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P1-Chp12-Differentiation (1).pptx

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P1-Chp12-Differentiation (1).pptx

  1. 1. P1 Chapter 12 :: Differentiation Last modified: 14th October 2020
  2. 2. www.drfrostmaths.com Everything is completely free. Why not register? Teaching videos with topic tests to check understanding. Register now to interactively practise questions on this topic, including past paper questions and extension questions (including MAT + UKMT). Teachers: you can create student accounts (or students can register themselves), to set work, monitor progress and even create worksheets. Dashboard with points, trophies, notifications and student progress. With questions by: Questions organised by topic, difficulty and past paper.
  3. 3. Chapter Overview 1:: Find the derivative of polynomials. 4:: Find and understand the second derivative 𝑑2𝑦 𝑑π‘₯2 or 𝑓′′ π‘₯ If 𝑦 = π‘₯4 βˆ’ 3π‘₯2 determine 𝑑2𝑦 𝑑π‘₯2 2:: Find equations of tangents and normal to curves. Those who have done either IGCSE Mathematics, IGCSE Further Mathematics or Additional Mathematics would have encountered this content. Otherwise it will be completely new! The point 𝑃 3,9 lies on the curve 𝐢 with equation 𝑦 = π‘₯2. Determine the equation of the tangent to 𝐢 at the point 𝑃. 3:: Identify increasing and decreasing functions. Find the range of values for which 𝑓 π‘₯ = π‘₯3 βˆ’ π‘₯is increasing. 5:: Find stationary points and determine their nature. Find the stationary points of 𝑦 = π‘₯3 βˆ’ π‘₯ and state whether each is a maximum or minimum point. If 𝑦 = 3π‘₯2 + π‘₯, find 𝑑𝑦 𝑑π‘₯ 6:: Sketch a gradient function. Draw 𝑦 = π‘₯3 and its gradient function on the same axes. 7:: Model real-life problems.
  4. 4. Gradient Function For a straight line, the gradient is constant: π‘š = 3 However, for a curve the gradient varies. We can no longer have a single value for the gradient; we ideally want an expression in terms of 𝒙 that gives us the gradient for any value of π‘₯ (unsurprisingly known as the gradient function). 𝒙 -3 -2 -1 0 1 2 3 Gradient -6 -4 -2 0 2 4 6 At GCSE, you found the gradient of a curve at a particular point by drawing a tangent. 𝑦 = π‘₯2 ? ? ? ? ? ? ? π‘š = βˆ’6 π‘š = βˆ’4 π‘š = βˆ’2 π‘š = 0 π‘š = 2 π‘š = 4 π‘š = 6 By looking at the relationship between π‘₯ and the gradient at that point, can you come up with an expression, in terms of π‘₯ for the gradient? πΊπ‘Ÿπ‘Žπ‘‘π‘–π‘’π‘›π‘‘ πΉπ‘’π‘›π‘π‘‘π‘–π‘œπ‘› = πŸπ’™ ? 12.2
  5. 5. Finding the Gradient Function The question is then: Is there a method to work out the gradient function without having to draw lots of tangents and hoping that we can spot the rule? 5,25 1 ? 𝑦 = π‘₯2 6,36 11 ? π‘š = Δ𝑦 Ξ”π‘₯ = 11 1 = 11 ? To approximate the gradient on the curve 𝑦 = π‘₯2 when π‘₯ = 5, we could pick a point on the curve just slightly to the right, then find the gradient between the two points: As the second point gets closer and closer, the gradient becomes a better approximation of the true gradient: 5,25 0.01 5.01,25.1001 0.1001 ? π‘š = 0.1001 0.01 = 10.01 ? The actual gradient when π‘₯ = 5 is 10, so this approximation is damn close! 12.2
  6. 6. Finding the Gradient Function This gives us a numerical method to get the gradient at a particular 𝒙, but doesn’t give us the gradient function in general. Let’s use exactly the same method, but keep π‘₯ general, and make the β€˜small change’ (which was previously 0.01) β€˜β„Žβ€™: π‘”π‘Ÿπ‘Žπ‘‘π‘–π‘’π‘›π‘‘ = lim β„Žβ†’0 π‘₯ + β„Ž 2 βˆ’ π‘₯2 β„Ž = lim β„Žβ†’0 2π‘₯β„Ž + β„Ž2 β„Ž = lim β„Žβ†’0 2π‘₯ + β„Ž = 2π‘₯ π‘₯, π‘₯2 π‘₯ + β„Ž, π‘₯ + β„Ž 2 As always, gradient is change in 𝑦 over change in π‘₯. The β„Ž disappears as β„Ž tends towards 0, i.e. we can effectively treat it as 0 at this point. ? ? ? ? The lim means β€œthe limit of the following expression as β„Ž tends towards 0”. For example, lim π‘₯β†’βˆž 1 π‘₯ = 0, because as π‘₯ β€œtends towards” infinity, the β€œlimiting” value of the expression is 0. And voila, we got the 2π‘₯ we saw earlier! ? ? ? ? π’š = π’™πŸ 12.2
  7. 7. Finding the Gradient Function ! The gradient function, or derivative, of the curve 𝑦 = 𝑓(π‘₯) is written as 𝑓′(π‘₯) or 𝑑𝑦 𝑑π‘₯ . 𝑓′ π‘₯ = lim β„Žβ†’0 𝑓 π‘₯ + β„Ž βˆ’ 𝑓 π‘₯ β„Ž The gradient function can be used to find the gradient of the curve for any value of π‘₯. We will soon see an easier/quicker way to differentiate expressions like 𝑦 = π‘₯3 βˆ’ π‘₯ without using β€˜limits’. But this method, known as differentiating by first principles, is now in the A Level syllabus, and you could be tested on it! Notation Note: Whether we use 𝑑𝑦 𝑑π‘₯ or 𝑓′(π‘₯) for the gradient function depends on whether we use 𝑦 = or 𝑓 π‘₯ = to start with: 𝑦 = π‘₯2 β†’ 𝑑𝑦 𝑑π‘₯ = 2π‘₯ 𝑓 π‘₯ = π‘₯2 β†’ 𝑓′ π‘₯ = 2π‘₯ β€œLagrange’s notation” There’s in fact a third way to indicate the gradient function, notation used by Newton: (but not used at A Level) 𝑦 = π‘₯2 β†’ 𝑦 = 2π‘₯ Advanced Notation Note: Rather than β„Ž for the small change in π‘₯, the formal notation is 𝛿π‘₯. So actually: 𝑓′ π‘₯ = lim 𝛿π‘₯β†’0 𝑓 π‘₯ + 𝛿π‘₯ βˆ’ 𝑓 π‘₯ 𝛿π‘₯ So we in fact have 3 symbols for β€œchange in”! β€’ Ξ”π‘₯: any change in π‘₯ (as seen in Chp5: π‘š = Δ𝑦 Ξ”π‘₯ ) β€’ 𝛿π‘₯: a small change in π‘₯ β€’ 𝑑π‘₯: an infinitesimally small change in π‘₯ So the estimated gradient using some point close by was 𝛿𝑦 𝛿π‘₯ , but in the β€˜limit’ as 𝛿π‘₯ β†’ 0, 𝛿𝑦 𝛿π‘₯ β†’ 𝑑𝑦 𝑑π‘₯ β€œLeibniz’s notation” 12.2
  8. 8. Example ! The point 𝐴 with coordinates 4,16 lies on the curve with equation 𝑦 = π‘₯2. At point 𝐴 the curve has gradient 𝑔. a) Show that 𝑔 = lim β„Žβ†’0 8 + β„Ž b) Deduce the value of 𝑔. 𝑔 = lim β„Žβ†’0 𝑓 4 + β„Ž βˆ’ 𝑓 4 β„Ž = lim β„Žβ†’0 4 + β„Ž 2 βˆ’ 42 β„Ž = lim β„Žβ†’0 16 + 8β„Ž + β„Ž2 βˆ’ 16 β„Ž = lim β„Žβ†’0 8β„Ž + β„Ž2 β„Ž = lim β„Žβ†’0 8 + β„Ž 𝑔 = 8 As β„Ž β†’ 0, clearly the limiting value of 8 + β„Ž is 8. Function is 𝑓 π‘₯ = π‘₯2 Use the β€œdifferentiation by first principles” formula. a b ? ? 2
  9. 9. Test Your Understanding Prove from first principles that the derivative of π‘₯4 is 4π‘₯3. 𝑓′ π‘₯ = lim β„Žβ†’0 𝑓 π‘₯ + β„Ž βˆ’ 𝑓 π‘₯ β„Ž = lim β„Žβ†’0 π‘₯ + β„Ž 4 βˆ’ π‘₯4 β„Ž = lim β„Žβ†’0 π‘₯4 + 4π‘₯3 β„Ž + 6π‘₯2 β„Ž2 + 4π‘₯β„Ž3 + β„Ž4 βˆ’ π‘₯4 β„Ž = lim β„Žβ†’0 4π‘₯3 β„Ž + 6π‘₯2 β„Ž2 + 4π‘₯β„Ž3 + β„Ž4 β„Ž = lim β„Žβ†’0 4π‘₯3 + 6π‘₯2 β„Ž + 4π‘₯β„Ž2 + β„Ž3 = 4π‘₯3 As β„Ž β†’ 0, any term involving a multiplication by β„Ž will become 0. ? Helping Hand: β€œRow 4” of Pascal’s Triangle is: 1 4 6 4 1 You’re welcome.
  10. 10. Exercise 12B Pearson Pure Mathematics Year 1/AS Pages 261-262 (Note that Exercise 12A was skipped in these slides) Don’t forget the β€˜Challenge’ question on Page 262! Reminder: 𝑓′(π‘₯) and 𝑑𝑦 𝑑π‘₯ both mean the gradient function, also known as the derivative of 𝑦. 𝑓′ π‘₯ = 𝑑𝑦 𝑑π‘₯ = lim β„Žβ†’0 𝑓 π‘₯ + β„Ž βˆ’ 𝑓 π‘₯ β„Ž
  11. 11. ME-WOW! Just for your interest… Why couldn’t we just immediately make β„Ž equal to 0 in lim β„Žβ†’0 π‘₯+β„Ž 2βˆ’π‘₯2 β„Ž ? God, not another one of these… If we just stick β„Ž = 0 in straight away: lim β„Žβ†’0 π‘₯ + 0 2 βˆ’ π‘₯2 0 = lim β„Žβ†’0 0 0 Wait, uh oh… 0 0 is known as an indeterminateform. Whereas we know what happens with expressions like 1 0 (i.e. its value tends towards infinity), indeterminate forms are bad because their values are ambiguous, and prevent π‘™π‘–π‘š expressions from being evaluated. Consider 0 0 for example: 0 divided by anything usually gives 0, but anything divided by 0 is usually infinity. We can see these conflict. Can you guess some other indeterminate forms, i.e. expressions whose value is ambiguous? 𝟎 𝟎 ∞ ∞ 𝟎 Γ— ∞ ∞ βˆ’ ∞ 𝟎𝟎 𝟏∞ ∞𝟎 Thankfully, when indeterminate forms appear in π‘™π‘–π‘š expressions, there are variety of techniques to turn the expression into one that doesn’t have any indeterminate forms. One simple technique, that worked in our example, is to expand and simplify. This gave us lim β„Žβ†’0 2π‘₯ + β„Ž and 2π‘₯ + 0 clearly is fine. Other techniques are more advanced. This is important when sketching harder functions: http://www.drfrostmaths.com/resources/resource.php?rid=163 ?
  12. 12. Differentiating π‘₯𝑛 Thankfully, there’s a quick way to differentiate terms of the form π‘₯𝑛 (where 𝑛 is a constant) with having to use first principles every time: ! If 𝑦 = π‘Žπ‘₯𝑛 then 𝑑𝑦 𝑑π‘₯ = π‘›π‘Žπ‘₯π‘›βˆ’1 (where π‘Ž, 𝑛 are constants) i.e. multiply by the power and reduce the power by 1 Examples: 𝑦 = π‘₯5 β†’ 𝑑𝑦 𝑑π‘₯ = 5π‘₯4 Power is 5, so multiply by 5 then reduce power by 5. 𝑓(π‘₯) = π‘₯ 1 2 β†’ 𝑓′ π‘₯ = 1 2 π‘₯βˆ’ 1 2 The power need not be an integer! Remember to use 𝑓′ π‘₯ not 𝑑𝑦 𝑑π‘₯ 𝑦 = 2π‘₯6 β†’ 𝑑𝑦 𝑑π‘₯ = 12π‘₯5 𝑓 π‘₯ = π‘₯ π‘₯4 = π‘₯βˆ’3 β†’ 𝑓′ π‘₯ = βˆ’3π‘₯βˆ’4 Why would it be incorrect to say that π’š = πŸπ’™ differentiates to π’…π’š 𝒅𝒙 = 𝒙 πŸπ’™βˆ’πŸ ? The rule only works when the base is π‘₯ and the power is a constant. Neither is true here! Note that π‘₯𝑛 is β€œa power of π‘₯” whereas 2π‘₯ is an exponential term (which you will encounter more in Chp14), and therefore differentiate differently. You will learn how to differentiate exponential terms in Year 2. ? ? ? ? ? 𝑦 = π‘₯6 = π‘₯3 β†’ 𝑑𝑦 𝑑π‘₯ = 3π‘₯2 ? ? ? 12.3
  13. 13. Test Your Understanding 𝑦 = π‘₯7 β†’ π’…π’š 𝒅𝒙 = πŸ•π’™πŸ” 𝑦 = 3π‘₯10 β†’ π’…π’š 𝒅𝒙 = πŸ‘πŸŽπ’™πŸ— 𝑓 π‘₯ = π‘₯ 1 2 π‘₯2 = π‘₯βˆ’ 3 2 β†’ π’…π’š 𝒅𝒙 = βˆ’ πŸ‘ 𝟐 π’™βˆ’ πŸ“ 𝟐 𝑦 = π‘Žπ‘₯π‘Ž β†’ π’…π’š 𝒅𝒙 = π’‚πŸπ’™π’‚βˆ’πŸ ? ? ? ? 𝑓(π‘₯) = 49π‘₯7 = 7π‘₯ 7 2 β†’ 𝒇′(𝒙) = πŸ’πŸ— 𝟐 𝒙 πŸ“ 𝟐 ? ? ? 1 2 3 4 5
  14. 14. Differentiating Multiple Terms Differentiate 𝑦 = π‘₯2 + 4π‘₯ + 3 First thing to note: If 𝑦 = 𝑓 π‘₯ + 𝑔 π‘₯ then 𝑑𝑦 𝑑π‘₯ = 𝑓′ π‘₯ + 𝑔′ π‘₯ i.e. differentiate each term individually in a sum/subtraction. 𝑑𝑦 𝑑π‘₯ = 2π‘₯ + 4 ? ? ? 𝑦 = 4π‘₯ = 4π‘₯1 Therefore applying the usual rule: 𝑑𝑦 𝑑π‘₯ = 4π‘₯0 = 4 Alternatively, if you compare 𝑦 = 4π‘₯ to 𝑦 = π‘šπ‘₯ + 𝑐, it’s clear that the gradient is fixed and π‘š = 4. 𝑦 = 3 = 3π‘₯0 Therefore applying the usual rule: 𝑑𝑦 𝑑π‘₯ = 0π‘₯βˆ’1 = 0 Alternatively, if you sketch 𝑦 = 4, the line is horizontal, so the gradient is 0.
  15. 15. Quickfire Questions 𝑦 = 2π‘₯2 βˆ’ 3π‘₯ β†’ π’…π’š 𝒅𝒙 = πŸ’π’™ βˆ’ πŸ‘ 1 2 3 4 5 𝑦 = 4 βˆ’ 9π‘₯3 β†’ π’…π’š 𝒅𝒙 = βˆ’πŸπŸ•π’™πŸ 𝑦 = 5π‘₯ + 1 β†’ π’…π’š 𝒅𝒙 = πŸ“ 𝑦 = π‘Žπ‘₯ β†’ π’…π’š 𝒅𝒙 = 𝒂 𝑦 = 6π‘₯ βˆ’ 3 + 𝑝π‘₯2 β†’ π’…π’š 𝒅𝒙 = πŸ” + πŸπ’‘π’™ ? ? ? ? ? (where π‘Ž is a constant) (where 𝑝 is a constant)
  16. 16. Harder Example Let 𝑓 π‘₯ = 4π‘₯2 βˆ’ 8π‘₯ + 3 a) Find the gradient of 𝑦 = 𝑓 π‘₯ at the point 1 2 , 0 b) Find the coordinates of the point on the graph of 𝑦 = 𝑓 π‘₯ where the gradient is 8. c) Find the gradient of 𝑦 = 𝑓 π‘₯ at the points where the curve meets the line 𝑦 = 4π‘₯ βˆ’ 5. 𝑓′ π‘₯ = 8π‘₯ βˆ’ 8 When 𝑓′ 1 2 = 8 1 2 βˆ’ 8 = βˆ’4 Remember that the β€˜gradient function’ allows you to find the gradient for a particular value of π‘₯. 8 = 8π‘₯ βˆ’ 8 π‘₯ = 2 𝑦 = 4 2 2 βˆ’ 8 2 + 3 = 3 Point is 2,3 This example is important! Previously you used a value of π‘₯ to get the gradient 𝑓′ π‘₯ . This time we’re doing the opposite: using a known gradient 𝑓′ π‘₯ to get the value of π‘₯. We therefore substitute 𝑓′ π‘₯ for 8. a b c First find point of intersection: 4π‘₯2 βˆ’ 8π‘₯ + 3 = 4π‘₯ βˆ’ 5 Solving, we obtain: π‘₯ = 1 or π‘₯ = 2 When π‘₯ = 1, 𝑓′ 1 = 0 When π‘₯ = 2, 𝑓′ 2 = 8 Once you have your π‘₯, you need to work out 𝑦. Ensure you use the correct equation! ? ? ? 12.4
  17. 17. Test Your Understanding Let 𝑓 π‘₯ = π‘₯2 βˆ’ 4π‘₯ + 2 a) Find the gradient of 𝑦 = 𝑓 π‘₯ at the point 1, βˆ’1 b) Find the coordinates of the point on the graph of 𝑦 = 𝑓 π‘₯ where the gradient is 5. c) Find the gradient of 𝑦 = 𝑓 π‘₯ at the points where the curve meets the line 𝑦 = 2 βˆ’ π‘₯. 𝑓′ π‘₯ = 2π‘₯ βˆ’ 4 When 𝑓′ 1 = 2 1 βˆ’ 4 = βˆ’2 5 = 2π‘₯ βˆ’ 4 π‘₯ = 9 2 𝑦 = 9 2 2 βˆ’ 4 9 2 + 2 = 17 4 Point is 9 2 , 17 4 a b c π‘₯2 βˆ’ 4π‘₯ + 2 = 2 βˆ’ π‘₯ Solving: π‘₯ = 0 or π‘₯ = 3 When π‘₯ = 0, 𝑓′ 0 = βˆ’4 When π‘₯ = 3, 𝑓′ 3 = 2 ? ? ?
  18. 18. Exercise 12D Pearson Pure Mathematics Year 1/AS Pages 265-266 (Note that Exercise 12C was skipped in these slides)
  19. 19. Differentiating Harder Expressions If your expression isn’t a sum of π‘₯𝑛 terms, simply manipulate it until it is! 1. Turn roots into powers: 𝑦 = π‘₯ = π‘₯ 1 2 β†’ π’…π’š 𝒅𝒙 = 𝟏 𝟐 π’™βˆ’ 𝟏 𝟐 𝑦 = 1 3 π‘₯ = π‘₯βˆ’ 1 3 β†’ π’…π’š 𝒅𝒙 = βˆ’ 𝟏 πŸ‘ π’™βˆ’ πŸ’ πŸ‘ 2. Split up fractions. 𝑦 = π‘₯2 + 3 π‘₯ = π‘₯2 + 3 π‘₯ 1 2 = π‘₯ 3 2 + 3π‘₯βˆ’ 1 2 β†’ π’…π’š 𝒅𝒙 = πŸ‘ 𝟐 𝒙 𝟏 𝟐 βˆ’ πŸ‘ 𝟐 π’™βˆ’ πŸ‘ 𝟐 3. Expand out brackets. 𝑦 = π‘₯2 π‘₯ βˆ’ 3 = π‘₯3 βˆ’ 3π‘₯2 β†’ π’…π’š 𝒅𝒙 = πŸ‘π’™πŸ βˆ’ πŸ”π’™ 4. Beware of numbers in denominators! 𝑦 = 1 3π‘₯ = 1 3 π‘₯βˆ’1 β†’ π’…π’š 𝒅𝒙 = βˆ’ 𝟏 πŸ‘ π’™βˆ’πŸ NOT 3π‘₯βˆ’1 !!! ? ? ? ? ? ? ? ? ? ?
  20. 20. Test Your Understanding Differentiate the following. 𝑦 = 1 π‘₯ = π’™βˆ’ 𝟏 𝟐 β†’ π’…π’š 𝒅𝒙 = βˆ’ 𝟏 𝟐 π’™βˆ’ πŸ‘ 𝟐 𝑦 = 2 + π‘₯3 π‘₯2 = πŸπ’™βˆ’πŸ + 𝒙 β†’ π’…π’š 𝒅𝒙 = βˆ’πŸ’π’™βˆ’πŸ‘ + 𝟏 𝑦 = 1 + 2π‘₯ 3π‘₯ π‘₯ = 𝟏 + πŸπ’™ πŸ‘π’™ πŸ‘ 𝟐 = 𝟏 πŸ‘ π’™βˆ’ πŸ‘ 𝟐 + 𝟐 πŸ‘ π’™βˆ’ 𝟏 𝟐 β†’ π’…π’š 𝒅𝒙 = βˆ’ 𝟏 𝟐 π’™βˆ’ πŸ“ 𝟐 βˆ’ 𝟏 πŸ‘ π’™βˆ’ πŸ‘ 𝟐 ? ? ?
  21. 21. Exercise 12E Pearson Pure Mathematics Year 1/AS Pages 267-268 Extension [MAT 2013 1E] The expression 𝑑2 𝑑π‘₯2 2π‘₯ βˆ’ 1 4 1 βˆ’ π‘₯ 5 + 𝑑 𝑑π‘₯ 2π‘₯ + 1 4 3π‘₯2 βˆ’ 2 2 is a polynomial of degree: A) 9 B) 8 C) 7 D) less than 7 Full expansion is not needed. The highest power term in the first polynomial is πŸπŸ”π’™πŸ’ Γ— βˆ’π’™ πŸ“ = βˆ’πŸπŸ”π’™πŸ—, differentiating twice to give βˆ’πŸπŸ” Γ— πŸ— Γ— πŸ– πŸ—π’™πŸ•. The highest power term in the second polynomial is πŸπŸ”π’™πŸ’ Γ— πŸ—π’™πŸ = πŸπŸ” Γ— πŸ— π’™πŸ– , differentiating once to give πŸπŸ” Γ— πŸ— Γ— πŸ– π’™πŸ• . These terms cancel leaving a polynomial of order (at most) 6. The answer is (D). This just means β€œdifferentiate twice”. We’ll be looking at the β€˜second derivative’ later in this chapter. ? 12.5
  22. 22. Finding equations of tangents (3, ? ) Find the equation of the tangent to the curve 𝑦 = π‘₯2 when π‘₯ = 3. Gradient function: π’…π’š 𝒅𝒙 = πŸπ’™ Gradient when π‘₯ = 3: π’Ž = πŸ” 𝑦-value when π‘₯ = 3: π’š = πŸ— So equation of tangent: π’š βˆ’ πŸ— = πŸ” 𝒙 βˆ’ πŸ‘ ? ? ? ? We want to use 𝑦 βˆ’ 𝑦1 = π‘š π‘₯ βˆ’ π‘₯1 for the tangent (as it is a straight line!). Therefore we need: (a) A point π‘₯1, 𝑦1 (b) The gradient π‘š.
  23. 23. Finding equations of normals Find the equation of the normal to the curve 𝑦 = π‘₯2 when π‘₯ = 3. Equation of tangent (from earlier): π’š βˆ’ πŸ— = πŸ”(𝒙 βˆ’ πŸ‘) Therefore equation of normal: π’š βˆ’ πŸ— = βˆ’ 𝟏 πŸ” 𝒙 βˆ’ πŸ‘ ? The normal to a curve is the line perpendicular to the tangent. (3,9) Fro Exam Tip: A very common error is for students to accidentally forget whether the question is asking for the tangent or for the normal.
  24. 24. Test Your Understanding Find the equation of the normal to the curve 𝑦 = π‘₯ + 3 π‘₯ when π‘₯ = 9. When π‘₯ = 9, 𝑦 = 9 + 3 9 = 18 𝑦 = π‘₯ + 3π‘₯ 1 2 ∴ 𝑑𝑦 𝑑π‘₯ = 1 + 3 2 π‘₯βˆ’ 1 2 π‘šπ‘‡ = 1 + 3 2 9βˆ’ 1 2 = 3 2 ∴ π‘šπ‘ = βˆ’ 2 3 Equation of normal: 𝑦 βˆ’ 18 = βˆ’ 2 3 π‘₯ βˆ’ 9 Fro Tip: I like to use π‘šπ‘‡ and π‘šπ‘ to make clear to the examiner (and myself) what gradient I’ve found. ? y ? gradient ? Final equation
  25. 25. Exercise 12F Pearson Pure Mathematics Year 1/AS Pages 269-270 Extension [STEP I 2005 Q2] The point 𝑃 has coordinates 𝑝2 , 2𝑝 and the point 𝑄 has coordinates π‘ž2 , 2π‘ž , where 𝑝 and π‘ž are non- zero and 𝑝 β‰  π‘ž. The curve 𝐢 is given by 𝑦2 = 4π‘₯. The point 𝑅 is the intersection of the tangent to 𝐢 at 𝑃 and the tangent to 𝐢 at 𝑄. Show that 𝑅 has coordinates π‘π‘ž, 𝑝 + π‘ž . The point 𝑆 is the intersection of the normal to 𝐢 at 𝑃 and the normal to 𝐢 at 𝑄. If 𝑝 and π‘ž are such that 1,0 lies on the line 𝑃𝑄, show that 𝑆 has coordinates 𝑝2 + π‘ž2 + 1, 𝑝 + π‘ž , and that the quadrilateral 𝑃𝑆𝑄𝑅 is a rectangle. Solutions on next slide. [STEP I 2012 Q4] The curve 𝐢 has equation π‘₯𝑦 = 1 2 . The tangents to 𝐢 at the distinct points 𝑃 𝑝, 1 2𝑝 and 𝑄 π‘ž, 1 2π‘ž , where 𝑝 and π‘ž are positive, intersect at 𝑇 and the normal to 𝐢 at these points intersect at 𝑁. Show that 𝑇 is the point 2π‘π‘ž 𝑝 + π‘ž , 1 𝑝 + π‘ž In the case π‘π‘ž = 1 2 , find the coordinates of 𝑁. Show (in this case) that 𝑇 and 𝑁 lie on the line 𝑦 = π‘₯ and are such that the product of their distances from the origin is constant. The β€˜difference of two cubes’: π‘Ž3 βˆ’ 𝑏3 = π‘Ž βˆ’ 𝑏 π‘Ž2 + π‘Žπ‘ + 𝑏2 will help for both of these. 1 2
  26. 26. Solutions to Extension Question 1 This is using something called β€˜implicit differentiation’ (Year 2), but you could easily do: 𝑦2 = 4π‘₯ β†’ 𝑦 = 2π‘₯ 1 2 𝑑𝑦 𝑑π‘₯ = π‘₯βˆ’ 1 2 = 1 π‘₯ = 2 𝑦
  27. 27. Increasing and Decreasing Functions What do you think it means for a function to be an β€˜increasing function’? ! An increasing function is one whose gradient is always at least 0. 𝑓′ π‘₯ β‰₯ 0 for all π‘₯. It would be β€˜strictly increasing’ if 𝑓 π‘₯ > 0 for all π‘₯, i.e. is not allowed to go horizontal. A function can also be increasing and decreasing in certain intervals. 2,3 4, βˆ’1 Increasing for π‘₯ ≀ 2 Decreasing for 2 ≀ π‘₯ ≀ 4 Increasing for π‘₯ β‰₯ 4 We could also write β€œπ‘“ π‘₯ is decreasing in the interval [2,4]” [𝒂, 𝒃] represents all the real numbers between 𝒂 and 𝒃 inclusive, i.e: π‘Ž, 𝑏 = π‘₯ ∢ π‘Ž ≀ π‘₯ ≀ 𝑏 ? ? ?
  28. 28. Examples Show that the function 𝑓 π‘₯ = π‘₯3 + 6π‘₯2 + 21π‘₯ + 2 is increasing for all real values of π‘₯. 𝑓′ π‘₯ = 3π‘₯2 + 12π‘₯ + 21 𝑓′ π‘₯ = 3 π‘₯2 + 4π‘₯ + 7 = 3 π‘₯ + 2 2 + 9 π‘₯ + 2 2 β‰₯ 0 for all real π‘₯, ∴ 3 π‘₯ + 2 2 + 9 β‰₯ 0 for all real π‘₯ ∴ 𝑓(π‘₯) is an increasing function for all π‘₯. Fro Tip: To show a quadratic is always positive, complete the square, then indicate the squared term is always at least 0. ? Find the interval on which the function 𝑓 π‘₯ = π‘₯3 + 3π‘₯2 βˆ’ 9π‘₯ is decreasing. 𝑓 π‘₯ = π‘₯3 + 3π‘₯2 βˆ’ 9π‘₯ 𝑓′ π‘₯ = 3π‘₯2 + 6π‘₯ βˆ’ 9 𝑓′ π‘₯ ≀ 0 3π‘₯2 + 6π‘₯ βˆ’ 9 ≀ 0 π‘₯2 + 2π‘₯ βˆ’ 3 ≀ 0 π‘₯ + 3 π‘₯ βˆ’ 1 ≀ 0 βˆ’3 ≀ π‘₯ ≀ 1 So 𝑓(π‘₯) is decreasing in the interval [βˆ’3,1] ?
  29. 29. Test Your Understanding Show that the function 𝑓 π‘₯ = π‘₯3 + 16π‘₯ βˆ’ 2 is increasing for all real values of π‘₯. Find the interval on which the function 𝑓 π‘₯ = π‘₯3 + 6π‘₯2 βˆ’ 135π‘₯ is decreasing. 𝑓′ π‘₯ = 3π‘₯2 + 16 π‘₯2 β‰₯ 0 for all real π‘₯ ∴ 3π‘₯2 + 16 β‰₯ 0 for all real π‘₯. Therefore 𝑓(π‘₯) is an increasing function for all real π‘₯. 𝑓′ π‘₯ = 3π‘₯2 + 12π‘₯ βˆ’ 135 𝑓′ π‘₯ ≀ 0 ∴ 3π‘₯2 + 12π‘₯ βˆ’ 135 ≀ 0 π‘₯2 + 4π‘₯ βˆ’ 45 ≀ 0 π‘₯ + 9 π‘₯ βˆ’ 5 ≀ 0 βˆ’9 ≀ π‘₯ ≀ 5 So 𝑓(π‘₯) is decreasing in the interval [βˆ’9,5] ? ?
  30. 30. Exercise 12G Pearson Pure Mathematics Year 1/AS Page 271
  31. 31. Second Order Derivatives When you differentiate once, the expression you get is known as the first derivative. Unsurprisingly, when we differentiate a second time, the resulting expression is known as the second derivative. And so on… 𝑦 = π‘₯4 𝑑𝑦 𝑑π‘₯ = 4π‘₯3 𝑦′ = 4π‘₯3 𝑦 = 4π‘₯3 𝑑2𝑦 𝑑π‘₯2 = 12π‘₯2 𝑦′′ = 12π‘₯2 𝑦 = 12π‘₯2 𝑓 π‘₯ = π‘₯4 𝑓′ π‘₯ = 4π‘₯3 𝑓′′ π‘₯ = 12π‘₯2 Lagrange’s Original Function First Derivative Second Derivative You can similarly have the third derivative ( 𝑑3𝑦 𝑑π‘₯3), although this is no longer in the A Level syllabus. We’ll see why might use the second derivative soon…
  32. 32. Just for your interest… How does the notation 𝑑2𝑦 𝑑π‘₯2 work? Why are the squareds where they are? Suppose that 𝑦 = π‘₯3 + 1. Then when we write 𝑑𝑦 𝑑π‘₯ , we’re effectively doing 𝑑 π‘₯3+1 𝑑π‘₯ (by substitution), although this would typically be written: 𝑑 𝑑π‘₯ π‘₯3 + 1 The 𝑑 𝑑π‘₯ (… ) notation is quite handy, because it behaves as a function and allows us to write the original expression and the derivative within a single equation: 𝑑 𝑑π‘₯ π‘₯3 + 1 = 3π‘₯2 Therefore, if we wanted to differentiate 𝑦 twice, we’d do: 𝑑 𝑑π‘₯ 𝑑 𝑑π‘₯ 𝑦 = 𝑑2 𝑑π‘₯2 𝑦 = 𝑑2 𝑦 𝑑π‘₯2 ME-WOW!
  33. 33. Examples If 𝑦 = 3π‘₯5 + 4 π‘₯2, find 𝑑2𝑦 𝑑π‘₯2. If 𝑓 π‘₯ = 3 π‘₯ + 1 2 π‘₯ , find 𝑓′′(π‘₯). 𝑦 = 3π‘₯5 + 4π‘₯βˆ’2 𝑑𝑦 𝑑π‘₯ = 15π‘₯4 βˆ’ 8π‘₯βˆ’3 𝑑2𝑦 𝑑π‘₯2 = 60π‘₯3 + 24π‘₯βˆ’4 This could also be written as: 60π‘₯3 + 24 π‘₯4 𝑓 π‘₯ = 3π‘₯ 1 2 + 1 2 π‘₯βˆ’ 1 2 𝑓′ π‘₯ = 3 2 π‘₯βˆ’ 1 2 βˆ’ 1 4 π‘₯βˆ’ 3 2 𝑓′′ π‘₯ = βˆ’ 3 4 π‘₯βˆ’ 3 2 + 3 8 π‘₯βˆ’ 5 2 ? ?
  34. 34. Test Your Understanding If 𝑦 = 5π‘₯3 βˆ’ π‘₯ 3 π‘₯ , find 𝑑2𝑦 𝑑π‘₯2. 𝑦 = 5π‘₯3 βˆ’ 1 3 π‘₯ 1 2 𝑑𝑦 𝑑π‘₯ = 15π‘₯2 βˆ’ 1 6 π‘₯βˆ’ 1 2 𝑑2 𝑦 𝑑π‘₯2 = 30π‘₯ + 1 12 π‘₯βˆ’ 3 2 ? (Note: For time reasons, I’m skipping Exercise 12H on Page 272) Also note an error on this page: β€œWhen you differentiate with respect to π‘₯, you treat any other letters as constants.” This is emphatically not true – it depends on whether the letter is a variable or a constant. The statement however is true of β€˜partial differentiation’ (which is not in the A Level syllabus). In fact in Year 2, you will learn that π‘₯𝑦, when differentiated with respect to π‘₯, gives π‘₯ 𝑑𝑦 𝑑π‘₯ + 𝑦, not to just 𝑦. What the textbook means is β€œWhen you differentiate with respect to π‘₯, you treat any letters, defined to be constants, as numbers.” So π‘Žπ‘₯ would differentiate to π‘Ž, just as 3π‘₯ differentiates to 3, but ONLY if you were told that π‘Ž is a constant! If π‘Ž was in fact a variable, we need to use something called the product rule (Year 2 content).
  35. 35. Stationary/Turning Points A stationary point is where the gradient is 0, i.e. 𝑓′ π‘₯ = 0. 𝑓′ π‘₯ = 0 𝑓′ π‘₯ = 0 Local maximum Local minimum Fro Note: It’s called a β€˜local’ maximum because it’s the function’s largest output within the vicinity. Functions may also have a β€˜global’ maximum, i.e. the maximum output across the entire function. This particular function doesn’t have a global maximum because the output keeps increasing up to infinity. It similarly has no global minimum, as with all cubics. Find the coordinates of the turning points of 𝑦 = π‘₯3 + 6π‘₯2 βˆ’ 135π‘₯ 𝑑𝑦 𝑑π‘₯ = 3π‘₯2 + 12π‘₯ βˆ’ 135 = 0 π‘₯2 + 4π‘₯ βˆ’ 45 = 0 π‘₯ + 9 π‘₯ βˆ’ 5 = 0 π‘₯ = βˆ’9 π‘œπ‘Ÿ π‘₯ = 5 When π‘₯ = 5, 𝑦 = 53 + 6 52 βˆ’ 135 5 = βˆ’400 β†’ 5, βˆ’400 When π‘₯ = βˆ’9, 𝑦 = βˆ’9 3 + 6 βˆ’9 2 βˆ’ 135 βˆ’9 = 972 β†’ (βˆ’9,972) ?
  36. 36. More Examples Find the least value of 𝑓 π‘₯ = π‘₯2 βˆ’ 4π‘₯ + 9 Method 1: Differentiation 𝑓′ π‘₯ = 2π‘₯ βˆ’ 4 = 0 π‘₯ = 2 𝑓 2 = 22 βˆ’ 4 2 + 9 = 5 So 5 is the minimum value. Method 2: Completing the square 𝑓 π‘₯ = π‘₯ βˆ’ 2 2 + 5 Therefore the minimum value of 𝑓(π‘₯) is 5, and this occurs when π‘₯ = 2. ? Method 1: Differentiation ? Method 2: Completing the Square Fro Note: Method 2 is only applicable for quadratic functions. For others, differentiation must be used. Find the turning point of 𝑦 = π‘₯ βˆ’ π‘₯ 𝑦 = π‘₯ 1 2 βˆ’ π‘₯ 𝑑𝑦 𝑑π‘₯ = 1 2 π‘₯βˆ’ 1 2 βˆ’ 1 = 0 1 2 π‘₯βˆ’ 1 2 = 1 π‘₯βˆ’ 1 2 = 2 β†’ 1 π‘₯ = 2 π‘₯ = 1 2 β†’ π‘₯ = 1 4 When π‘₯ = 1 4 , 𝑦 = 1 4 βˆ’ 1 4 = 1 4 So turning point is 1 4 , 1 4 ?
  37. 37. Points of Inflection There’s a third type of stationary point (that we’ve encountered previously): 𝑓′ π‘₯ = 0 A point of inflection is where the curve changes from convex concave (or vice versa). convex concave (the same terms used in optics!) i.e. the line curves in one direction before the point of inflection, then curves in the other direction after. Technically we could label these either way round depending on where we view the curve from. What’s important is that the concavity changes. Fro Side Note: Not all points of inflection are stationary points, as can be seen in the example on the right. A point of inflection which is a stationary point is known as a saddle point.
  38. 38. How do we tell what type of stationary point? Local Minimum Gradient just before Gradient at minimum Gradient just after -ve 0 +ve Point of Inflection Gradient just before Gradient at p.o.i Gradient just after +ve 0 +ve Local Maximum Gradient just before Gradient at maximum Gradient just after +ve 0 +ve ? ? ? ? ? ? ? ? ? Method 1: Look at gradient just before and just after point.
  39. 39. How do we tell what type of stationary point? Method 1: Look at gradient just before and just after point. Find the stationary point on the curve with equation 𝑦 = π‘₯4 βˆ’ 32π‘₯, and determine whether it is a local maximum, a local minimum or a point of inflection. Strategy: Find the gradient for values just before and after π‘₯ = 2. Let’s try π‘₯ = 1.9 and π‘₯ = 2.1. 𝑑𝑦 𝑑π‘₯ = 4π‘₯3 βˆ’ 32 = 0 π‘₯ = 2 ∴ 𝑦 = βˆ’48 Stationary point is 2, βˆ’48 𝒙 = 𝟏. πŸ— 𝒙 = 𝟐 𝒙 = 𝟐. 𝟏 Gradient βˆ’4.56 0 5.04 Shape Looking at the shape, we can see that 2, βˆ’48 is a minimum. ? Turning Point ? Determine point type ?
  40. 40. Method 2: Using the second derivative The method of substituting values of π‘₯ just before and after is a bit cumbersome. It also has the potential for problems: what if two different types of stationary points are really close together? Recall the gradient gives a measure of the rate of change of 𝑦, i.e. how much the 𝑦 value changes as π‘₯ changes. Thus by differentiating the gradient function, the second derivative tells us the rate at what the gradient is changing. Thus if the second derivative is positive, the gradient is increasing. If the second derivative is negative, the gradient is decreasing. +ve gradient 0 gradient -ve gradient At a maximum point, we can see that as π‘₯ increases, the gradient is decreasing from a positive value to a negative value. ∴ 𝑑2 𝑦 𝑑π‘₯2 < 0
  41. 41. Method 2: Using the second derivative ! At a stationary point π‘₯ = π‘Ž: β€’ If 𝑓′′ π‘Ž > 0 the point is a local minimum. β€’ If 𝑓′′ π‘Ž < 0 the point is a local maximum. β€’ If 𝑓′′ π‘Ž = 0 it could be any type of point, so resort to Method 1. I will eventually do a β€˜Just for your Interest…’ thingy on why we can’t classify the point when 𝑓′′ π‘₯ = 0, and how we could use the third derivative! The stationary point of 𝑦 = π‘₯4 βˆ’ 32π‘₯ is 2, βˆ’48 . Use the second derivative to classify this stationary point. 𝑑𝑦 𝑑π‘₯ = 4π‘₯3 βˆ’ 32 𝑑2 𝑦 𝑑π‘₯2 = 12π‘₯2 When π‘₯ = 2, 𝑑2𝑦 𝑑π‘₯2 = 12 2 2 > 0 Therefore the stationary point is a minimum point. Find 𝑑2𝑦 𝑑π‘₯2 for the π‘₯ at the stationary point. ?
  42. 42. Test Your Understanding Edexcel C2 May 2013 Q9 ? ?
  43. 43. Sketching Graphs All the way back in Chapter 4, we used features such as intercepts with the axes, and behaviour when π‘₯ β†’ ∞ and π‘₯ β†’ βˆ’βˆž in order to sketch graphs. Now we can also find stationary/turning points! [Textbook] By first finding the stationary points, sketch the graph of 𝑦 = 1 π‘₯ + 27π‘₯3 𝑦 = π‘₯βˆ’1 + 27π‘₯3 𝑑𝑦 𝑑π‘₯ = βˆ’π‘₯βˆ’2 + 81π‘₯2 = 0 βˆ’ 1 π‘₯2 + 81π‘₯2 = 0 β†’ 81π‘₯2 = 1 π‘₯2 81π‘₯4 = 1 β†’ π‘₯ = 1 3 π‘œπ‘Ÿ βˆ’ 1 3 𝟏 πŸ‘ , πŸ’ , βˆ’ 𝟏 πŸ‘ , βˆ’πŸ’ As π‘₯ β†’ ∞, π’š β†’ ∞ As π‘₯ β†’ βˆ’βˆž, π’š β†’ βˆ’βˆž π‘₯ not defined at 0 (due to 1 π‘₯ term) ? Vertical Asymptotes ? As π‘₯ β†’ ∞, π‘₯ β†’ βˆ’βˆž ? Turning Points π‘₯ 𝑦 1 3 4 βˆ’4 βˆ’ 1 3 ? Graph
  44. 44. Exercise 12I Pearson Pure Mathematics Year 1/AS Page 276 Extension [MAT 2014 1C] The cubic 𝑦 = π‘˜π‘₯3 βˆ’ π‘˜ + 1 π‘₯2 + 2 βˆ’ π‘˜ π‘₯ βˆ’ π‘˜ has a turning point, that is a minimum, when π‘₯ = 1 precisely for A) π‘˜ > 0 B) 0 < π‘˜ < 1 C) π‘˜ > 1 2 D) π‘˜ < 3 E) all values of π‘˜ [MAT 2004 1B] The smallest value of the function: 𝑓 π‘₯ = 2π‘₯3 βˆ’ 9π‘₯2 + 12π‘₯ + 3 In the range 0 ≀ π‘₯ ≀ 2 is what? [MAT 2001 1E] The maximum gradient of the curve 𝑦 = π‘₯4 βˆ’ 4π‘₯3 + 4π‘₯2 + 2 in the range 0 ≀ π‘₯ ≀ 2 1 5 occurs when: A) π‘₯ = 0 B) π‘₯ = 1 βˆ’ 1 3 C) π‘₯ = 1 + 1 3 D) π‘₯ = 2 1 5 [STEP I 2007 Q8] A curve is given by: 𝑦 = π‘Žπ‘₯3 βˆ’ 6π‘Žπ‘₯2 + 12π‘Ž + 12 π‘₯ βˆ’ (8π‘Ž + 16) where π‘Ž is a real number. Show that this curve touches the curve with equation 𝑦 = π‘₯3 at 2,8 . Determine the coordinates of any other point of intersection of the two curves. (i) Sketch on the same axes these two curves when π‘Ž = 2. (ii) … when π‘Ž = 1 (iii) when π‘Ž = βˆ’2 Hint: When two curves touch, their 𝑦 values must match, but what else must also match? 1 2 3 4
  45. 45. Solutions to Extension Questions [MAT 2014 1C] The cubic 𝑦 = π‘˜π‘₯3 βˆ’ π‘˜ + 1 π‘₯2 + 2 βˆ’ π‘˜ π‘₯ βˆ’ π‘˜ has a turning point, that is a minimum, when π‘₯ = 1 precisely for A) π‘˜ > 0 B) 0 < π‘˜ < 1 C) π‘˜ > 1 2 D) π‘˜ < 3 E) all values of π‘˜ π’…π’š 𝒅𝒙 = πŸ‘π’Œπ’™πŸ βˆ’ 𝟐 π’Œ + 𝟏 𝒙 + (𝟐 βˆ’ π’Œ) When 𝒙 = 𝟏, π’…π’š 𝒅𝒙 = πŸ‘π’Œ βˆ’ πŸπ’Œ βˆ’ 𝟐 + 𝟐 βˆ’ π’Œ ≑ 𝟎 Therefore there is a turning point for all values of π’Œ. However, this must be a minimum. π’…πŸ π’š π’…π’™πŸ = πŸ”π’Œπ’™ βˆ’ 𝟐(π’Œ + 𝟏) When 𝒙 = 𝟏, π’…πŸπ’š π’…π’™πŸ = πŸ’π’Œ βˆ’ 𝟐 If minimum: πŸ’π’Œ βˆ’ 𝟐 > 𝟎 β†’ π’Œ > 𝟏 𝟐 [MAT 2004 1B] The smallest value of the function: 𝑓 π‘₯ = 2π‘₯3 βˆ’ 9π‘₯2 + 12π‘₯ + 3 In the range 0 ≀ π‘₯ ≀ 2 is what? 𝒇′ 𝒙 = πŸ”π’™πŸ βˆ’ πŸπŸ–π’™ + 𝟏𝟐 = 𝟎 π’™πŸ βˆ’ πŸ‘π’™ + 𝟐 = 𝟎 𝒙 βˆ’ 𝟏 𝒙 βˆ’ 𝟐 = 𝟎 𝒙 = 𝟏 β†’ 𝒇 𝟏 = πŸ– 𝒙 = 𝟐 β†’ 𝒇 𝟐 = πŸ• At start of range: 𝒇 𝟎 = πŸ‘ (0,3) (1,8) (2,7) Therefore answer is 3. 1 2
  46. 46. Solutions to Extension Questions [MAT 2001 1E] The maximum gradient of the curve 𝑦 = π‘₯4 βˆ’ 4π‘₯3 + 4π‘₯2 + 2 in the range 0 ≀ π‘₯ ≀ 2 1 5 occurs when: A) π‘₯ = 0 B) π‘₯ = 1 βˆ’ 1 3 C) π‘₯ = 1 + 1 3 D) π‘₯ = 2 1 5 Gradient: π’…π’š 𝒅𝒙 = πŸ’π’™πŸ‘ βˆ’ πŸπŸπ’™πŸ + πŸ–π’™ So find max value of πŸ’π’™πŸ‘ βˆ’ πŸπŸπ’™πŸ + πŸ–π’™ π’…πŸ π’š π’…π’™πŸ = πŸπŸπ’™πŸ βˆ’ πŸπŸ’π’™ + πŸ– = 𝟎 πŸ‘π’™πŸ βˆ’ πŸ”π’™ + 𝟐 = 𝟎 𝒙 = 𝟏 Β± 𝟏 πŸ‘ Due to the shape of a cubic, we have the a local maximum gradient at 𝒙 = 𝟏 βˆ’ 𝟏 πŸ‘ and a local minimum at 𝒙 = 𝟏 + 𝟏 πŸ‘ In range 𝟎 ≀ 𝒙 ≀ 𝟐 𝟏 πŸ“ , answer must either be 𝒙 = 𝟏 + 𝟏 πŸ‘ or 𝒙 = 𝟐 𝟏 πŸ“ . Substituting these into π’š yields a higher value for the latter, so answer is (D). 3
  47. 47. [STEP I 2007 Q8] A curve is given by: 𝑦 = π‘Žπ‘₯3 βˆ’ 6π‘Žπ‘₯2 + 12π‘Ž + 12 π‘₯ βˆ’ (8π‘Ž + 16) where π‘Ž is a real number. Show that this curve touches the curve with equation 𝑦 = π‘₯3 at 2,8 . Determine the coordinates of any other point of intersection of the two curves. (i) Sketch on the same axes these two curves when π‘Ž = 2. (ii) … when π‘Ž = 1 (iii) … when π‘Ž = βˆ’2 Solutions to Extension Questions 4
  48. 48. Sketching Gradient Functions The new A Level specification specifically mentions being able to sketch 𝑦 = 𝑓′(π‘₯). If you know the function 𝑓′(π‘₯) explicitly (e.g. because you differentiated 𝑦 = 𝑓(π‘₯)), you can use your knowledge of sketching straight line/quadratic/cubic graphs. But in other cases you won’t be given the function explicitly, but just the sketch. π‘₯ 𝑦 Click to Sketch 𝑦 = 𝑓′(π‘₯) Click to Sketch 𝑦 = 𝑓′(π‘₯) Click to Sketch 𝑦 = 𝑓′(π‘₯) The gradient at the turning point is 0. The gradient of 𝑦 = 𝑓 π‘₯ is negative, but increasing. The gradient of 𝑦 = 𝑓 π‘₯ is positive, and increasing.
  49. 49. A Harder One π‘₯ 𝑦 > > > > > > > 𝑦 = 𝑓 π‘₯ 𝑦 = 𝑓′(π‘₯) Gradient is negative, but increasing. Fro Tip: Mentally describe the gradient of each section of the curve, starting your mental sentence with β€œThe gradient is…” and using terms like β€œpositive”, β€œbut increasing”, β€œbut not changing”, etc. Gradient is 0. Gradient is positive: initially increases but then decreases. Gradient is 0, but positive before and after. Gradient is positive: initially increases but then decreases. Gradient is 0. Gradient is negative. Initially decreases but then tends towards 0. Fro Pro Tip: The gradient is momentarily not changing so the gradient of the gradient is 0. We get a turning point in 𝑦 = 𝑓′ π‘₯ whenever the curve has a point of inflection. Again, on 𝑦 = 𝑓(π‘₯) this is a point of inflection so the gradient function has a turning point. It is also a stationary point, so 𝑓′ π‘₯ = 0. i.e. The 𝑓′(π‘₯) curve touches the π‘₯-axis.
  50. 50. Test Your Understanding π‘₯ 𝑦 Solution > 𝑦 = 𝑓 π‘₯ 𝑦 = 𝑓′(π‘₯) Gradient is negative, but increasing. Gradient is 0. Gradient is positive; initially increases but then decreases. Gradient is positive, but tending towards 0.
  51. 51. Exercise 12J Pearson Pure Mathematics Year 1/AS Page 278 Extension [MAT 2015 1B] 𝑓 π‘₯ = π‘₯ + π‘Ž 𝑛 where π‘Ž is a real number and 𝑛 is a positive whole number, and 𝑛 β‰₯ 2. If 𝑦 = 𝑓(π‘₯) and 𝑦 = 𝑓′(π‘₯) are plotted on the same axes, the number of intersections between 𝑓(π‘₯) and 𝑓′ π‘₯ will: A) always be odd B) always be even C) depend on π‘Ž but not 𝑛 D) depend on 𝑛 but not π‘Ž E) depend on both π‘Ž and 𝑛. 1 (Official solution) If 𝒇(𝒙) is an even function then 𝒇′ 𝒙 will be an odd function, and vice versa. 𝒇(𝒙) and 𝒇′ 𝒙 will cross the 𝒙-axis at βˆ’π’‚, 𝟎 and will have one further intersection when π’š and 𝒙 are greater than 0. The answer is (b). Recall that: Odd function: 𝑓 βˆ’π‘₯ = βˆ’π‘“ π‘₯ Even function: 𝑓 βˆ’π‘₯ = 𝑓(π‘₯) ?
  52. 52. These are examples of optimisation problems: we’re trying to maximise/minimise some quantity by choosing an appropriate value of a variable that we can control. We have a sheet of A4 paper, which we want to fold into a cuboid. What height should we choose for the cuboid in order to maximise the volume? π‘₯ 𝑦 We have 50m of fencing, and want to make a bear pen of the following shape, such that the area is maximised. What should we choose π‘₯ and 𝑦 to be? Optimisation Problems/Modelling β„Ž
  53. 53. β€˜Rate of change’ Up to now we’ve had 𝑦 in terms of π‘₯, where 𝑑𝑦 𝑑π‘₯ means β€œthe rate at which 𝑦 changes with respect to π‘₯”. But we can use similar gradient function notation for other physical quantities. Real Life Maths! A sewage container fills at a rate of 20 cm3 per second. How could we use appropriate notation to represent this? 𝑑𝑉 𝑑𝑑 = 20 π‘π‘š3/𝑠 β€œThe rate at which the volume 𝑉 changes with respect to time 𝑑.”
  54. 54. Example Optimisation Problem Optimisation problems in an exam usually follow the following pattern: β€’ There are 2 variables involved (you may have to introduce one yourself), typically lengths. β€’ There are expressions for two different physical quantities: β€’ One is a constraint, e.g. β€œthe surface area is 20cm2”. β€’ The other we wish to maximise/minimise, e.g. β€œwe wish to maximise the volume”. β€’ We use the constraint to eliminate one of the variables in the latter equation, so that it is then just in terms of one variable, and we can then use differentiation to find the turning point. 𝒙 𝒙 π’š [Textbook] A large tank in the shape of a cuboid is to be made from 54m2 of sheet metal. The tank has a horizontal base and no top. The height of the tank is π‘₯ metres. Two of the opposite vertical faces are squares. a) Show that the volume, V m3, of the tank is given by 𝑉 = 18π‘₯ βˆ’ 2 3 π‘₯3 . We need to introduce a second variable ourselves so that we can find expressions for the surface area and volume. b) Given that π‘₯ can vary, use differentiation to find the maximum or minimum value of 𝑉. 2π‘₯2 + 3π‘₯𝑦 = 54 𝑉 = π‘₯2 𝑦 But we want 𝑉 just in terms of π‘₯: 𝑦 = 54 βˆ’ 2π‘₯2 3π‘₯ β†’ 𝑉 = π‘₯2 54 βˆ’ 2π‘₯2 3π‘₯ 𝑉 = 54π‘₯2 βˆ’ 2π‘₯4 3π‘₯ = 18π‘₯ βˆ’ 2 3 π‘₯3 These are the two equations mentioned in the guidance: one for surface area and one volume. 𝑑𝑉 𝑑𝑑 = 18 βˆ’ 2π‘₯2 = 0 ∴ π‘₯ = 3 𝑉 = 18 3 βˆ’ 2 3 3 3 = 36 Once we have the β€˜optimal’ value of π‘₯, we sub it back into 𝑉 to get the best possible volume. ? ?
  55. 55. Test Your Understanding Edexcel C2 May 2011 Q8 ? a ? b ? c
  56. 56. Exercise 12K Pearson Pure Mathematics Year 1/AS Page 281 Extension [STEP I 2006 Q2] A small goat is tethered by a rope to a point at ground level on a side of a square barn which stands in a large horizontal field of grass. The sides of the barn are of length 2π‘Ž and the rope is of length 4π‘Ž. Let 𝐴 be the area of the grass that the goat can graze. Prove that 𝐴 ≀ 14πœ‹π‘Ž2 and determine the minimum value of 𝐴. 1 ?

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