Expert: Ahmed Salami - 5/12/2009

Question
Suppose the height of the tide is given by the equation
h(t) = 4 sin {2Pi(t+8)/P}
T is measured in hour since midnight of 29 december 2008 and P is the length of a day,ie 24 hours
a) When did the first high tide for 2009 occur and how high was it?
b)At what rate was the sea level changing at 13pm on january 16 this year?Was is going up or down?

To be honest i still couldn't understand the question and i must learn how to do this for my final exam.Thanks for your help.

Answer
Hi Jack,
h(t) = 4sin {2Pi(t+8)/P}
P = 24 hrs
a)Starting from midnight 29 december 2008, there are exactly 2 days i.e 48hrs before 2009.
High tides occur when dh/dt = 0 and h is positive
dh/dt = 4cos {2Pi(t+8)/P} x 2pi/P
     = (8pi/P)cos {2Pi(t+8)/P}
     = (8pi/24)cos {2Pi(t+8)/24}
     = (pi/3)cos {Pi(t+8)/12}
equating to zero,
(pi/3)cos {Pi(t+8)/12} = 0
High tides occur when
{Pi(t+8)/12} = pi/2 or 5pi/2 or 9pi/2 or 13pi/2 or .......
t+8 = 6 or 30 or 54 or 78 or .....
t = -2 or 22 or 46 or 70 or .....
Notice that we have an infinite solution because the sin function is periodical i.e it keeps on repeating itself at regular intervals.

The first high tide in 2009 only occurs after t = 48 i.e t = 70
h = 4sin {2Pi(70+8)/24}
 = 4sin (13pi/2)
 = 4 x 1
 = 4

If the question was asking for the height of the tide immediately it was 2009, then t = 48 and
h = 4sin {2Pi(48+8)/24}
 = 4sin (14pi/3)
 = 4 x (sqrt3)/2
 = 2(sqrt3)
 = 3.46

b)The rate of change at any time is the value of dh/dt at that time.
dh/dt = (pi/3)cos {Pi(t+8)/12}
At 1pm on january 16 2009, there have been (15 x 24) + 13 hours in 2009 which means that
t = 48 + (15 x 24) + 13
 = 421 hours
dh/dt = (pi/3)cos {Pi(421+8)/12}
     = (pi/3)cos {143Pi/4}
     = (pi/3)(1/sqrt2)
     = (sqrt2)pi/6
     = 0.74
dh/dt is positive and so the sea level in going up.

You can always get back to me.

Regards