University:
Streetsville Secondary School
Course:
MCR3U (Functions)
Academic year:

2024
Views:

16

Pages:

12
Author:

Andy N.

5.5/5.6 Application of Sine and Cosine Functions Worksheet #2 MCRU Jensen SOLUTIONS 1) A motion sensor recorded the motion of a child on a swing. The data was graphed, as shown. a) Find the max and min values. max = 2.25 min = 0,25 b) Find amplitude a = 2-25-0.25 = 1 c) Determine the vertical shift of the function. C= Max-amp = 2.25-1 = 1.25 d) Find the period of the function period = 8-3=5 e) Determine the phase shift, if the motion were to be modelled using a sine function. riskg midline is l" las to the left of the max. ão d= 3-1.25 = 1.75 2) The height of the blade of a wind turbine as it turns through an angle of 0 is given by the function h(0) - 8.5 sin (0 + 180º) + 40, with height measured in metres. a) Find the maximum and minimum positions of the blade. Max = 40+8-5 min = 40-8.5 = 44.5m =31.5m b) Explain what the value of 40 in the equation represents. verticad shift; of height of the center of the turbine. c) Explain what the value of the amplitude represents. Leigth of the blade. d) Sketch the function over two cycles. 3) The height, h, in meters, of the tide in a given location on a given day at t hours after midnight can be modeled using the sinusoidal function h(t) = 5 sin[30(t - 5)] + 7. a) Find the max and min values for the depth of water. max: 7+5 =12 m Min = 7-5 = 2 m b) What time is high tide? What time is low tide? rising midline for sine furchen is moved 5 unts right, Mars 5+ k=5+%=8 (8:00 an) Because of the 12 hour A c) What is the depth of the watêr at 9:00 am? 1, there is also a Min = 5 - 12 = 5-10 = 2 (2:00 am) h19) = 5 sin{3019-5)] +7 = 5 Sin(120) +7 = 113 m d) Find all the times during a 24-h period when the depth of the water is 3 meters. 3=5 51(30/+-5))+7 t- 5=-1.77 .: Sin (20 (t-s)] t= 3.23 30(4-5) = sla" (3) about 3 hours 14 minutes •53.13 This is 1 hour 14 wins ofter the min; do the height 3 0 will be the seune 1 low and My wins before Hat; at 12:46 au. ON BACK h(3) at 12:46 an and p.m. (because of 12 how perriod) and 3:14 am and 3:14 p.m. 4) The population, P, of a lakeside town with a large number of seasonal residents can be modeled using the function P(t) = 5000 sin[30(t - 7)] + 8000, where t is the number of months after New Year's Day. a) Find the max and min values for the population over a whole year. MaX = 8000 + 5000 = 13000 MiR = 8000 - 5000 = 3000 b) When is the population a maximum? When is it a minimum? max of sine at 1 90=3 but you must shift T to the right. min at - 40: -90 = -3 do maX 10 months after New Years and min 4 morths astes New tears c) What is the population on September 30th? P(9) = 5000 sin {30 (9-7) 3+8000 = 5000 Sin (60) +8000 = 12330 5) The population of prey in a predator-prey relation is shown. Time is in years since 1985. a) Determine the max and min values of the population, to the nearest 50. Use these to find the amplitude. Max: 850 a = 850-250 міл = 250 a = 300 b) Determine the vertical shift, c. C = Max - amp C = 850.300 c: 550 c) Determine the phase shift, d. d= 0 d) Determine the period. Use the period to determine the value of k. Period = 8-2 = 6 K= 360 • = 60 e) Model the population versus time with a sinusoidal function. P = 300 sin(60t) + 550 6) The number of millions of visitors that a tourist attraction gets can be modeled using the equation y = 2.3sin [30(x + 1)] + 4.1, where x = 1 represents January, x = 2 represents February, and so on. a) Determine the period of the function and explain its meaning. Period: 350 = 12 12 maths. b) Graph the function for 12 months. Max = 4.1 + 2.3 = 6.4 MIn = 4.) - 2.3 = 1.8 Risug midline at - b) Graph the function for 12 months. Max = 4.1 + 2,3 = 6.4 Min = 4.1 - 2.3 = 1.8 Rising midline at - l Fisst max at -1 + 9 = -1+3 = 2 first min is 30 = 6 after max. c) Which month has the most visitors? math 2: February d) Which month has the least visitors? Math 8: August Answers 2) apatinum 225, minimum 0.25 c) vertical shift up 1.25 8) horizontal shift 175 to the right. 2) a) maximum 48.5, minimum 31.5 b) The height of the center of the turbine c) The amnlitude of 8 5 conrocontethe lonath oftho blodo l 3) a) From the equation, c = 7 and a = 5, so the function has a midline value of 7 and an amplitude of 5. The maximum height is 12 m and the minimum height is 2 m. b) From the equation, k = 30 and d = 5, so the period is 12 h and the phase shift is 5 h right. The first midline value occurs at 5:00 a.m. The first maximum occurs one-quarter period, or 3 h after this, at 8:00 a.m. The previous minimum is 3 h prior to 5:00 a.m., at 2:00 a.m. Because of the 12-h period, there will also be a maximum at 8:00 p.m. and a minimum at 2:00 p.m. c) 11.3 m d) The solution gives a time of approximately 3:14 a.m. This time is 1 h 14 min after the first minimum so the depth should also occur 1 h 14 min before 2:00 a.m, at 12:46 a.m. Because of the 12-h period, the depth will also occur at 12:46 p.m. and 3:14 p.m. 4) a) From the equation, c = 8000 and a = 5000, so the function has a midline value of 8000 and an amplitude of 5000. The maximum population is 13 000 and the minimum population is 3000. b) From the equation, k = 30 and d = 7, so the period is 12 months and the phase shift is 7 months right. The initial midline value occurs at t = 7. The maximum occurs 3 months later at t = 10 (October) and the minimum 3 months earlier at t = 4 (April). c) 12 330 5) a) From the graph the maximum population is approximately 850 animals and the minimum population is approximately 250 animals.The amplitude is approximately 300 animals, so a = 300. b) The vertical shift is the maximum value minus the amplitude, so c = 550. c) The midline intersects the graph at t = 0 so no horizontal shift is necessary, so d = 0. d) The pattern repeats every 6 years, so the period is 6 years. k=60. e) A sine function that models the population of prey, N, with respect to time, t, is N = 300 sin 60t + 550. 6) a) 12 months c) February d) August