sketch the graph of y  8 x  3 x 3 4  24 x 2  12 x 3 dx SP’s are (0,0) and (2,16) Solution:   2 0 0 2 X 3. Find Stationary Points and their dy nature (dy/dx = 0 ) -ve 0 0 +ve +ve dx dy Slo  24(1) 2  12(1)3  36 d (1) pe dy  24(1) 2  12(1)3  12 d (1) (0,0) is a rising point of inflexion dy (2,16) is a maximum tp  24(3) 2  12(3)3  108 d (3)
  • 87. Example 1 Biggest power Sketch the graph of y  8 x  3 x 3 4 dominates Solution: y  8 x3  3x 4 4. Behaviour as x gets very big and very small x ᆴ ᆬ As x ᆴ ᆬ -3x will dominate 4 x ᆴ ᆬ and so large x’s will be -ve Even Power - +ve Odd Power – get what As x ᆴ ᆬ -3x4 dominates you started with and so small x’s will be -ve 5. Plot the SP’s, the intercepts and draw curve.
  • 88. x-intercepts (0,0) & ( 2 2 Solution: y-intercept is (0,0) ,0) 3 (0,0) is a rising point of inflexion (2,16) is a maximum tp (2,16) (0,0) 22 3
  • 89. x-intercepts (0,0) & ( 2 2 Solution: y-intercept is (0,0) ,0) 3 (0,0) is a rising point of inflexion (2,16) is a maximum tp (2,16) (0,0) 22 3 y  8 x3  3x 4
  • 90. Heinemann , p.107, EX 6N Q1, 2, 5, 7 & 8
  • 91. Max /Min on Closed interval 10.
  • 92. What is a closed interval? y = f(x) Closed interval Up to now we have looked at graphs without any restrictions i.e we have considered the function and its graph for all possible values of x. Sometimes we may wish to just look at a certain part of the function or graph i.e restrict the interval.
  • 93. Where are the maximum and minimum values in a closed interval? 1 ᆪ x ᆪ 2 Lets consider f(x) = 8 + 2x – x2 on the interval f(x) x Smallest x f(x) x f(x) Largest 1 9 & SP -1 5 0 8 1.1 8.99 -0.9 5.39 0.1 8.19 1.2 8.96 -0.8 5.76 0.2 8.36 1.3 8.91 -0.7 6.11 0.3 8.51 1.4 8.84 -0.6 6.44 0.4 8.64 1.5 8.75 -0.5 6.75 0.5 8.75 1.6 8.64 -0.4 7.04 0.6 8.84 1.7 8.51 -0.3 7.31 0.7 8.91 1.8 8.36 -0.2 7.56 0.8 8.96 1.9 8.19 -0.1 7.79 0.9 8.99 2 8
  • 94. The proof
  • 95. Maximum & Minimum Values on a Closed Interval Copy the following: A closed interval is often written as { −2 < x < 3} or [−2 , 3] This interval would refer to the values of the function from −2 to 3 In a closed interval the maximum and minimum values of a function y are either at a stationary point or at Local maximum an end point of the interval. x Local minimum
  • 96. Example 1 y  4  3x 2  x3 Find the maximum and minimum values of within the interval 3 ᆪ x ᆪ 1 y  4  3x 2  x3 Solution: dy  6 x  3 x 2 dx 1. Find Stationary Points At SP’s dy/dx = 0 0  6 x  3 x 2 0  3x(2  x) 3 x  0 or (2  x)  0 x  2 x=0 y = 4 y  4  3(2)  (2) 2 3 y=0 SP’s are (0,4) & (-2,0)
  • 97. Example 1 y  4  3x 2  x3 Find the maximum and minimum values of within the interval 3 ᆪ x ᆪ 1 y  4  3x 2  x3 Solution: At x = -3 2. Find value of f(x) at each end point y  4  3(3) 2  (3)3 of given interval y  4  3(9)  (27) y  4  27  27 (-3,4) y=4
  • 98. Example 1 y  4  3x 2  x3 Find the maximum and minimum values of within the interval 3 ᆪ x ᆪ 1 y  4  3x 2  x3 Solution: At x = 1 2. Find value of f(x) at each end point y  4  3(1) 2  (1)3 of given interval y  4  3(1)  (1) y  4  3 1 (1,0) y=0 3. Compare Y-COORDINATES to (0,4) & (-2,0) (-3,4) (1,0) find smallest and biggest Maximum = 4, Minimum = 0 4. Make statement
  • 99. The Proof
  • 100. Heinemann , p.109, EX 6O Q2 (a), (b), (e) & (f)
  • 101. Graph of the Derived 11. Function
  • 102. What will the graph of a derivative look like? y = x3 – 6x2 + 9x Gradient –ve Gradient to SP 2 +ve Gradient +ve to SP 1 Imagine you were asked to sketch a graph of f ' ( x) What would the key points be?
  • 103. Is the gradient constant? y = x3 – 6x2 + 9x Lets consider dy/dx = 3x2 – 12x + 9 dy/dx x x dy/dx x dy/dx 3.25 1.688 -1 24 1.25 -1.31 3.5 3.75 -0.75 19.7 1.5 -2.25 Min Moving 3.75 6.188 -0.5 15.8 to zero = -3 1.75 -2.81 4 9 -0.25 12.2 2 -3 4.25 12.19 0 9 2.25 -2.81 4.5 15.75 SP at 0.25 6.19 2.5 -2.25 (3,0) 4.75 19.69 0.5 3.75 SP at 2.75 -1.31 5 24 (1,0) 0.75 1.69 3 0 1 0 Decreasing to (2,-3) then increasing
  • 104. The original and the derived y  f ' ( x) y = f(x)
  • 105. Example 1 Sketch the derived function for f(x) y = f(x) x SP occurs 5 at x = 5 - 0 + New y-values f '(x)
  • 106. Example 1 Negative but approaching zero x SP occurs 5 at x = 5 - 0 + New y-values f '(x)
  • 107. Example 1 y  f ' ( x) Positives Negative but approaching zero x SP occurs 5 at x = 5 - 0 + New y-values f '(x)
  • 108. Example 2 Sketch the derived function for f(x) y = f(x) (2,-4) x 0 2 SPs occur at f'(x) + 0 - 0 + New y-values x=0 &x=2
  • 109. Example 2 Sketch the derived function for f(x) (2,-4) x 0 2 SPs occur at f'(x) + 0 - 0 + New y-values x=0 &x=2
  • 110. Example 2 Sketch the derived function for f(x) Positive but approaching zero (2,-4) x 0 2 SPs occur at f'(x) + 0 - 0 + New y-values x=0 &x=2
  • 111. Example 2 Sketch the derived function for f(x) Goes into –ve’s but returns to zero by 2 (2,-4) x 0 2 SPs occur at f'(x) + 0 - 0 + New y-values x=0 &x=2
  • 112. Example 2 Sketch the derived function for f(x) Into +ve’s x 0 2 SPs occur at f'(x) + 0 - 0 + New y-values x=0 &x=2
  • 113. Heinemann , p.111, EX 6P
  • 114. Graph of the Derived 11. Function
  • 115. What will the graph of a derivative look like? y = x3 – 6x2 + 9x Gradient –ve Gradient to SP 2 +ve Gradient +ve to SP 1 Imagine you were asked to sketch a graph of f ' ( x) What would the key points be?
  • 116. Is the gradient constant? y = x3 – 6x2 + 9x Lets consider dy/dx = 3x2 – 12x + 9 dy/dx x x dy/dx x dy/dx 3.25 1.688 -1 24 1.25 -1.31 3.5 3.75 -0.75 19.7 1.5 -2.25 Min Moving 3.75 6.188 -0.5 15.8 to zero = -3 1.75 -2.81 4 9 -0.25 12.2 2 -3 4.25 12.19 0 9 2.25 -2.81 4.5 15.75 SP at 0.25 6.19 2.5 -2.25 (3,0) 4.75 19.69 0.5 3.75 SP at 2.75 -1.31 5 24 (1,0) 0.75 1.69 3 0 1 0 Decreasing to (2,-3) then increasing
  • 117. The original and the derived y  f ' ( x) y = f(x)
  • 118. Example 1 Sketch the derived function for f(x) y = f(x) x SP occurs 5 at x = 5 - 0 + New y-values f '(x)
  • 119. Example 1 Negative but approaching zero x SP occurs 5 at x = 5 - 0 + New y-values f '(x)
  • 120. Example 1 y  f ' ( x) Positives Negative but approaching zero x SP occurs 5 at x = 5 - 0 + New y-values f '(x)
  • 121. Example 2 Sketch the derived function for f(x) y = f(x) (2,-4) x 0 2 SPs occur at f'(x) + 0 - 0 + New y-values x=0 &x=2
  • 122. Example 2 Sketch the derived function for f(x) (2,-4) x 0 2 SPs occur at f'(x) + 0 - 0 + New y-values x=0 &x=2
  • 123. Example 2 Sketch the derived function for f(x) Positive but approaching zero (2,-4) x 0 2 SPs occur at f'(x) + 0 - 0 + New y-values x=0 &x=2
  • 124. Example 2 Sketch the derived function for f(x) Goes into –ve’s but returns to zero by 2 (2,-4) x 0 2 SPs occur at f'(x) + 0 - 0 + New y-values x=0 &x=2
  • 125. Example 2 Sketch the derived function for f(x) Into +ve’s x 0 2 SPs occur at f'(x) + 0 - 0 + New y-values x=0 &x=2
  • 126. Heinemann , p.111, EX 6P
  • 127. Optimisation : Maxima & 12. Minima
  • 128. Problem Solving Differentiation can be used to solve problems which require maximum or minimum values. Problems typically cover topics such as areas, volumes and rates of change. They often involve having to establish a suitable formula in one variable and then differentiating to find a maximum or minimum value. This is known as Optimisation. It is important to check the validity of any solutions as often an answer is either nonexistent (e.g. a negative length or time) or outside an acceptable interval.
  • 129. Optimisation Optimisation is the term we use to describe a simple process of finding a maximum or minimum value for a given situation. Normally we would have to graph functions and then use our graph to establish a maximum or minimum. As we have learned previously, differentiation allows us to quickly find the required value and is the expected manner of solving problems like this in Higher mathematics. Drawing graphs would not be an acceptable solution.
  • 130. Optimisation Problems posed will often involve more than one variable. The process of differentiation requires that we rewrite or re- arrange formulae so that there is only one variable – typically x. Questions are therefore multi-part where the first part would involve establishing a formula in x from a given situation, part 2 would involve the differentiation and validation of an acceptable answer, with part 3 the solution to the problem set. You would do well to note that although the first part is important, you can normally expect to complete the rest of the question even when you cannot justify a formula in part 1.
  • 131. Copy the following: Optimization : Maxima and Minima •Differentiation is most commonly used to solve problems by providing a “best fit” solution. •Maximum and minimum values can be obtained from the Stationary Points and their nature. •In exams you may be asked to “prove” a particular formula is valid. Even if you cannot prove this USE THIS FORMULA TO ANSWER THE REST OF THE QUESTION.
  • 132. Example 1 (Formula Given) The “Smelly Place” Garden Centre has a model train which is used to show visitors around the many displays of plants and flowers. It has been established that the cost per month, C, of running the train is given by: 400 C  50  25V  V where V is the speed in miles per hour. Calculate the speed which makes the cost per month a minimum and hence calculate this cost.
  • 133. 400 C  50  25V  Example 1 V Solution: C  50  25V  400V 1 1. Find derivative, remembering to prepare for differentiation. dC  25  400V 2 dV dC 400  25  2 dV V
  • 134. 400 C  50  25V  Example 1 V Solution: dC 400  25  2 dV V 2. Make statement then At SP’s dC/dV = 0 set derivative equal to zero. 400 0  25  2 V 400 25 Cross Multiply  V2 1 400  25V 2 Reject –ve speed V 2  16 V 4
  • 135. 400 C  50  25V  Example 1 V Solution: dC 400  25  2 V 4 dV V 3. Justify nature of each SP using a nature table. 4 X 4 4 dC 400  25  2  375 dC -ve 0 d (1) (1) +ve dV dC 400 Slope  25  2  17 d (5) (5) A minimum tp occurs and so cost 4. Make statement is minimised at V = 4.
  • 136. 400 C  50  25V  Example 1 V A minimum tp occurs and so cost Solution: is minimised at V = 4. 400 4. Sub x-value found into C  50  25V  V original expression to find corresponding minimum value. For V = 4: (DO NOT USE dy/dx) 400 C  50  25(4)  4 C  50  100  100 C = £250 Minimum cost is £250 per month 5. Answer Question and occurs when speed = 4mph
  • 137. Heinemann , p.115, EX 6R Q3
  • 138. Example 2 (Perimeter and area) A desk is designed which is rectangular in shape and which is required in the design brief to have a perimeter of 420 cm. If x is the length of the base of the desk: (a) Find an expression for the area of the desk, in terms of x. (b) Find the dimensions which will give the maximum area. (c) Calculate this maximum area.
  • 139. Example 2 y x Solution to (a): Area = length x breadth 1. Start with what you know Area = x x y Lets call breadth y, so: 2. In order to differentiate we Perimeter = 2x + 2y need expression only involve X’s. Use other information to 420  2 x  2 y find an expression for Y. Divide 2 y  420  2 x by 2 3. Change subject to y. y  210  x
  • 140. Example 2 y  210  x Area = x x y Solution to (a): 4. Now substitute this expression Area = x x (210 – x) for y into our original expression for AREA. Area  A( x)  210 x  x 2 Solution to (b): Aᄁ x)  210  2 x ( 5. To find dimensions giving maximum or minimum first find At SP’s A’(x) = 0 derivative. 210  2 x  0 6. Make statement and set 2 x  210 derivative equal to zero. x  105
  • 141. Example 2 y  210  x Area = x x y Solution to (a): x  105 7. Now use your expression for y y  210  105  105 to find other dimension.
  • 142. Aᄁ x)  210  2 x Example 2 ( X = 105 and Y = 105 Solution: 8. Must justify maximum values using nature table. 105 X Aᄁ (100)  210  2(100)  10 Aᄁ x) +ve 0 ( -ve Aᄁ (110)  210  2(110)  10 Slope The area is maximised when 9. Make statement length is 105cm and breadth is 105 cm.
  • 143. Example 2 The area is maximised when length is 105cm and breadth is 105 cm. Solution to (c): Area  A( x)  210 x  x 2 10. To find actual maximum area sub maximum x-value just Areamax  210(105)  (105) 2 found into expression for area. (DO NOT USE dy/dx) Areamax  22, 050  11, 025 Areamax  11, 025 cm 2 PROOF
  • 144. Heinemann , p.112, EX 6Q Q1, 2, 5
  • 145. Example 3 (Surface Area and Volume) A cuboid of volume 51.2 cm3 is made with a length 4 times its breadth. If x cm is the breadth of the base of the cuboid: (a) Find an expression for the surface area of the cuboid, in terms of x. (b) Find the dimensions which will give the minimum surface area and calculate this area. xcm
  • 146. Example 3 h x Solution to (a): 4x 1. Start with what you know Surface Area = sum of area of faces = 2(4x x h) + 2(x x h)+ 2(4x x x) Lets call height h, so: = 10xh + 8x2 2. In order to differentiate Volume = 4x x x x h expression must only involve X’s. Use other information to ᄡ 51.2  4x 2 h find an expression for h. 51.2 12.8 h 2 3. Change subject to h. 2 4x x
  • 147. Surface Area = sum of area of faces Example 2 = 10xh + 8x2 12.8 Solution to (a): h 2 x 4. Now substitute this expression for y into our original expression ₩ 12.8 ( ) A( x)  10 ᄡ x ᄡ 2  8 x 2 for SURFACE AREA. x │ ₩x 128 ( ) A( x)   8x2 x2 │ 128 A( x)  8 x  2 x
  • 148. 128 Example 3 A( x)  8 x  2 x Solution to (b): A( x)  8 x 2  128 x 1 5. To find dimensions giving 128 Aᄁ x)  16 x  2 maximum or minimum first find ( x derivative. At SP’s A’(x) = 0 6. Make a statement and set 128 Multiply 16 x  2  0 derivative equal to zero. by x2 x 16 x 3  128  0 16 x 3  128 x3  8 x 382
  • 149. 12.8 Example 3 h 2 x Solution to (b): 12.8 12.8 h   3.2 2 (2) 4 7. To find corresponding height substitute x value just found into So dimensions minimising area: expression for h found in step 2. Breadth = 2 cm Length = 8 cm Height = 3.2 cm
  • 150. 128 Example 3 Aᄁ x)  16 x  2 ( x x 382 Solution: 8. Must justify minimum values using nature table. X 2 Aᄁ  16  128  112 (1) Aᄁ x) -ve 0 ( +ve 128 Aᄁ  16(4)   56 (4) 16 Slope The surface area is minimised when 9. Make statement the breadth is 2 cm
  • 151. The surface area is minimised when Example 3 the breadth is 2 cm 128 Solution to (c): A( x)  8 x  2 x 10. To find actual maximum area 128 Areamin  8(2)  2 sub maximum x-value just 2 found into expression for area. (DO NOT USE dy/dx) Areamin  32  64 Areamin  96 cm 2 PROOF
  • 152. Heinemann , p.112, EX 6R Q1, 2, 5
  • 153. Example 4 ( ADDITIONAL - NOT ESSENTIAL) A channel for carrying cables is being dug out at the side of the road. A flat section of plastic is bent into the shape of a gutter and placed Into the channel to protect the cables. 100 cm x 4 0 c x m The dotted line represents the fold in the plastic, x cm from either end. (a) Show that the volume of each section of guttering is V ( x)  200 x(20  x) (b) Calculate the value of x which gives the maximum volume of gutter and find this volume.
  • 154. Example 2 (a) Show that the volume of each section of guttering is V ( x)  200 x(20  x) Solution to part (a): Require volume so: Length = 100 cm V  l ᄡ bᄡ h Breadth = (40 – 2x) cm V  100 ᄡ (40  2 x) ᄡ x Height = “folded bit” = x cm V  100 x(40  2 x) Factorise bracket V  100 x ᄡ 2(20  x) V  200 x(20  x) as required.
  • 155. (b) Calculate the value of x which gives the maximum Example 2 volume of gutter. V ( x)  200 x(20  x) Maximum and Solution to part (b): minimum will V ( x)  200 x(20  x) occur at SP’s V ( x)  4000 x  200 x 2 Must prepare for differentiation V ' ( x)  4000  400 x At SP’s V ' ( x)  0 Must make this statement 0  4000  400x 0  400(10  x) 0  (10  x) Now prove this is a maximum tp x = 10
  • 156. (b) Calculate the value of x which gives the maximum Example 2 volume of gutter. V ( x)  200 x(20  x) Solution to part (b): x = 10 Now prove this is a maximum tp V ' ( x)  4000  400 x   X 10 10 10 dV -ve +ve 0 dx Slope So when x = 10 we have a maximum
  • 157. (b) Calculate the value of x which gives the maximum Example 2 volume of gutter. V ( x)  200 x(20  x) Solution to part (b): So when x = 10 we have a maximum V ( x)  200 x(20  x) Now calculate maximum volume V (10)  200 ᄡ 10 ᄡ (20  10) V (10)  20000cm3 So maximum volume of guttering Make statement is 20,000 cm3.

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