`Chapter21A B C D E F Rates of change Limits The derivative function Simple rules of differentiation Tangents to curves The second derivativeDifferential calculusSyllabus reference: 7.1, 7.2, 7.3 Contents:626DIFFERENTIAL CALCULUS(Chapter 21)OPENING PROBLEMValentino is riding his motorbike around a racetrack. A computer chip on his bike measures the distance Valentino has travelled as time goes on. This data is used to plot a graph of Valentino's progress.distance travelled (m) 400 300 200100 Things to think about: a What is meant by a rate? 0 0 30 60 90 b What do we call the rate at which Valentino time (s) is travelling? c What is the difference between an instantaneous rate and an average rate? d How can we read a rate from a graph? e Which is the fastest part of the racetrack?ARates are used every day to measure performance. For example, we measure:RATES OF CHANGEA rate is a comparison between two quantities of different kinds.² the speed at which a car is travelling in km h¡1 or m s¡1 . ² the fuel efficiency of the car in km L¡1 or litres per 100 km travelled. ² the scoring rate of a basketballer in points per game.Example 1Self TutorJosef typed 213 words in 3 minutes and made 6 errors, whereas Marie typed 260 words in 4 minutes and made 7 errors. Compare their performance using rates. 213 words 3 minutes 6 errors Josef's error rate = 213 words 260 words Marie's typing rate = 4 minutes 7 errors Marie's error rate = 260 words Josef's typing rate = = 71 words per minute. ¼ 0:0282 errors per word. = 65 words per minute. ¼ 0:0269 errors per word.) Josef typed at a faster rate but Marie typed with greater accuracy.DIFFERENTIAL CALCULUS(Chapter 21)627EXERCISE 21A.11 Karsten's pulse rate was measured at 67 beats per minute. a Explain exactly what this rate means. b How many heart beats would Karsten expect to have each hour? 2 Jana typed a 14 page document and made eight errors. If an average page of typing has 380 words, find Jana's error rate in: a errors per word b errors per 100 words. 3 Niko worked 12 hours for \$148:20 whereas Marita worked 13 hours for \$157:95. Who worked for the better hourly rate of pay? 4 New tyres have a tread depth of 8 mm. After driving for 32 178 km, the tread depth on Joanne's tyres was reduced to 2:3 mm. What was the wearing rate of the tyres in: a mm per km travelled b mm per 10 000 km travelled? 5 We left Kuala Lumpur at 11:43 am and travelled to Penang, a distance of 350 km. We arrived there at 3:39 pm. What was our average speed in: a km h¡1 b m s¡1 ?INVESTIGATION 1CONSTANT AND VARIABLE RATES OF CHANGEWhen water is added at a constant rate to a cylindrical container, the depth of water in the container is a linear function of time. This is because the volume of water added is directly proportional to the time taken to add it. If water was not added at a constant rate the direct proportionality would not exist. The depth-time graph for a cylindrical container is shown alongside. In this investigation we explore the changes in the graph for different shaped containers such as the conical vase.DEMOwaterdepthdepth time depthtimeWhat to do: 1 What features of the graph indicate a rate of change in water level that is: a constant b variable?628DIFFERENTIAL CALCULUS(Chapter 21)2 For each of the following containers, draw a depth-time graph as water is added: a b c d e3 Use the water filling demonstration to check your answers to question 2. 4 Write a brief report on the connection between the shape of a vessel and the corresponding shape of its depth-time graph. You may wish to discuss this in parts. For example, first examine cylindrical containers, then conical, then other shapes. Gradients of curves must be included in your report. 5 Draw possible containers as in question 2 which have the following depth-time graphs: adepthbdepthcdepthddepthtimetimetimetimeAVERAGE RATE OF CHANGEIf the graph which compares two quantities is a straight line, there is a constant rate of change in one quantity with respect to the other. This constant rate is the gradient of the straight line. If the graph is a curve, we can find the average rate of change between two points by finding the gradient of the chord or line segment between them. The average rate of change will vary depending on which two points are chosen, so it makes sense to talk about the average rate of change over a particular interval.Example 2Self Tutor350 300 250 200 150 100 50 population sizeThe number of mice in a colony was recorded on a weekly basis. a Estimate the average rate of increase in population for: i the period from week 3 to week 6 ii the seven week period. b What is the overall trend with regard to population increase over this period?123456weeks 7DIFFERENTIAL CALCULUS (Chapter 21)629aipopulation growth rate increase in population = increase in time (240 ¡ 110) mice = (6 ¡ 3) weeks ¼ 43 mice per week population growth rate = (315 ¡ 50) mice (7 ¡ 0) weeksThe average rate of change between two points on the graph is the gradient of the chord between them.ii¼ 38 mice per week b The graph is increasing over the period by larger and larger amounts, so the population is increasing at an ever increasing rate.EXERCISE 21A.21 For the travel graph given alongside, estimate the average speed: a in the first 4 seconds b in the last 4 seconds c in the 8 second interval.7 6 5 4 3 2 1 time (s) 1 2 3 4 5 6 7 8 distance (m)2 The numbers of lawn beetles per m2 of lawn which are left surviving after various doses of poison are shown in the graph alongside. a Estimate the rate of beetle decrease when: i the dose increases from 0 to 10 g ii the dose increases from 4 to 8 g. b Describe the effect on the rate of beetle decline as the dose goes from 0 to 14 g.beetles (per m2 ) 40 30 20 10 2 4 6 8 10 12 14 dose (g)INSTANTANEOUS RATES OF CHANGEThe speed of a moving object such as a motor car, an aeroplane, or a runner, will vary over time. The speed of the object at a particular instant in time is called its instantaneous speed. To examine this concept in greater detail, consider the following investigation.630DIFFERENTIAL CALCULUS (Chapter 21)INVESTIGATION 2INSTANTANEOUS SPEEDWhen a ball bearing is dropped from the top of a tall building, the distance it has fallen after t seconds is recorded, and the following graph of distance against time obtained. In this investigation we will try to measure the speed of the ball at the instant when t = 2 seconds.D 80 60 40 20 2 F(2,¡20) t 4 6 M(4,¡80)chord curveThe average speed in the time interval 2 6 t 6 4 is distance travelled = time taken (80 ¡ 20) m = (4 ¡ 2) s = = 30 m s¡160 2m s¡1DEMOWhat to do: 1 Click on the icon to start the demonstration. F is the point where t = 2 seconds, and M is another point on the curve. To start with M is at t = 4 seconds. The number in the box marked gradient is the gradient of the chord FM. This is the average speed of the ball bearing in the interval from F to M. For M at t = 4 seconds, you should see the average speed is 30 m s¡1 . 2 Click on M and drag it slowly towards F. Copy and complete the table alongside with the gradient of the chord when M is at various times t. 3 Observe what happens as M reaches F. Explain why this is so. 4 When t = 2 seconds, what do you suspect the instantaneous speed of the ball bearing is? 5 Move M to the origin, and then slide it towards F from the left. Copy and complete the table alongside with the gradient of the chord when M is at various times t. 6 Do your results agree with those in 4? t 3 2:5 2:1 2:01 gradient of FMt 0 1:5 1:9 1:99gradient of FMDIFFERENTIAL CALCULUS(Chapter 21)631From the investigation you should have discovered that: The instantaneous rate of change of a variable at a particular instant is given by the gradient of the tangent to the graph at that point. For example, the graph alongside shows how a cyclist accelerates away from an intersection. The average speed over the first 8 seconds is 100 m = 12:5 m s¡1 . 8 sec Notice that the cyclist's early speed is quite small, but it increases as time goes by.distance (m) tangent P100 80 60 40 20 0 012345time (t sec) 6 7 8To find the instantaneous speed at any time instant, for example t = 4, we draw the tangent at the point and find its gradient.distance (m) 100 80 60 40 20 0 0 1 2 point where t=4This tangent passes through (2, 0) and (7, 40). ) the instantaneous speed at t = 4 = gradient of tangent =tangent at t = 4 5 6 7 8 time (t sec)PRINTABLE GRAPHS(40 ¡ 0) m (7 ¡ 2) s40 5=m s¡1= 8 m s¡134EXERCISE 21A.31 For each of the following graphs, estimate the rate of change at the point shown by the arrow. Make sure your answer has the correct units. a b4 3 2 1 1 t=1 time (s) 2 3 4 distance (m) 8 6 4 2 1 2 3 distance (km)t=3 time (h) 4c4 3 2 1profit (\$¡thousands) x = 30 number of items sold 10 20 30 40 50d120 100 80 60 40 20population of bat colonyt=5 time (weeks) 1 2 3 4 5 6 7 8 9 10632DIFFERENTIAL CALCULUS(Chapter 21)2 Water is leaking from a tank. The number of thousands of litres of water left in the tank after x hours is given in the graph alongside. a How much water was in the tank originally? b How much water was in the tank after 1 hour? c How quickly was the tank losing water initially? d How quickly was the tank losing water after 1 hour?volume (thousands of litres) 8 6 4 20.51.52.53.5 time (x hours)BWe have seen that the gradient of the chord AB measures the average rate of change of the function for the given change in x-values. The gradient of the tangent at point A measures the instantaneous rate of change of the function at point A. We say that in the limit as B approaches A, the gradient of the chord AB will be the gradient of the tangent at A.chord (secant)LIMITScurve BAtangentThe gradient of the tangent at x = a is defined as the gradient of the curve at the point where x = a, and is the instantaneous rate of change in f (x) with respect to x at that point.INVESTIGATION 3THE GRADIENT OF A TANGENTGiven a curve f (x), we wish to find the gradient of the tangent at the point (a, f(a)). For example, the point A(1, 1) lies on the curve f (x) = x2 . What is the gradient of the tangent at A?DEMOyf(x) = x 2A(1, 1) xDIFFERENTIAL CALCULUS(Chapter 21)633What to do: 1y f(x) = x 2 B(x, x 2) A(1, 1) xSuppose B lies on f (x) = x2 and B has coordinates (x, x2 ). a Show that the chord AB has gradient f (x) ¡ f (1) x2 ¡ 1 or . x¡1 x¡1b Copy and complete the table alongside: 2 Comment on the gradient of AB as x gets closer to 1. 3 Repeat the process as x gets closer to 1, but from the left of A. 4 Click on the icon to view a demonstration of the process. 5 What do you suspect is the gradient of the tangent at A?x 5 3 2 1:5 1:1 1:01 1:001Point B (5, 25)gradient of AB 6Fortunately we do not have to use a graph and table of values each time we wish to find the gradient of a tangent. Instead we can use an algebraic and geometric approach which involves limits.LIMIT ARGUMENTFrom the investigation, the gradient of AB = ) the gradient of AB =y = x2 B1 B2 B3 B4x2 ¡ 1 x¡1 (x + 1)(x ¡ 1) = x + 1 provided that x 6= 1 x¡1In the limit as B approaches A, x ! 1 and the gradient of AB ! the gradient of the tangent at A. So, the gradient of the tangent at the point A is mT = lim x2 ¡ 1 x!1 x ¡ 1x!1= lim x + 1, x 6= 1 =2As B approaches A, the gradient of AB approaches or converges to 2.Atangent at ALimit arguments like that above form the foundation of differential calculus. However, formal treatment of limits is beyond the scope of this course.634DIFFERENTIAL CALCULUS(Chapter 21)Example 3Self Tutorax!2Evaluate:lim x2bx!0lim5x + x x2cx!3limx ¡9 x¡32a x2 can be made as close as we like to 4 by making x sufficiently close to 2. ) lim x2 = 4.x!2b5x + x2 x!0 x x(5 + x) = lim x!0 x = lim 5 + x provided x 6= 0 limx!0cx2 ¡ 9 x!3 x ¡ 3 (x + 3)(x ¡ 3) = lim x!3 x¡3 1 = lim (x + 3) provided x 6= 3 limx!3=5=3+3 =6EXERCISE 21B1 Evaluate: a lim (x + 4)x!3 x!2 x!0bx!¡1lim (5 ¡ 2x)c lim (3x ¡ 1)x!4d lim 5x2 ¡ 3x + 2 g lim (x2 + 5)e lim h2 (1 ¡ h)h!0fx!¡1lim1 ¡ 2x x2 + 1hx!¡2lim4 x2 Evaluate the following limits by looking for a common factor in the numerator and denominator: x2 ¡ 3x 2h2 + 6h h3 ¡ 8h b lim c lim a lim x!0 h!0 h!0 x h hCTHE DERIVATIVE FUNCTIONWe have seen how the rate of change of a function at a particular instant is given by the gradient of the tangent at that point. We can hence describe a gradient function which, for any given value of x, gives the gradient of the tangent at that point. We call this gradient function the derived function or derivative function of the curve. If we are given y in terms of x, we represent the derivative function by `dee y by dee x'. If we are given the function f(x), we represent the derivative function by f 0 (x). We say this as `f dashed x'. dy . We say this as dxDIFFERENTIAL CALCULUS (Chapter 21)635INVESTIGATION 4THE DERIVATIVE OF y = x2The graphs below show y = x2 with tangents drawn at the points where x = ¡3, ¡2, ...., 3.y 8 6 4 2 -4 -3 -2 -1 y 8 6 4 2 -2 -1 D 1 2 x y = x2 8 6 4 2 E 1 2 3 4 y 8 6 4 2 1 2 3 4 x G y = x2 x x -4 -3 -2 -1 y y = x2 8 6 4 2 1 2 3 4 B y 8 6 4 2 x C -4 -3 -2 -1 y y = x2 F x y 8 6 4 2 xAy = x2y = x2y = x2What to do: 1 Use the shaded triangles to find the gradients of the tangents to y = x2 at the seven different points. Hence complete the following table: x-coordinate ¡3 gradient of tangent2 Use your table to help complete: &quot;the gradient of the tangent to y = x2 at (x, y) is m = ::::&quot;DEMO3 Click on the icon to check the validity of your statement in 2. Click on the bar at the top to drag the point of contact of the tangent along the curve. You should have found that the gradient of the tangent to y = x2 at the point (x, y) is given by 2x. dy So, y = x2 has the derivative function = 2x, dx or alternatively, if f (x) = x2 then f 0 (x) = 2x.636DIFFERENTIAL CALCULUS (Chapter 21)INVESTIGATION 5THE DERIVATIVE OF y = xnDERIVATIVE DETERMINERIn this investigation we seek derivative functions for other functions of the form y = xn where n 2 Z .What to do: 1 Click on the icon to run the derivative determiner software. 2 Choose the function y = x. a Use the software to complete the table: x dy dx b Graph ¡3 ¡2 ¡1 0 1 2 3x dy dxdy against x. dx Hence predict a formula for the dy . derivative function dx3 Repeat 2 for the functions: a y = x3 b y = x4 c y = x¡1 d y = x¡2 Hint: When x = 0, the derivatives of both y = x¡1 and y = x¡2 are undefined. 4 Use your results from 2 and 3 to complete: Function x x2 x3Derivative function 2xx4 x¡1 dy 5 Predict the form of where y = xn . dx x¡2You should have discovered that:If y = xn thendy = nxn¡1 . dxDISCUSSIONDoes the rule &quot;if y = xn then dy = nxn¡1 &quot; work when n = 0? dxDIFFERENTIAL CALCULUS (Chapter 21)637FINDING THE DERIVATIVE FUNCTION USING LIMITSConsider a general function y = f (x) where A is (x, f(x)) and B is (x + h, f (x + h)).y f(x + h) B y = f(x)The chord AB has gradient = = f (x + h) ¡ f (x) x+h¡x f (x + h) ¡ f (x) : hf(x)Ah x x+h xIf we now let B approach A, then the gradient of AB approaches the gradient of the tangent at A.So, the gradient of the tangent at the variable point (x, f (x)) is the limiting value of f (x + h) ¡ f (x) h as h approaches 0, or lim f (x + h) ¡ f (x) . hh!0This formula gives the gradient of the tangent for any value of the variable x. Since there is only one value of the gradient for each value of x, the formula is actually a function. The gradient or derivative function is defined as f 0 (x) = lim f (x + h) ¡ f (x) . hh! 0When we evaluate this limit to find the derivative function, we call this the method of first principles. Example 4 Use the definition of f 0 (x) to find the gradient function of f (x) = x2 . f 0 (x) = limh!0Self Tutorf (x + h) ¡ f (x) h (x + h)2 ¡ x2 h x2 + 2hx + h2 ¡ x2 h h(2x + h) h fas h 6= 0g= lim = lim = limh!0h!0h!0 h!0= lim (2x + h) = 2x638DIFFERENTIAL CALCULUS (Chapter 21)EXERCISE 21C1 Find, from first principles, the gradient function of f (x) where f (x) is: a x b 5 c x3 d x4 Hint: (a + b)3 = a3 + 3a2 b + 3ab2 + b3 (a + b)4 = a4 + 4a3 b + 6a2 b2 + 4ab3 + b4 2 Find, from first principles, f 0 (x) given that f(x) is: a 2x + 5 b ¡x + 4 c x2 ¡ 3x d 2x2 + x ¡ 1DSIMPLE RULES OF DIFFERENTIATIONDifferentiation is the process of finding a derivative or gradient function.There are a number of rules associated with differentiation. These rules can be used to differentiate more complicated functions without having to resort to the tedious method of first principles.INVESTIGATION 6SIMPLE RULES OF DIFFERENTIATIONIn this investigation we attempt to differentiate functions of the form cxn where c is a constant, and functions which are a sum or difference of terms of the form cxn . What to do: 1 Find, from first principles, the derivatives of: a x2 b 4x2 c x3 d 2x3 Hence copy and complete: a f (x) = x2 + 3x Hence copy and complete: &quot;If f (x) = cxn , then f 0 (x) = ::::&quot; b f (x) = x3 ¡ 2x2 &quot;If f (x) = u(x) + v(x) then f 0 (x) = ::::&quot;2 Use first principles to find f 0 (x) for:We have now determined the following rules for differentiating: Function a constant xnf (x) a xnf 0 (x) 0 nxn¡1 anxn¡1 u0 (x) + v0 (x)a constant multiple of xn multiple termsaxn u(x) + v(x)DIFFERENTIAL CALCULUS (Chapter 21)639Using the rules we have now developed we can differentiate sums of powers of x. For example, if f(x) = 3x4 + 2x3 ¡ 5x2 + 7x + 6 then f 0 (x) = 3(4x3 ) + 2(3x2 ) ¡ 5(2x) + 7(1) + 0 = 12x3 + 6x2 ¡ 10x + 7 Example 5 Find f 0 (x) for f (x) equal to: a 5x3 + 6x2 ¡ 3x + 2 a ) b 7x ¡ 4 3 + 3 x x c x2 + 4x ¡ 5 xSelf Tutorf (x) = 5x3 + 6x2 ¡ 3x + 2 f 0 (x) = 5(3x2 ) + 6(2x) ¡ 3(1) = 15x2 + 12x ¡ 3 4 3 + 3 x x = 7x ¡ 4x¡1 + 3x¡3bf (x) = 7x ¡ )f 0 (x) = 7(1) ¡ 4(¡1x¡2 ) + 3(¡3x¡4 ) = 7 + 4x¡2 ¡ 9x¡4 4 9 =7+ 2 ¡ 4 x x x2 + 4x ¡ 5 f (x) = x x2 5 = +4¡ x x = x + 4 ¡ 5x¡1 f 0 (x) = 1 + 5x¡2Remember that 1 = x¡n . xnc)Example 6 Find the gradient function of f (x) = x2 ¡Self Tutor4 . x Hence find the gradient of the tangent to the function at the point where x = 2. f (x) = x2 ¡ 4 x = x2 ¡ 4x¡1 Now f 0 (2) = 4 + 1 = 5. So, the tangent has gradient = 5.) f 0 (x) = 2x ¡ 4(¡1x¡2 ) = 2x + 4x¡2 4 = 2x + 2 x640DIFFERENTIAL CALCULUS (Chapter 21)EXERCISE 21D1 Find f 0 (x) given that f (x) is: a x3 d x2 + x g x3 + 3x2 + 4x ¡ 1 j 2x ¡ 3 x2 b 2x3 e 4 ¡ 2x2 h 5x4 ¡ 6x2 k x3 + 5 x c 7x2 f x2 + 3x ¡ 5 i 3 ¡ 6x¡1 l x3 + x ¡ 3 x2 Suppose f (x) = 4x3 ¡ x. Find: a f 0 (x) 3 Suppose g(x) = a g 0 (x) b f 0 (2) x2 + 1 . Find: x b g0 (3) c f 0 (0)c g0 (¡2) 8 at x = 9 x2 d y = 2x ¡ 5x¡1 at x = 2 b y= f y= x3 ¡ 4x ¡ 8 at x = ¡1 x24 Find the gradient of the tangent to: a y = x2 at x = 2 c y = 2x2 ¡ 3x + 7 at x = ¡1 e y= 5 Find x2 ¡ 4 at x = 4 x2dy for y = (3x + 1)2 . dx Hint: Start by expanding the brackets.Example 7 If y = 3x2 ¡ 4x, find As y = 3x2 ¡ 4x, dy is: dx dy and interpret its meaning. dxSelf Tutordy = 6x ¡ 4. dx² the gradient function or derivative function of y = 3x2 ¡ 4x, from which the gradient at any point can be found ² the instantaneous rate of change in y as x changes.6dy 3 , find and interpret its meaning. x dx b The position of a car moving along a straight road is given by S = 2t2 + 4t metres dS where t is the time in seconds. Find and interpret its meaning. dt c The cost of producing and selling x toasters each week is given by dC and interpret its meaning. C = 1785 + 3x + 0:002x2 dollars. Find dx a If y = 4x ¡DIFFERENTIAL CALCULUS (Chapter 21)641Ey = f(x) tangent gradient m point of contactTANGENTS TO CURVESConsider a curve y = f (x).If A is the point with x-coordinate a, then the gradient of the tangent at this point is f'(a) = m.The equation of the tangent is y ¡ f (a) = f 0 (a) fequating gradientsg x¡a or y ¡ f (a) = f 0 (a)(x ¡ a).A(a, f(a))x=aAlternatively, the equation of the tangent at the point A(a, b) is y¡b = f 0 (a) x¡a y ¡ b = f 0 (a)(x ¡ a).orThe equation can also be written in the form y = mx + c: For example: If f (x) = x2 then f 0 (x) = 2x. At x = 3, m = f 0 (3) = 6, so the tangent has gradient 6. Since f (3) = 9, the tangent has equation y ¡ 9 = 6(x ¡ 3) or y = 6x ¡ 9. You can also find the equations of tangents at a given point using your graphics calculator. For assistance with this, consult the graphics calculator instructions at the start of the book. Example 8Self TutorFind the equation of the tangent to f (x) = x2 + 1 at the point where x = 1.y f(x) = x 2 + 1Since f (1) = 1 + 1 = 2, the point of contact is (1, 2). Now f 0 (x) = 2x, so m = f 0 (1) = 2 ) the tangent has equation1(1, 2) xy¡2 =2 x¡1 which is y ¡ 2 = 2x ¡ 2 or y = 2x.642DIFFERENTIAL CALCULUS (Chapter 21)Example 9 Find the equations of any horizontal tangents to y = x3 ¡ 12x + 2. Since y = x3 ¡ 12x + 2, dy = 3x2 ¡ 12 dx so 3x2 ¡ 12 = 0 ) 3(x2 ¡ 4) = 0 3(x + 2)(x ¡ 2) = 0 ) x = ¡2 or 2Self TutorHorizontal tangents have gradient 0, ) When x = 2, y = 8 ¡ 24 + 2 = ¡14 When x = ¡2, y = ¡8 + 24 + 2 = 18 ) the points of contact are (2, ¡14) and (¡2, 18)) the tangents are y = ¡14 and y = 18.EXERCISE 21E1 Find the equation of the tangent to: a y = x2 at x = 4 c y = 3x¡1 at x = ¡1 e y = x2 + 5x ¡ 4 at x = 1 g y = x3 + 2x at x = 0 i y = x + 2x¡1 at x = 2 2 Find the coordinates of the point(s) on: a f (x) = x2 + 3x + 5 where the tangent is horizontal b f (x) = x3 + x2 ¡ 1 where the tangent has gradient 1 c f (x) = x3 ¡ 3x + 1 where the tangent has gradient 9 d f (x) = ax2 + bx + c where the tangent has zero gradient. b y = x3 at x = ¡2 4 d y = 3 at x = 2 x f y = 2x2 + 5x + 3 at x = ¡2 h y = x2 + x¡1 at x = 0 j y= x2 + 4 at x = ¡1 xFTHE SECOND DERIVATIVEGiven a function f (x), the derivative f 0 (x) is known as the first derivative. The second derivative of f (x) is the derivative of f 0 (x), or the derivative of the first derivative. We use f 00 (x) or y00 or d2 y to represent the second derivative. dx2DIFFERENTIAL CALCULUS (Chapter 21)643f 00 (x) is read as &quot;f double dashed of x&quot;. µ ¶ d2 y d dy is read as &quot;dee two y by dee x squared &quot;. = dx2 dx dxTHE SECOND DERIVATIVE IN CONTEXTMichael rides up a hill and down the other side to his friend's house. The dots on the graph show Michael's position at various times t.t=0 Michael's place t=5 t = 15 t = 17 t = 19 friend's houseDEMOt = 10The distance Michael has travelled at various times is given in the following table: Time (t min) Distance travelled (s m) 0 0 2:5 5 7:5 10 12:5 15 17 19498 782 908 989 1096 1350 1792 2500The model s ¼ 1:18t3 ¡ 30:47t2 + 284:52t ¡ 16:08 metres fits this data well, although the model gives s(0) ¼ ¡16:08 m whereas the actual data gives s(0) = 0. This sort of problem often occurs when modelling from data. A graph of the data points and the model is given below:2500 2000 1500 1000 500 5 10 15 t s s = 1.18t 3 - 30.47t 2 + 284.52t - 16.08Nowds ¼ 3:54t2 ¡ 60:94t + 284:52 metres per minute is the instantaneous rate of change dt in displacement per unit of time, or instantaneous velocity. The instantaneous rate of change in velocity at any point in time is Michael's acceleration, µ ¶ d2 s d ds = 2 is the instantaneous acceleration. so dt dt dt In this case d2 s ¼ 7:08t ¡ 60:94 metres per minute per minute. dt2We see that when t = 12, s ¼ 1050 m, ds ¼ 63 metres per minute, dt and d2 s ¼ 24 metres per minute per minute. dt2644DIFFERENTIAL CALCULUS (Chapter 21)Example 10Self Tutor3 . xFind f 00 (x) given that f (x) = x3 ¡ Now f (x) = x3 ¡ 3x¡1 ) ) f 0 (x) = 3x2 + 3x¡2 f 00 (x) = 6x ¡ 6x¡3 6 = 6x ¡ 3 xEXERCISE 21F1 Find f 00 (x) given that: a f (x) = 3x2 ¡ 6x + 2 c f (x) = 2 Find 2 ¡ 3x x2 b f (x) = 2x3 ¡ 3x2 ¡ x + 5 d f (x) = 2x ¡ x3 x2d2 y given that: dx2 b y = x2 ¡a y = x ¡ x35 x2c y=4¡x x3 The position of a bicycle rider from his starting point is given by s = 2t3 ¡ 15t2 + 100t metres, where t is the time in minutes. a Find his position after: i 5 minutes ii 10 minutes. b For t = 5, find the cyclist's instantaneous: i velocity ii acceleration. 4 The position of Kelly from her home is given by s = 10t3 ¡ 100t2 + 200t metres, where t is the time in minutes, 0 6 t 6 7:24. a When is Kelly at home? b When is Kelly furthermost from her home? How far is she from home at this time? c For t = 4, find Kelly's: i position ii instantaneous velocity iii instantaneous acceleration. Explain what these results mean.DIFFERENTIAL CALCULUS (Chapter 21)645REVIEW SET 21A1 The total number of televisions sold over many months is shown on the graph alongside. Estimate the rate of sales: a from 40 to 50 months b from 0 to 50 months c at 20 months.8000 6000 4000 2000 time (months) 10 20 30 40 50 number of televisions sold2 Use the rules of differentiation to find f 0 (x) for f (x) equal to: a 7x3 b 3x2 ¡ x3 c (2x ¡ 3)2 a f 0 (x) 1 x d b f 0 (2) 7x3 + 2x4 x2 c f 0 (0).3 Consider f(x) = x4 ¡ 3x ¡ 1. Find:4 Find the equation of the tangent to y = ¡2x2 at the point where x = ¡1. 5 Find f 00 (x) for: a f (x) = 3x2 ¡ b f (x) = (x + 4)26 Find the derivative of f(x) = 3x ¡ 1 from first principles.REVIEW SET 21B1 Consider the function f (x) = x2 +2x, with graph shown alongside. a Find the gradient of the line through (1, 3) and the point on f (x) with x-coordinate: i 2 ii 1:5 iii 1:1 0 b Find f (x). c Find the gradient of the tangent to f (x) at (1, 3). Compare this with your answers to a. 2 Find dy for: dx b y= x3 ¡ x x2 c y = 2x + x¡1 ¡ 3x¡2y = x2 + 2x y 8 6 4 2 -1 -4 -3 -2 -2 1 2 3 (1, 3) xa y = 3x2 ¡ x43 Find the equation of the tangent to y = x3 ¡ 3x + 5 at the point where x = 2.646DIFFERENTIAL CALCULUS (Chapter 21)4 Find all points on the curve y = 2x + x¡1 which have a tangent parallel to the x-axis. 5 If f (x) = 7 + x ¡ 3x2 , find: a f (3) b f 0 (3) c f 00 (3). 6 Find, using the limit method, the derivative function of f (x) = x3 ¡ 2x.REVIEW SET 21C1 Use the rules of differentiation to find f 0 (x) for f (x) equal to: 4 1 ¡ 2 b 2x¡3 + x¡4 c a x4 + 2x3 + 3x2 ¡ 5 x x 2 Find the gradient of f (x) at the given point for the following functions: a f (x) = x2 ¡ 3x, at x = ¡1 2 c f (x) = x + , at x = 3 x 3 Find the equation of the tangent to y = b f (x) = ¡3x2 + 4, at x = 2 d f (x) = x3 ¡ x2 ¡ x ¡ 2, at x = 012 at the point (1, 12). x2 4 Sand is poured into a bucket for 30 seconds. After t seconds, the weight of sand is S(t) = 0:3t3 ¡ 18t2 + 550t grams. Find and interpret S 0 (t). 5 Find d2 y for: dx2 6 Evaluate: a lim x2 + 3x!1a 7x3 ¡ 4xb 2x2 + 3¡x x5 x c lim h2 ¡ 3h hbx!¡4limh!07 Find, from first principles, the derivative of f (x) = x2 + 2x.`

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