Because of the gravitational attractions of the moon and the sun on the Earth, water in seas and oceans tend to rise and fall periodically corresponding to what is called high and low tides. In a a typical situation, the time between two high tides is close to 12 hours. In a certain costal area, the depth of water may be approximated by a sinusoidal function of the form d(t) = - 2.5 cos[ b(t - 2) ] + 3.5, where d is in meters and t in in hours where t = 0 corresponds to 12 am.
a) Find b (b > 0) if d has a period of 12 hours.
b) From t =0 to t = 12, at what time is d the smallest (low tide) and at what time it is highest (high tide)?
c) From t = 0 to t = 12, what are the interval of time during which the depth of the water 4.5 meters or more?
Answers
Solution
a) Using the period, we have
12 = 2π/b
b = π/6
b) d(t) is now given y
d(t) = - 2.5 cos[ (π/6)(t - 2) ] + 3.5
the smallest value of d = -2.5 + 3.5 = 1
hence d is smallest for t such that - 2.5 cos[ (π/6)(t - 2) ] + 3.5 = 1
Solve to get : cos[ (π/6)(t - 2) ] = 1
(π/6)(t - 2) = 0
t = 2 , corresponding to 2 am
the highest value of d = 2.5 + 3.5 = 6
hence d is largest for t such that - 2.5 cos[ (π/6)(t - 2) ] + 3.5 = 6
Solve to get : cos[ (π/6)(t - 2) ] = - 1
(π/6)(t - 2) = π
t = 8 , corresponding to 10 am
NOTE We could have answered part b) using the fact that the distance between a minimum and the following maximum in a sinusoidal function is half a period and therefore d is maximum at t = 2 + (1/2)12 = 8
c) We first need to find t for which h(t) = 4.5 by solving the equation
Graph of y = d(t) and y = 4.5
- 2.5 cos[ (π/6)(t - 2) ] + 3.5 = 4.5
t1 = 6 arccos(-1/2.5) + 2 = 5.8 hours
t2 = 8 + (8 - 5.8) = 10.2 hours (use symmetry with respect to position of maximum)
total number of hours for which d(t) is more than 4.5 m is : 10.2 - 5.8 = 4.4 hours.
