a) period = 2pi/.5 = 12.663 hrs
so we need the height to be 2.5
5 + 4.6sin(.5t)= 2.5
sin(.5t) = (2.5-5)/4.6 = -.543478
make sure your calculator is set to radians and take
inverse sine(+ .543478)
to get .574575 radians
But, the sine was negative so the "angle" must have been in quadr III or quadrant IV
so .5t = pi + .574575 or .5t = 2pi - .574575
t = 7.432336 or t = 11.41722
or t = 7:26 am or t = 11:25 am
remember that midnight was t=0
so the water was below 2.5 m between
7:25 am and 11:25 am
and the period was 12.663 hrs or 12:40 hrs
so the next time the water will go below 2.5 m
= 7:25 + 12:40 = 17:65 = 18:05 or 6:05 pm
For a boat to float in a tidal bay, the water must be at least 2.5 meters deep. The depth of the water around the boat, d(t), in meters, where t is measured in hours since midnight, is d(t)=5+4.6sin(0.5t)
Use three decimal places in all intermediate calculations.
(a) What is the period of the tides in hours?
Round your answer to three decimal places.
(b) If the boat leaves the bay at midday, what is the latest time it can return before the water becomes too shallow?
Round your answer to three decimal places.
1 answer
2 answers