Question

(+20) In a tidal river, the time between high tide and low tide is 6 hours. At high tide the depth of the water is 18 feet, while at low tide the depth is 6 feet. Assume the water depth is a trigonometric function of time. Assume there is a high tide at noon. a. (+4) Fill in the table for one period assuming a high tide at 12:00 noon: Estimate the t, time elapsed in h, depth of the time of day hours river in feet Noon - 12:00 0 15

Answer 1

7 Time between high tide and low ticle=6 hr. high tide height = 18 feet low tide height = 60 7 Time of the day to time by depth of elapsed Chr) the rines. Noon = 12:00 evening = 6.00 midnige (= 12:00 morning - 6:00 woon = 12:00 18 24

D. Height in meter feet w=2n H = 60 2 Period=labour of 6 12 18 24 30 36 42 48 I sketched that. 4 periods. time in hour Sorry for ề you H= f(t) = 12 + 0 6 cos (25 t) He fet) = 12+ 6 Los (M/Gt) where t is in hour. Scannea with

So y = 4(4) =15 15 = 12 +6 cos (ott) 3= 6 cos (7) cos( 176 t = 1/2 = cos (773) t = 42 sail al So it can no. 2:00 18 to ( If you t= all 2 draw Ct in a graph you can obtain th]

Graph for 12+6*cos(TT/6*x) x: 2.00731284 y: 14.980082 + 1 -20 -15 -10 5 10 15 20 More info