#$&*
course Mth 271
1/24 6
The Introductory ModelDepth vs. time for water flowing from a uniform cylinder through a uniform hole near the bottom
When water is allowed to flow freely from a uniform hole in the side of a uniform cylinder, with no water added to the cylinder after the flow begins, the depth of water in the cylinder will obviously decrease. What is not completely obvious is whether the depth increases at a constant rate, at a decreasing rate or at an increasing rate.
The majority of the class predicted that water would flow out at a decreasing rate, so that depth would change more and more slowly. A significant number of participants predicted that the water would flow out at a constant rate, while a few predicted an increasing rate. These behaviors are depicted on the graph below. Each graph depicts depth vs. time, with depth on the vertical axis and time on the horizontal axis. One function predicts that the water would flow at a constant rate, while on another in would flow at an increasing rate and on a third it would flow at a decreasing rate.
The actual flow was observed using a simple computerized timer. Depths of 90, 80, 70, ..., 10, 9, 8, ..., 2 centimeters (above the hole) were marked on a clear uniform cylinder and the clock times and time intervals were recorded. It was observed that the flow for each 10-cm interval required a longer time than for the preceding interval, which was seen to clearly indicate a decreasing rate of flow.
Participants were then asked to pick three representative data points from their depth vs. time tables, and to plot these points. Typical graphs looked like the following:
It is clear that this graph depicts a depth that is decreasing at a decreasing rate.
Of course the remaining points could have been plotted, and will be plotted as homework.
A Mathematical Model
It would be useful to have a mathematical rule, or function, that allows us to compute the depth for any given time, or the time for any given depth. Such a rule would be called a mathematical model of the flow depth vs. time.
For example, suppose that the three data points were (30, 49), (60,16) and (90,1), where the coordinates of a point represent time and depth, in that order. A mathematical model for these three data points would be
depth = .01 t^2 - 2t + 100.
When t = 30 is substituted into this model, we obtain depth = .01(30^2) - 2(30) + 100 = 49, which coincides with the data point (30,49). You should check for yourself that when t = 60 and t = 90 are substituted into the above model, we obtain depths of 16 and 1, respectively. Note that this is consistent with the data points (60,16) and (90,1).
You should note also that the depth function is of the form depth = a t^2 + b t + c, which is a quadratic function. Any time we have three data points with different t values, we can find a quadratic function that fits these three points exactly, in the above sense: all three data points can be exactly validated as above. We will see soon how this is done.
The above model was based on three data points. Suppose that other data points included (20,63) and (50,27). We do not expect that these points will necessarily fit the model exactly, since the model was obtained from the original three points. We can check to see how close (20,63) is to the model by substituting t = 20 to see how close our result is to 63. Check for yourself that when t = 20, we obtain depth = 64, which differs from the observed depth of 63. The observed depth differs from a model by -1, since 63 is one unit less than 64. You should check that the other data point (50,27) differs from the model by +2.
The quadratic formula
The quadratic formula is contained in the following statement, which you are required to know verbatim:
If y = a x^2 + bx + c, then y = 0 if, and only if, x = [ -b +- `sqrt(b^2 - 4ac) ] / (2a).
This formula is more familiar when written out in more standard mathematical notation. However, it is not presently feasible to use standard mathematical notation on Internet documents, and in addition to the disadvantages (like being harder to read) there are advantages to the linear notation used above. For example it is essential to understand why the denominator 2a must be in parentheses, as must the numerator. This notation forces the user to direct careful attention to the order of operations.
This statement of the quadratic formula can be used to analyze a quadratic mathematical model of the flow depth function. Specifically we will be able to use the statement to determine the time when the depth reaches any given level, or to determine those depths which will never be reached.
Simultaneous linear equations
You should recall having seen systems of simultaneous linear equations something like the following:
2x + 7y = 9
3x - 4y = 5.
Hopefully you recall that if we multiply the first equation by 3 and the second by -2, we obtain
6x + 21y = 27
-6x + 8y = -10.
Adding these two equations we obtain
0x + 29y = 17,
which can be written simply as
29y = 17.
Of course we can easily find y if we divide both sides by 29. However you know that very well, and that is not the point here. The point is that we have combined the two equations to eliminate one of the variables, which is what permits us to solve for the other variable.
We will look more closely at how to solve systems of equations shortly. However, right now we are more interested in obtaining a system of equations to be solved in order to model the depth vs. time behavior of the water in the cylinder.
Obtain simultaneous linear equations in order to find a quadratic model of three data points
Suppose you hadn't been told above what the model was for the depth function that fits the points (30,49), (60,16) and (90,1), but that you did somehow know that a quadratic function
depth function: depth = a t^2 + bt + c
would fit the points, if only you could find the values of a, b and c. You could obtain an equation for each data point, as follows:
For the first data point (30,49), we see that when t = 30, depth = 49. If we substitute these values into the general quadratic function depth = a t^2 + bt + c, we obtain
49 = a (30^2) + b(30) + c,
or, with a little rearrangement,
900 a + 30 b + c = 49.
Using the second data point (60,16) in the same way we obtain
3600 a + 60 b + c = 16.
Similarly the third data point (90,1) gives us the equation
8100 a + 90 b + c = 1.
These three equations give us the 3-equation system
900 a + 30 b + c = 49
3600 a + 60 b + c = 16
8100 a + 90 b + c = 1.
Once the equations have been found we can solve them to obtain the values of a, b and c, which are then substituted back into the form depth = a t^2 + b t + c. You might or might not know how to solve this system; the solution process will be discussed later. In any case, it turns out that for this system we obtain a = .01, b = -2 and c = 100, giving us the previously stated function
depth function for three chosen points: depth = .01 t^2 - 2t + 100.
We aren't worried about actually solving this system just yet (in case you are curious, we use elimiination much as in the earlier example; although this one has messier numbers and requires a bit more work, it's not that much harder). The process of substituting data points into the form of a function is simple enough. You should check your understanding by obtaining the linear equations for a, b and c that correspond to the data points (20,63), (50,27) and (70,10).
Exercises:
Here are some data for the temperature of a hot potato vs. time:
Time (minutes) Temperature (Celsius)
0 95
10 75
20 60
30 49
40 41
50 35
60 30
70 26
Graph these data below, using an appropriate scale:
Pick three representative points and circle them.
(10,75),(20,60),(30,49)
Write the equations that result from the assumption th at the appropriate mathematical model is a quadratic function y = a t^2 + b t + c.
75=a(10^2)+b(10)+c
60=a(20^2)+b(20)+c
49=a(30^2)+b(30)+c
Eliminate c from your equations to obtain two equations in a and b.
15=-300a-10b
26=-800a-20b
Solve for a and b.
-2(15=-300a-10b)
26=-800a-20b
-30=600a
26=-800a
-4=-200a
.02=a
26=-800(.02)-20b
26=-16-20b
42=-20b
-2.1=b
Write the resulting model for temperature vs. time.
(.02) t^2-2.1t+94
Make a table for this function:
Time (minutes) Model Function's Prediction of Temperature
0 94
10 75
20 60
30 49
40 42
50 39
60 40
70 45
Sketch a smooth curve representing this function on your graph.
Expand your table to include the original temperatures and the deviations of the model function for each time:
Time (minutes) Temperature (Celsius) Prediction of Model Deviation of Observed Temperature from Model
0 95 94
10 75 75
20 60 60
30 49 49
40 41 42
50 35 39
60 30 40
70 26 45
Find the average of the deviations.
7? I don’t really understand this part
Comment on how well the function model fits the data. (Note: the model might or might not do a good job of fitting the data. Some types of data can be fit very well by quadratic functions, while some cannot).
@& Your first line is
0 95 94 ,
representing clock time, temperature, and the prediction of the model.
The prediction of the model deviates from the observed temperature by 1 degree. So the deviation is 1 degree.
The deviations of your next four lines are 0, 0, 0 and 1 degree.
The model fits very well up to this point. Beyond this point the deviations rapidly grow.
The average deviation is what you get when you calculate the deviations, then average them.
Note that your model doesn't fit the last four data points well. This is because you based it on three consecutive data points. You would have obtained a more accurate model had you spread your selected data points out more, for example choosing the t = 10, 30 and 60 points instead of the t = 10, 20 and 30 points.
No need to redo anything. Everything to this point is fine, and you can simply continue.*@