MATHEMATICAL METHODS UNITS 3 AND Applications of modelling periodic behaviour

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1 MATHEMATICAL METHODS UNITS 3 AND 0 Applications of modelling periodic behaviour The sine and cosine functions are used to model periodic behaviour, i.e. behaviour that repeats itself in a cycle. If y = a sin n(x b) + c Amplitude = a [or y = a cos n(x b) + c], then Period = n 2 Translation = b units in the positive x-direction Vertical translation = c units in the positive y-direction. Example 1 It is suggested that the height h(t) metres of a tide above the mean sea level can be approximated with the rule h(t) = sin t, where t is the number of hours after midnight. 6 A Draw the graph of y = h(t) for 0 t 2 h hours - B When was high tide? High tide occurs when h(t) =. This implies that sin t 6 5 t =, = 1, because this is when sin is a maximum. t = 3, 15 hours. i.e. the high tide occurs at 0300 and 1500 hours. C What was the height of the high tide? The high tide height was metres above the mean height. D What was the height of the tide at 8 am?

2 h(8) = sin 8 6 The water was 2 = sin 3 3 = 2 = metres below the mean height at 8 am. E A boat can only cross the harbour bar when the tide is at least 1 metre above the mean sea level. When could a boat cross the harbour? i.e. h(t) = 1 sin = 1 3 sin 3 = 0.25 From the unit circle, sin is positive in the first and second quadrants. Make sure your calculator is in radians, and solve this equation, by finding sin -1 = This will give you an answer of You then need to find the other times that this is true, i.e. when it is , , and t = , 2.889, , t = 0.82, 5,5176,,2, You now need to convert this into hours and minutes, by multiplying by 60. t = 00:29, 05:31, :29, 17:31. We need to look at the graph to see what this means. h hours - Thus the boat can pass across the harbour bar between 00:29 and 05:31 and between :29 and 17:31

3 Example 2 The height of the sea level (above a fixed point) on a given day varies over time according to the effect of the tides. The height, y cm, is given by the equation y = 2 t 5 sin + 1, 6 where t represents the number of hours after midnight on the 21 st September. a) Sketch, on the set of axes below, a graph representing the height of the sea level on the 21 st September from midnight to midday y (midday) t -1-2 Ans. b) State (i) the period of the function (ii) the amplitude of the function Ans (i) From the graph, the period is hours. This also comes from period = the period is 2 6 = hours. 2, where n = n 6, Ans (ii) from the equation the amplitude is 5 2 cm.

4 Example 3 In the depths of a jungle in Brazil, an Indian tribe keeps its treasure in a stone chest which is on a rock ledge on the banks of the piranha-infested Amazon River. The chest cannot be moved and can only be opened by using a metal key which is kept in a hole in the rock just near the chest. Because the piranha eat human flesh, the key can only be safely taken from the hole when the water level in the river falls below the level of the key. The ledge that the chest is on is 6. metres above the river bed, and the hole that the key is in is 6.1 metres above the river bed. chest key d metres 6. m 6.1 m Tasmania Jones, the intrepid adventurer, is keen to pillage the treasure. He is aware that the depth of water in the river could be modelled by the relation t d(t) = cos where d metres is the depth at time t hours after noon on a given Monday. 1 a) Write down the minimum and maximum depths of the river. Ans. The amplitude of the cos function is.0 m. the minimum height is = 6.0 and the maximum is = 1.0 m 6.0m and 1.0 m (You must include units) b) Show that the depth of water is the same at noon every Monday. Ans. 2 period = n = 2 1 = 28 hours. Time from Monday to Monday is 2 7 = 168 hours. 168 = 6 there are 6 complete cycles every 7 days. 28 same depth at noon every Monday. Alternative solution.

5 d(0) = 1m d(168) = 1m d(336) = 1m same depth every Monday. This is one of the show that type of question, and it is insufficient to just consider one or two cases. c) Find the day and time when the water first reaches its minimum level. t Ans. The minimum level is 6.0m 6.0 = cos 1 t -.0 =.0cos 1 t -1.0 = cos 1 t The first time this occurs is when = t = 1 hours. This has not yet answered the question, the time and day is Tuesday 2am. d) Find the depth of the river at (i) noon on Monday (t = 0) (ii) 2am on Tuesday (t = 1) t Ans. (i) By substituting t = 0 into d(t) = cos (ii) Sub t = 1 the previous question) 1 becomes d(0) = cos 0 = 1 m. d(1) = cos( ) 1 = 6.0m (This should be obvious from e) On the set of axes provided below, sketch the graph of the depth of the river versus time showing three complete cycles of the graph. Ans. 3 cycles will be 3 28 =8 hours f) Determine the first time after noon on a Monday when

6 (i) the chest is completely uncovered. (ii) the key is able to be taken from the rock t Ans (i) 6. = cos 1 t -3.6 =.0cos 1 t -0.9 = cos 1 t = 2.69, t =. Midnight on Monday. 1 t Ans (ii) 6.1 = cos 1 t -3.9 =.0cos 1 t = cos 1 t = 2.92, t = 13. 1am on Tuesday. 1 g) Find out the length of time that Tasmania Jones will have to try to pillage the chest, by taking the key from the rock, stealing the treasure, and returning the key, in a futile attempt to avoid detection. State your answer in minutes. The key is exposed when the water level drops to 6.1 metres. Its lowest point is 6.0 metres. If it is at 6.1 m at 1am, and 6.0m at 2am (from before), then it will take another hour to return to 6.1 m. to carry out this dastardly deed Tasmania Jones has from 1am to 3am, i.e. 0 minutes. Unfortunately for Tasmania Jones, the Indians capture him before he is able to steal the loot. They tie him by a vine 20 metres above the bottom of the river at pm on a Tuesday and slowly lower him towards the water at a rate of 1 metre per hour. h) draw a graph, on the same set of axes as part (e), showing Tasmania s height above the bottom of the river versus time. (i) Using your graph, determine the day and time at which Tasmania Jones will first touch the water of the Amazon river. This is best done on the calculator, to find the intersection of the two equations.

7 The equation for the straight line is y = 50 x. From the calculator t = 2 hours. 6am Wednesday Example The position x of a particle at time t is given by the equation x = 2cos2(t - 2 ) + 1. (i) Find the position of the particle when t = 0 Ans. p(0) = 2cos2(0-2 ) + 1 = 2cos + 1 = = -1 (ii) What is the maximum value of x? Ans. The maximum value occurs when the cos function is a maximum. This cos function has a maximum of 2. the maximum value for x = = 3. (iii) What is the minimum value of x? Ans. The minimum value occurs when the cos function is a minimum. This cos function has a minimum of -2. the minimum value for x = = -1. (iv) At what time did the particle first reach the position x = 0. Give your answer to one decimal place. Ans. x = 2cos2(t - 2 ) + 1 becomes 0 = 2cos2(t - 2 ) = 2cos2(t - 2 ) -0.5 = cos2(t - 2 ) 2(t - 2 ) = - 3 or + 3 2t - = 2 3 2t = 5 3 t = 5 6 or 6 or 3 or 7 3 = 3 Use t = 6 the first time, t = (we don t know the units, so we can t include them) (v) Sketch the graph of x for 0 t 2

8 Example 5 Two pistons A and B move backwards and forwards in a cylinder as shown. A B O x y The distance x centimetres of the right hand end of piston A from the point O at time t seconds is modelled by the formula x = 3sin(2t) + 3 and the distance y centimetres of the left hand end of piston B from the point O at time t seconds is modelled by the formula y = 2sin 3 t + 8. The pistons are set in motion at time t = 0. a) (i) State the amplitude of the motion of piston A. (ii) Show that the maximum and minimum values of x are 6 and 0 respectively (iii) Show that when t =, the right hand end of piston A is at its maximum distance from O. (iv) Find the next four t values (t > ) fro which x = 6 Ans. (i) From the equation the amplitude of piston A is 3. This comes from x = 3sin. Ans (ii) You could either draw a graph, or do a calculation to show your answer. The word show effectively means prove, so just quoting the answer is not sufficient. If you use your calculator to sketch the graph, you must draw a graph to show your answer. The other way to do it is to explain the equation. x = 3sin(2t) + 3 means that the amplitude is 3. Since the graph has had a translation of 3 in the y- direction. The range of 3sin(2t) = [-3, 3], the translation changes this to [0, 6]. Since 1 sin(2t) 1 3 3sin(2t) 3 0 3sin(2t) minimum is 0, maximum is 6 You need to make this last statement to answer the question. Ans (iii) x = 3sin(2t) + 3 becomes x = 3sin( 2 ) + 3 when t =.

9 sin 2 = 1 x = x = 6 metres. This is the maximum value of x, from previous question. Ans. (iv) The first answer is, the next four answers must be 5, 9, 13 17, from the graph. The algebraic solution is x = 3sin(2t) + 3 becomes 6 = 3sin(2t) = 3sin(2t) 1 = sin(2t) So the next four solutions are 5, 9, 13 17, (= c, c, c, c ) 2t = 2, 5 2, 9 2, 13 17, 2 2 t =, 5, 9, 13 17, b) (i) The maximum value of y is 10. State the minimum value of y. (ii) 7 Show that y attains its minimum value when t = (iii) Ans. (i) y = 2sin t Find all other values of t, 0 t, for which y attains its minimum value The minimum occurs when sin t Ans(ii) If y = 6 then y = 2sin t y = = becomes 6 = 2sin t 3 = , -2 = 2sin 3 t -1 = sin 3 t 3 3 t = 2 3 3t = t = t = 7 The alternative solution is found by substituting t = into the function.

10 y = 2sin 3 t + 8 becomes y = 2sin ( 7 ) = 2sin( 3 2 ) + 8 = = 6. Ans.(iii) The best way to do this is to sketch the graph. 2 2 The period of the function is given by period =, here n = 3, period = n 3 2 The minimums occur at intervals of 3 the next minimums occur when t = 8 = = 15, 15 8 and + = , and + = 31 and 39 7, 15, 23, 31, 39 7, (= c, c, c, c,.305 c )

11 c) On the set of axes below, draw the graph of x(t) and y(t). y(t) = 2sin 3 t + 8 x = 3sin(2t) + 3 d) (i) State the time when the pistons first touch each other. (ii) How many seconds are there between the first and second times the pistons touch? Ans. (i) From the solutions, they touch when y is at its second minimum and when x is at its first maximum 5 13 Ans (ii) The second time they touch is when t = the time between is 2 = 2.28 secs. Your answer must be in seconds. e) Let T n seconds be the time at which the pistons meet for the nth time. Then T n = a + bn, where a and b are constants. Find the values of a and b. Ans. 5 They meet at when n = They meet at when n = They meet at when n = 3 etc. The change from n =1 to n =2, involves an increase of 2. b = When n = 1, = a + 2. a = - 2 = a = and b = 2 T n = +2n

12 f) At what time is the right hand end of piston A first cm from O? Give your answer to the nearest one-hundredth of a second. x = 3sin(2t) + 3 becomes = 3sin(2t) = 3sin(2t) 3 1 = sin(2t) t = 2 1 sin -1 ( 3 1 ) t = = 0.17 secs (correct to 2 dp) g) (i) Find the average rate of change of position with respect to time (average speed) of piston B in the first 0.2 seconds. (ii) Use calculus to find the speed of piston B at time t = 0.2. Ans.(i) The average rate of change is given by y(0.2) y(0) = 2 si n3(0.2) 8 (2sin(3(0) ) 8) 0.2 2sin sin( 0.785) = = = 0.2 = cm/s dy Ans (ii) = 6cos dt 3 t at t = 0.2 dy = 6cos dt = cm/s.

F.IF.C.7: Graphing Trigonometric Functions 4

F.IF.C.7: Graphing Trigonometric Functions 4 Regents Exam Questions www.jmap.org Name: 1 In the interval 0 x 2π, in how many points will the graphs of the equations y = sin x and y = 1 2 intersect? 1) 1 2) 2 3) 3 4) 4 3 A radio wave has an amplitude