Solving the system of simultaneous equationsSolving the system of simultaneous equations
Solving the system for the parameters a, b, c of the flow model
Eliminating Variables when the numbers are messy
A default procedure when you can't find easy multipliers
Assessing the function model
Looking at the model vs. the data to estimate the accuracy of the model
The basic questions we can ask of a function model
Determining y from x (depth from time), x from y (time from depth )
Solving the quadratic equation using the quadratic formula
Solving the quadratic equation to determine time at a given flow depth
Exercises
Analyzing your flow data; Applying the procedure to two data sets
In the first assigment we obtained a system of equations for three selected points. Here we will see how to solve these equations.
Recall that ss a result of our assumption that y = a t^2 + b t + c and our selection of three points, we obtained the equations
900 a + 30 b + c = 49
3600 a + 60 b + c = 16
8100 a + 90 b + c = 1.
These equations can be solved for the values of a, b and c, which are then substituted into the form y = a t^2 + b t + c to obtain our function model. a, b and c are referred to as the parameters of the model. By varying a, b and c, we can obtain different quadratic functions. When we solve the equations we obtain the values of these parameters that ensure that the model will exactly fit the three points we have chosen.
To solve the system of equations we begin by numbering the equations as follows:
`1: 900 a + 30 b + c = 49
`2: 3600 a + 60 b + c = 16
`3: 8100 a + 90 b + c = 1.
We wish to solve these equations by eliminating variables until we have an equation with just one variable, which we will then easily solve for that variable.
We begin by noting that for these equations, c always has the same coefficient, namely 1. Therefore if we subtract any equation from any other, we will obtain an equation for a and b; c will have been eliminated. By going from three variables to two, we are making progress toward our goal of obtaining an equation in one variable. We easily obtain two equations in the two variables a and b as follows:
First we subtract equation `1 from equation `3:
`3: 8100 a + 90 b + c = 1.
`1: 900 a + 30 b + c = 49
`3 - `1: 7200 a + 60 b = -48 (we don't bother to write 0 c).
Then we subtract equation `2 from equation `3:
`3: 8100 a + 90 b + c = 1.
`2: 3600 a + 60 b + c = 16
`3 - `2: 4500 a + 30 b = -15.
The two equations we have just obtained form a system of two simultaneous equations in two unknowns. As noted above, this is progress.
The system of two equations is now solved by elimination. We begin by numbering the equations:
`1a: 7200 a + 60 b = - 48
`2a: 4500 a + 30 b = - 15
We note that 60 b is a multiple of 30 b. We can easily eliminate b if we multiply the second equation by -2 and add the resulting equations:
`1a: 7200 a + 60 b = - 48
-2 * `2a: -9000 a - 60 b = 30.
Adding: -1800 a = -18
This equation is easily solved:
-1800 a = -18
a = 18 / -1800 = .01.
Having obtained a value for a, we can substitute this value into either `1a or `2a to obtain a value for b. Here we substitute a = .01 into `1a:
7200 ( .01 ) + 60 b = - 48
72 + 60 b = - 48
60 b = - 48 - 72 = -120
b = -120 / 60 = -2
Now we know that a = .01 and b = -2. We can substitute this into any of the three original equations to obtain a value for c. We choose to substitute into equation `1:
900 a + 30 b + c = 49
900 (.01) + 30 (-2) + c = 49
9 - 60 + c = 49
c - 51 = 49
c = 49 + 51 = 100
We therefore see that a = .01, b = 2 and c = 100. We can (and you should) validate these values by substituting them into equations `2 and `3.
Our original model depth = a t^2 + b t + c now becomes
depth vs. clock time model: y = depth = -.01 t^2 - 2t + 100.
The above strategy will always work, provided the equations do have a solution (which they do for a quadratic model of any three selected points with different t coordinates). However, the numbers don't always come out so nicely.
If we are fitting a quadratic equation, we can always eliminate c easily enough, since the coefficients of c are always 1.
However, the system of two equations that we get for a and b can have inconvenient numbers. For example we could get a system like
1.53 a - .07 b = .93
2.11 a + 1.3 b = .71.
In this case we don't have a variable that can be eliminated using easy numbers. We therefore use a default strategy that will always work. We multiply the first equation by 2.11, which is the coefficient of a in the second equation, and multiply the second equation by - 1.53, which is the coefficient of a in the first:
2.11 * [ 1.53 a - .07 b = .93 ]
-1.53 * [ 2.11 a + 1.3 b = .71 ]
This strategy will give us equal and opposite coefficients for a, so that when the resulting equations are added the variable a will be eliminated:
3.2283 a - .1477 b = 1.9623
-3.2283 a - 1.989 b = 1.0863
The rest of the calculation is simple enough and is left to the reader. Just add the two equations to get an equation for b. Then solve for b. Then substitute the result back into one of the original equations and solve for a. (If the original equations are known, we can then substitute a and b and solve the result for c).
As we have seen previously, we take our values of a, b and c and put them into the assumed form y = a t^2 + b t + c. For the values a = .01, b = -2 and c = 100 obtained above we have the model:
Function model: depth = y = .01 t^2 - 2 t + 100
We compare this model with our data points. Recall that the model was based on the three points (30, 49), (60,16) and (90,1), and that these points fit the model perfectly. The model deviated slightly from the points (20,63) and (50,27), since
for t = 20 the model predicts depth = y = 64, compared to the observed 63, and
for t = 50 the model predicts depth = y = 25, compared to the predicted 27.
We can make a table of the observed and predicted depths. We place all our original data in two columns and our predicted results in a third:
t observed
y = depth observed
y = depth predicted
= .01t^2 - 2t + 100deviation
0
99
100
1
10
83
81
2
20
63
64
1
* 30 * 49
* 49
* 0
40
37
36
1
50
27
25
2
* 60 * 16
* 16
* 0
70
8
9
1
80
5
4
1
* 90 * 1
* 1
* 0
The predicted values are obtained by plugging the respective t values income the model depth = y = .01 t^2 - 2t + 100. Note again that the predicted values and the observed values are identical for the three chosen points, which are indicated by asterisks (*).
The fourth column consists of the deviations between observed values and predicted values.
The deviations are obviously 0 at the three points on which we based the model (indicated by asterisks). The other points deviated by 1 or 2 from the model, some higher and some lower than the model.
The average of the deviations is easily found to be .9, which is pretty low when the range of y values is from 1 to 100, a range of 99 from lowest to highest. The model is therefore seen to be pretty close to the data, so that it is very plausible that the flow depth is actually a quadratic function of time, and furthermore that it is pretty close to the function model we have obtained.
As we saw above, we can use the depth vs. time model y = .01 t^2 - 2 t + 100 to predict the depth at any given time. We simply plug the time in for t and calculate the depth y. So if, for example, we need to know the depth at time t = 17 seconds, we calculate the predicted depth: depth = y = .01 (17^2) - 2 (17) + 100 = 68.5 cm.
If we have a graph of depth vs. time, we can estimate the depth at a given time by first locating the time on the t axis. We then find the corresponding depth by moving vertically upward to the graph, then horizontally to the y axis to read the depth. The y axis point corresponds to the depth at the given time.
Alternatively, we might want to know the time at which a certain depth is achieved. For example, we might want to know just when the depth reaches 31 cm. We can of course estimate this depth on a graph: we locate the desired depth on the y axis, move horizontally to the graph then vertically to the t axis.
(For practice, look at your depth vs. time graph for the actual data taken in class, and estimate the time at which the depth was 27 cm, and the depth when the time was 46 seconds).
Our graph can give us a good estimate of depth or time. If we want more precise results, we have to set up and solve equations.
Setting up an equation for finding the time when we reach a given depth is easy enough: we just plug our depth into the form of the function. In the present case, if the depth is 31 cm we have
depth = .01 t^2 - 2 t + 100 and depth = 31, so
equation for time when depth = 31: 31 = .01 t^2 - 2 t + 100.
If we can solve this equation for t, we will have the clock time we are looking for. This equation can be solved, but only if we use the right techniques.
Students at this point usually try the wrong techniques.
Trying to isolate t on one side by adding or subtracting quantities on both sides, or dividing or multiplying both sides by the same quantity, do not by themselves lead to a solution.
We could for example get the equation into the form .01 t^2 - 2 t = -69, but what then? We were looking for somthing of the form a t = b, and we have that t^2 interfering with our plans. We simply can't isolate t in this way.
To solve the equation it is just about necessary to recognize that it is in fact a quadratic equation. The quadratic formula gives us a means of solving such an equation.
The standard form of a quadratic equation is
standard form of a quadratic equation: a t^2 + b t + c = 0.
The present equation 31 = .01 t^2 - 2 t + 100 is not in standard form, since it doesn't have 0 as one side. However this is easily remedied. If we subtract 31 from both sides we easily obtain
equation in standard form: .01 t^2 - 2t + 69 = 0.
We can now apply the quadratic formula to our equation.
This formula states that if y = at^2 + bt + c, then y = 0 if and only if
t = [-b +- `sqrt(b^2-4ac)] / (2a).
This means that the expression .01 t^2 - 2t + 69 = 0, for which a = .01, b = -2 and c = 69, will equal 0 if and only if t takes a value
t = [- (-2) + - `sqrt[ (-2)^2 - 4(.01)(69 ) ] / (2 * .01).
This expression looks pretty horrendous. You should write it out in more traditional mathematical form. You get something like this:
We will write this last result here as t = ( 2 +- `sqrt(1.24) ) / .02.
The + case of the +- gives us t = (2 + `sqrt(1.24) ) / .02 = 155.7, while
the - case gives us t = (2 - `sqrt(1.24) ) / .02 = 44.3. We therefore have solutions
t = 155.7 or 44.3.
The quadratic formula gives us two solutions for this problem. Clearly there is only one answer to the question of when the depth reaches 31 centimeters, since depth passes through the 31-cm mark only once.
Since the flow lasted for less than 100 seconds, and since during that time the depth went from greater than 31 cm to less than 31 cm, we could conclude that the t = 155.7 sec. solution is irrelevant to the situation and that only the 44.3 sec. solution is relevant. This solution is easily confirmed, since from the data we can see that the 31 cm depth occurs between t = 40 and t = 50 sec. So we choose the solution t = 44.3 sec, discarding the solution t = 155.7 sec.
Conclusion: depth = 31 cm at t = 44.3 sec. (The t = 155.7 sec solution is discarded).
When submitting your work electronically, show the details of your work and give a good verbal description of your graphs.
One very important goal of the course is to learn to communicate mathematical thinking and logical reasoning. If you can effectively communicate mathematics, you will be able to effectively communicate a wide range of important ideas, which is extremely valuable in your further education and in your career.
When writing out solutions, self-document. That is, write your solution so it can be read without reference by the reader to the problem statement. Use specific and descriptive statements like the following:
Complete the modeling process for your own flow depth vs. time data.
Use your model to predict depth when clock time is 46 seconds, and the clock time when the water depth first reaches 14 centimeters.
Comment on whether the model fits the data well or not.
Even though you probably understand the process at this point, it can be challenging to get through these problems without making mistakes. An error on one step can throw the entire problem off, and result in a model that doesn't work at all.
There is a limit to how much time you should spend worrying about finding and correcting arithmetic mistakes. Of course it is important that you follow the solution procedures correctly, that you recognize when your model doesn't work, and that you do spend significant time trying to track down arithmetic errors. However if you believe you are doing the procedures correctly (other than a pesky arithmetic error or two), and have spent more than a couple of hours trying to track the error down, you should proceed to the Query and submit your responses through Exercise 1. Wait for your work to be posted before moving on to Exercise 2 below.
2. Follow the complete modeling procedure for the two data sets below, using a quadratic model for each. Note that your results might not be as good as with the flow model. It is even possible that at least one of these data sets cannot be fit by a quadratic model.
Data Set 1
In a study of precalculus students, average grades were compared with the percent of classes in which the students took and reviewed class notes. The results were as follows:
Percent of Assignments Reviewed
Grade Average
0
1
10
1.790569
20
2.118034
30
2.369306
40
2.581139
50
2.767767
60
2.936492
70
3.09165
80
3.236068
90
3.371708
100
3.5
It's best to obtain and use your own model. However if after reasonable effort (an hour or so) you fail to get a model that appears to make sense, you may use the model y = - 0.0003·x^2 + 0.041·x + 1.41 to answer the questions below. When you do the Query, you will be expected to show the work you have done up to this point. You should then indicate that your model doesn't seem to work, and state that you are using the y = - 0.0003·x^2 + 0.041·x + 1.41 model. This model isn't based on a very good selection of points, so it's possible to get a much better model, but this one will suffice to answer the questions.
Quadratic equations can't always be solved, so it is possible that some of the questions asked below will have no answer.
Determine from your model the percent of classes reviewed to achieve grades of 3.0 and 4.0.
Determine also the projected grade for someone who reviews notes for 80% of the classes.
Comment on how well the model fits the data. The model may fit or it may not.
Comment on whether or not the actual curve would look like the one you obtained, for a real class of real students.
Data Set 2The following data represent the illumination of a comet by the sun at various distances from the sun:
Distance from Sun (AU)
Illumination of Comet (W/m^2)
1
935.1395
2
264.4411
3
105.1209
4
61.01488
5
43.06238
6
25.91537
7
19.92772
8
16.27232
9
11.28082
10
9.484465
Obtain a model.
It's best to obtain and use your own model. However if after reasonable effort (an hour or so) you fail to get a model that appears to make sense, you may use the model 256·x^2 - 1439·x + 2118 to answer the questions below. When you do the Query, you will be expected to show the work you have done up to this point. You should then indicate that your model doesn't seem to work, and state that you are using the y = 256·x^2 - 1439·x + 2118 model. This model isn't based on a very good selection of points, so it's possible to get a much better model, but this one will suffice to answer the questions.
Quadratic equations can't always be solved, so it is possible that some of the questions asked below will have no answer.
Determine from your model what illumination would be expected at 1.6 Earth distances from the sun.
At what range of distances from the sun would the illumination be comfortable for reading, if reading comfort occurs in the range from 25 to 100 Watts per square meter?
Analyze how well your model fits the data and give your conclusion. The model might fit, and it might not. You determine whether it does or doesn't.
