POSSIBLE SOLUTION(S)
PART A:
We enter the information from the table as a data set. Orient the time values so that the day begins at t = 0 (i.e., t = 0 at midnight ).
Input :=
givendata = {{0, 15.63},{1, 18.04},{2, 19.53},{3, 20.35},
{4, 19.74},{5, 17.82},{6, 15.63},{7, 12.54},{8, 11.19},
{9, 10.20},{10, 10.96},{11, 13.14},{12, 15.17},{13, 18.04},
{14, 19.49},{15, 20.00},{16, 19.87},{17, 17.77},{18, 15.55},
{19, 13.15},{20, 10.82},{21, 10.38},{22, 10.70},{23, 12.89},
{24, 15.21}};
A graphical view of this data will assist the the investigation of the relationship of time and water level.
Input :=
points=ListPlot[ givendata, PlotStyle->{PointSize[.02]}];
Input :=
joined=ListPlot[ givendata, PlotStyle->{PointSize[.02]},
PlotJoined->True];
Input :=
Show[points,joined];
The data points seem to form a sinusoidal wave. Thus, an equation of the form h = a sin b(t + c) + d can be used to model this situation. We must determine values for the amplitude, period, phase shift, and vertical shift that will "best" fit this data.
The water level seems to have a maximum value of 20.35 feet and a minimum value of 10.2 feet. Since the height varies by approximately 10 feet, we can estimate the amplitude of the sine function to be a = 10/2=5.
The water level on this day begins at 15.63 feet. It returns to roughly this value at 6 a.m. and again at noon, showing a cyclic behavior. This can also be seen on the plot, where the curve completes one full cycle in a time period of approximately 12 hours. The period of the sine function is
2 Pi / b, where b is the coefficient of t. Therefore, the period is
2 Pi / b = 12, which gives that b = Pi / 6 .
The curve that fits the points does not appear to have a horizontal phase shift from the graph of y = sin x, so c = 0 for our model. There is a vertical shift which moves the "mean" height from 0 to somewhere around 15, so we will start with d = 15 in our equation.
Input :=
Model[t_] =5 Sin[ Pi/6 t ] + 15.25;
curve = Plot[ Model[t], {t,0,24}];
Input :=
Show[curve,points];
This did fairly well, but maybe we can raise the graph just a little bit to get a better fit.
Input :=
Model2[t_] :=5 Sin[ Pi/6 t ] + 15.25;
curve2 = Plot[ Model2[t], {t,0,24}];
Input :=
Show[curve2,points];
PART B
Using the time scale we have established, 1:30 p.m. is equivalent to 13.5 hours. Thus, we can simply use the model equation that we have developed with t = 13.5.
Input :=
Model2[13.5]//N
Output =
18.7855
This value seems reasonable because it is between the heights of 18.04 and 19.49 which were recorded on the table at 1 p.m. and 2 p.m. respectively.
PART C
The water clearly is rising. This question is somewhat trivial. Its value lies in that it can be answered graphically (by examining the plot), numerically (by examining the table), or symbolically (by examining the derivative). Here we illustrate the symbolic approach, which shows that the slope is positive for this value of time.
Input :=
Model2'[13.5]//N
Output =
1.8512
PART D
At 3:30 p.m., our model time scale gives t = 15.5 hours.
Input :=
Model2[15.5]//N
Output =
20.0796
By the model, the height of the water would be 20.0796 feet. Is this reasonable? From 3 p.m. to 4 p.m., the water level decreases from 20.0 ft to 19.87 ft. Under the assumption that the water level is strictly decreasing during this time interval, the water could not have reached a height of more than 20 feet.
Looking at the plot of the actual data points with the model function, it can be observed that the model curve does not fit the data points as neatly in this interval. This would explain the unusual result given by the model. Which is right, the model or the table?
PART E
This will again require using the derivative, this time evaluated when t= 7. The function appears to be decreasing there, so it is not surprising that the result is negative. The rate of decrease is 2.27 ft / hour
Input :=
Model2'[7]//N
Output =
-2.26725
PART F
To determine all the times within the first 12 hours at which the water level will be 18 feet we plot our Model2.
Input :=
Plot[Model2[t],{t,0, 12}]
Output =
-Graphics-
It appears as if there are two times at which the water level will be 18 feet in the first 12 hours - around 2 AM and around 4:30 AM.
Input :=
sol1 = FindRoot[Model2[t]==18,{t,2}]
Output =
{t -> 1.11223}
Input :=
sol2 = FindRoot[Model2[t]==18,{t,4}]
Output =
{t -> 4.88777}
Thus we have the two times at which the water level will be 18 feet in the first 12 hours:
Input :=
t1 = t/.sol1[[1]]
Output =
1.11223
Input :=
t2 = t/.sol2[[1]]
Output =
4.88777
And as a check we see that
Input :=
Model2[t1]//N
Output =
18.
Input :=
Model2[t2]//N
Output =
18.