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On a particular Labor Day, the high tide in southern California occurs at 7:12 am. At that time you measure the water at the end of Santa Monica Pier to be 11 ft deep. At 1:24 pm, it is low tide, and you measure the water to only be 7 ft deep. Assume the depth of the water is a sinusoidal function of time with a period of half a lunar day (12 hr, 24 min)

A) Write an equation that represents this function.

B) At what time does the first low tide occur?

C) What was approximate depth of the water at 9 pm?

D) What is the first time on that Labor Day that the water is 9 ft Deep?

Thanks if anyone attempts this beast.


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Are you able to find the amplitude and offset of the sinusoid? – Unreasonable Sin Nov 22 '11 at 0:33

Let me first say that, while this may appear to be a long and complicated question, if you break it into parts it is not.

Let's see what information we can learn from the problem.

Assume the depth of the water is a sinusoidal function of time

OK, so that gives us a huge leap. What are some examples of sinusoidal functions? Certainly $f(t)=\sin(t)$ qualifies, but this is but the simplest one. We also have: $$f(t)=23\sin(t)\ ,\qquad f(t)=\sin(68t)\ , \qquad f(t)=\sin(t+4)\ ,\qquad f(t)=\sin(t)+95$$ or any combination of the above, using any constants in place of the explicit numbers. How can we write this most generally? Well, let $A,\ B,\ ,C$ and $D$ be constants. Then, in general, a sinusoidal function will be given by: $$f(t)=A\sin\left(B(t+C)\right)+D\ .$$ We've exhausted the information we can glean from that statement, let's move on.

High Tide: 11 ft deep

Low Tide: 7 ft deep

This means that the largest value of $f(t)$ is 11, and the smallest is 7. Since sine oscillates between 1 and -1, we know that this information will not help us find $B$ or $C$. On the other hand, since we know how sine behaves, we know that the smallest value of $f(t)$ will be $A(-1)+D$ (when the sine function is at its lowest point) and the largest value of $f(t)$ will be $A(1)+D$, (when the sine function is at its highest point). We now need to solve

$$A+D=11\ ,$$ $$D-A=7\ .$$

You can do this a number of ways; for instance, add the first equation to the second and you will find that $2D=18$, so we have $D=9$. Substitute this value of $D$ into either equation to arrive at $A=2$.

It is important to tie this result to the graphical idea of what we are doing. The value of $A$ is called the amplitude, and it will "stretch" the sine function vertically, as shown here:


The value of $D$ will move the sine function vertically:


So we now have pinned down $A$ and $D$, and we have $$f(t)=2\sin(B(t+C)+9\ .$$ We are given that

The period is half a lunar day (12hr, 24min)

It will do us no good in mixed form like that, so we'll convert it to 744 minutes. We had not previously mentioned what value $t$ describes, but I think at this point it makes sense to remark that it will be the number of minutes after the (literal) start of Labor Day.

Now, surely you are aware that the period of the sine function is $2\pi$, a beautiful number mathematically, but fairly ugly in this context. We need the period to be 744 minutes, and one way to think about this is that we want to stretch the sine function horizontally so that the value that used to be at $2\pi$ is now at 744. Let's look at just $g(t)=\sin(Bt)$. If we wanted the value of $g(744)$ to be the same as the value of $\sin(2\pi)$, we can just try to figure out how to make $B(744)=2\pi$. But this is easy, we can just solve for $B=\frac {2\pi} {744}$. The same thing happens with our more complicated function (that's why I used $B$ again), so we now have $$f(t)=2\sin\left(\frac {2\pi} {744} (t+C)\right) + 9\ .$$

There's only one thing left to address, our function needs to be oriented horizontally. This is called the "offset". Conceptually in our problem, you can think of the offset as describing the position of the waves at a specific time. Right now, we are starting with the waves at 9 feet, since $f(0)=9$ (as you can verify), but nothing in the problem tells us what height the waves started at at the beginning of the day. We do have the height of the waves at 7:12am, though, which is 432 minutes after the beginning of the day. We are told

At 7:12am, [...] the water is 11 feet deep.

So we need $f(432)=11$. This is a little complicated to calculate, but you can think about it like this: the high tide needs to happen at $t=432$. The sine function will describe a "high tide" 1/4 of the way through its period (remember that $\sin(\pi/4)=1$?), and our function $f$ has a period of 744, so (without adjusting C) it will have a "high tide" at 186, which is not what we want. We want $432+C=186$, which implies that $C=-246$. We now have the whole function:

$$f(t)=2\sin\left(\frac {2\pi} {744} (t-246)\right) + 9$$

Again, the geometric notion of what $C$ does is important - as you probably know, changing $C$ is moving the graph horizontally to the left or the right. If $C$ is negative, the graph moves to the right, so we have shifted the graph to the right in an effort to make the high-tide line up with 7:12am.


Now that you have the function, I leave it to you to answer the rest of the questions.


The analysis you are expected to make is likely to differ from the one below. But perhaps what is written here will be of some use.

Low tide is $7$ feet, and high tide is $11$ feet, which is $4$ feet more than the low of $7$ feet. Halfway between the two is $9$ feet. If we are going to use a sinusoidal model, the formula can be taken to look like
$$y=9+2\sin(k(t+a)).$$ (Since the sine function bounces between $-1$ and $1$, the above function bounces between $7$ and $11$.)

Further details depend on how you measure time, and whether you measure "angles" in radians or in degrees. It is quite likely that a "plug in" formula has been provided in your course. But we will do an analysis that uses only basic knowledge about the sine function.

We use radians, so the calculator will have to be set to radian mode. Adjusting the formula so we can use the calculator in degree mode is not difficult. I do not know whether your course uses radian mode or degree mode.

In $6$ hours and $12$ minutes, we go from high tide to low tide, and vice-versa. Minutes are a little inconvenient. Since $12$ minutes are one-fifth of an hour, $6$ hours and $12$ minutes is $6.2$ hours. Thus, as time changes by $6.2$ hours, the "angle" should change by $\pi$. That means that $k$ should be $\frac{\pi}{6.2}$.

Finally, we want to decide when to measure time from. We will make a mathematically unattractive decision about time, and decide that $0$ hour is at the stroke of midnight at the beginning of Labor Day. At time $t=7.2$, it is high tide, corresponding say to $k(t+a)=\frac{\pi}{2}$. So we have $$\frac{\pi}{6.2}(7.2+a)=\frac{\pi}{2}.$$ Solve for $a$: we get $a=-4.1$. We have derived the formula $$y=9+2\sin\left(\frac{\pi}{6.2}(t-4.1)\right).\qquad\qquad(\ast)$$ Let's check whether this gives the right answers at the times we know about. Look first at time $7$:$12$, that is, $t=7.2$. So substitute $7.2$ for $t$ in $(\ast)$. We get $y=11$. Good!

At $1$:$24$ pm, it is $13.4$ hours after midnight. Substitute $13.4$ for $t$ in formula $(\ast)$. We get $7$. Good!

Now that we have a formula, we can answer the various questions. Actually, we really don't need the formula to answer most of the questions.

For example, to answer B), note that the first low tide occurs $6.2$ hours before the $7$:$12$ high tide, so at $1$:$00$ AM. To answer D) note that $9$ feet is halfway (in height) between low tide and high tide. The first time on Labor Day that the water depth is $9$ feet is halfway between the time of the first low tide and the time of the first high tide, at $4$:$06$ AM.

For the approximate depth at $9$:$00$ PM, we will use the formula $(\ast)$. Put $t=21$ in the formula. The calculator says that $y$ is approximately $10.5175$. This much precision is a little silly, since our input data of $11$ feet and $7$ feet are likely approximate. Let's say about $10.5$ feet.