Trigonometry Word Problem

Question

The water at a boat dock is 7 feet deep at low tide, and 11 feet deep at high tide. One day, low tide occurred at 4 am and high tide at 10 am. Find a function for the height of the tide as a function of time t .

Think about the characteristics: amplitude, period, midline, and try to sketch a figure reflecting how the tide height moves up and down.

Answer 1

In order to build a trig function for something like the tide, it helps to pull out the four pieces one at a time rather than trying to write the whole formula at once. The general shape we are aiming for is

h(t) = A cos(B(t - C)) + D

and each of those letters is one of the characteristics the problem mentioned.

Start with the midline, D. The tide swings between 7 feet and 11 feet, so the middle of that swing is the average of the two:

D = (7 + 11)/2 = 9

Next the amplitude, A, which is how far the tide travels above or below that midline. From 9 up to 11 is 2 feet, and from 9 down to 7 is also 2 feet:

A = (11 - 7)/2 = 2

Now the period. Low tide was at 4 am and high tide at 10 am, which is 6 hours apart. Here is the part to be careful with: going from a low to a high is only HALF of a full cycle, since the tide still has to come back down. So the full period is 12 hours, and since B = 2pi divided by the period:

B = 2pi/12 = pi/6

Last is the horizontal shift, C. We know low tide happens at t = 4. A regular cosine starts at its maximum, so if we use negative cosine instead it starts at its minimum, which is exactly what we want. Putting the shift at 4 lines that minimum up with 4 am:

h(t) = -2cos(pi/6 (t - 4)) + 9

Let us check it at both times we were given. At t = 4, the inside is 0 and cos(0) = 1, so we get -2(1) + 9 = 7, which is low tide. At t = 10, the inside is pi/6 times 6 which is pi, and cos(pi) = -1, so we get -2(-1) + 9 = 11, which is high tide. Both match.

So, our final answer is h(t) = -2cos(pi/6 (t - 4)) + 9, where t is hours after midnight and h is in feet.

Answer 2

Tide comes in twice a day and its 6 hours between highs and lows

Thus using a sinusoidal model, the Period T = 12 hours

The Amplitude = 1/2 Peak to Peak = 1/2 ( 11 - 7 ) = 2

The midline is at 1/2 (11 + 7) = 9 feet

The sine is zero at 9 ft so it = 0 at midpoints between 4 and 10 which is t = 13 and 7

OR = 1 and 7 if the period T = 12 hours

Using a sine function, zero would be at midpoint between 4 and 10 = 6pm/am

sin ( wt + phi ) w = 2 pi f = 2 pi / T = pi / 6

For Phase shift ( pi/6 )t + phi = 0 at t =1 and 7 If t=7 => phi = - 7pi/6

So the function would be

******* H( t ) = 2 sin ( pi/6 (t) - 7pi/6 ) + 9 ********

To check H = 11 at t = 10

H(10) = 2 sin ( pi/6 (10) - 7 pi/6 ) + 9

= 2 sin (3 pi/6) + 9

= 2 (1) + 9 = 11 check ok

And ... H = 7 at t = 4

H(4) = 2 sin ( pi/6 (4) - 7 pi/6 ) + 9

= 2 sin ( - 3 pi/6 ) +9 = 7 check ok