First Day 040113
164.106.222.236
Dave Smith
dsmith@vhcc.edu AND dasmith@naxs.com
1. y vs. x means that y is on the vertical (up-and-down) axis and x on the horizontal (right-and-left) axis. A table of y vs. x has x in the first column, y in the second.
2. Predict the shape of graphs of depth vs. clock time, temperature vs. clock time, weight vs. length and frequency vs. length for the three systems described.
3. Write down the temperature vs. time of day at every reading.
4. From the data obtained for the flow from the cylinder determine the depth vs. clock time, where clock time is the total time since the first reading.
5. Write down the weight supported vs. the vertical position of the weights.
6. Write down the frequency (the number of complete cycles in a minute) vs. the length of your pendulum.
6. Sketch graphs of depth vs. clock time, temperature vs. clock time, weight vs. vertical position, and frequency vs. length using a reasonable scale for each graph.
7. Pick three representative points from your curve on the depth vs. clock time graph.
8. Plug the coordinates of these three points into the form y = a t^2 + b t + c of a quadratic function in order to get three equations in the parameters a, b and c.
9. Solve the resulting system of three equations by elimination.
10. Replace a, b and c in the form y = a t^2 + b t + c by the values you obtained for a, b and c.
11. Evaluate the depth y at each of the observed clock times and compare with the actual depth.
12. Solve y = a t^2 + b t + c for the clock time at which a specified depth is attained (you will be given a specified depth in class).
13. From the curve for your frequency vs. length pick two representative points and plug them into the power-function form y = a x^p, obtaining two equations.
14. Solve the two equations simultaneously for a and p (hint: first divide one equation by the other to eliminate a and use trial and error to estimate p).
15. Replace a and p in the model by the values you obtained.
16. Plug in the lengths you obtained and find the frequencies predicted by the model; compare with the actual frequencies.
17. Plug in the frequencies you obtained and solve for the lengths.
18. Observe the models obtained by the instructor using Excel. The instructor will obtain linear, quadratic, exponential and power-function models for each set of data. Note that except for the fact that linear functions are also quadratic, a set of data that is modeled well by one function type cannot be modeled by another.
First-week Goals Include:
This course is a study of functions and that the main function types are, in the order to be studied, quadratic, linear, power, polynomial and exponential functions.
Data which can be modeled accurately by a quadratic function cannot be modeled accurately by an exponential function, and vice versa.
Linear, quadratic, power and exponential functions have the basic forms y = x, y = x^2, y = x^p and y = 2^x.
The general forms of linear and quadratic and power functions, as used at this point of the course, are y = m x + b, y = a x^2 + b x + c and y = a x^p.
The symbols a, b, c, p and m are called the parameters of the functions.
Quadratic functions have three parameters; linear functions have two. Power functions in the form y = a x^p have two parameters.
There is a more general form of power functions to be studied later, and the general form of the exponential function will be studied later.
After graphing y vs. x data we fit a smooth curve -- not a series of point-to-point line segments and not a wobbly line that tries to actually hit the data points -- to the curve. The curve shows the trend of the data and does not necessarily pass through any data point, but passes as close as possible on the average to our data points.
If we plug the coordinates of an appropriate number of points from our best-fit curve into the form of a function we can solve for the unknown parameters to obtain a model.
You don’t use data points to get a model.
The number of data points required to solve the model is equal to the number of parameters of the function.
A model can be evaluated by using the function to predict data points and looking at the discrepancies between data values and predictions.
Print out grids to use for graphs.
Strings to ceiling, set up for washers
Ice water and thermometer into bottle
Water for flow expt into a 3-liter bottle
Tie strings to washers.
Distribute rulers.
Set up spreadsheets
3-ring circus
flow data
temperature of water in bottle
length vs. weight for rbs (hang big washers from a series combination; tie strings to ceiling before class)
Water depth vs. t, pendulum freq vs. length both decrease at a decreasing rate
Quadratic function works for water but power fn for pendulum
Get data then demonstrate with Excel
Maybe get to thermometer and exponential fn (observe temp of a bottle of water initially at 0 C vs. t for the first hour or so).
(spring and linear fn project a mass hanging from a series of rbs on screen with spreadsheet cells to measure lengths (linear or poly?))
Work through the details of getting the models
Show overview of function types
Next time: Idea of a rate, rates and graphs.
Sketch a smooth curve through your depth vs. clock time data points.
Estimate the coordinates of three points on your curve but not coinciding with actual data points.
From your curve estimate the clock time at which depth is 53, as well as the depth at clock time t = 10.
Suppose the points are (5, 70), (11,50) and (14, 44).
Plug the coordinates of each point into the form y = a t^2 + b t + c. You'll get three equations.
The equations are
For the first two equations subtract the second from the first
25 a + 5 b + c = 70
- [ 121 a + 11 b + c = 50 ]
_____________________
-96 a - 6 b = 20
Subtracting the second equation from the third we get
75 a + 3 b = -6
This gives us the system
-96 a - 6 b = 20
75 a + 3 b = -6
Doubling the second equation gives us the system
-96 a - 6 b = 20
150 a + 6 b = -12
which we add to get
54 a = 8
We easily solve this to get
a = 8 / 54 = 4 / 27
We plug this into either of the two equations in a and b. Picking the first of these equations
-96 a - 6 b = 20
we obtain
-96 (4 / 27) - 6 b = 20.
Multiplying both sides by 27 to clear the fraction and doing the rest of the obvious stuff we get, if my arithmetic is correct,
b = - 154 / 27.
Plugging our values of a and b into the first of the original equations we get
25 (4 / 27) + 5 (-154 / 27) + c = 70
which we solve by familiar but irritating details to get
c = 2560 / 27.
Our values of a, b and c give us the model
y = 4 / 27 t^2 - 154 / 27 t + 2560 / 27.
We can approximate this equation by
y = 0.148·t^2 - 5.70·t + 94.8
The Excel best-fit model for our actual data is
y = 0.131 t^22 - 6.74 t + 95.4
which should be better than our hand-graphed result.
Class estimates of the clock time at which depth is 53, as well as the depth at clock time t = 10, give us respective averages of t = 7.4 and y = 46.
How do these estimates compare with the predictions of our function
y(t) = 0.148·t^2 - 5.70·t + 94.8?
y stands for depth, so if we plug in y(t) = 53 we obtain the equation
53 = 0.148·t^2 - 5.70·t + 94.8.
If we can solve this equation for t then we have a prediction based on the model.
We can solve this equation by putting it into the form a t^2 + b t + c = 0 and using the quadratic formula.
We end up with two solutions:
t = 9.86 or
t = 28.7.
The t = 28.7 doesn't correspond to anything that happens in the real world, but the t = 9.86 does. Our graph isn't overly accurate, and our t = 7.4 estimate doesn't compare badly with the t = 9.86 from the equation.
To get the model's prediction of the depth when t = 10 we simply plug t = 10 into
y(t) = 0.148·t^2 - 5.70·t + 94.8
to get
y(10) = 0.148·10^2 - 5.70·10 + 94.8 = 52.6.
Again this compares reasonably well with our sloppy-graph estimate of y = 46.
Write down the full statement of the quadratic formula.
What three equations would you get if you plugged the points (2, 88), (5, 37) and (9, 11) into the model y = a t^2 + b t + c of a quadratic?
We get
4 a + 2 b + c = 88
25 a + 5 b + c = 37
81 a + 9 b + c = 11.
If you obtained the model y = .05 t^2 - 3 t + 72 what are the first two steps you would take to determine the clock time when depth is 15?
We first substitute 15 for y to get the equation 15 = .05 t^2 - 3 t + 72.
Then we subtract 15 from both sides to get into the standard form of the quadratic:
.05 t^2 - 3 t + 57 = 0.
From here we can factor, use the quadratic formula, or whatever to solve for t.
What are the forms of the basic functions we have looked at so far?
We've looked at least briefly at the following:
quadratic, linear, polynomial, power and exponential functions.
The basic forms are
y = a t^2 + b t + c (quadratic)
y = m x + b (linear)
y = an x^n + a(n-1) x^(n-1) + a(n-2) x^(n-2) + ... + a1 x + a0. (polynomial)
y = A x^n (power)
y = A b^t (exponential)
We want to look at the how quickly depth is changing over different time intervals for the function y(t) = .05 t^2 - 3 t + 72:
t = 0 gives us y(0) = 72 and
t = 10 gives us y(10) = .05 * 10^2 - 3 * 10 + 72 = 47
so the graph goes from (0, 72) to (10, 47) between t = 0 and t = 10.
The average rate of change is the slope between the points, which is
slope = rise / run = (47 - 72) / (10 - 0) = -25 / 10 = -2.5.
We follow the same process and get ave rate = -1.5.
We again follow the same process and get ave rate = -.5.
What do you think would be the average rate on each interval?
What was the average rate between t = 10 and t = 20, and interval of length 10?
Slope was -1.5.
What is the average rate between t = 10 and t = 11, and interval of length 1?
t = 11 point is (11, 45.05), found by substituting t = 11.
Slope is thus rise / run = (45.05 -47) / (11-10) = -1.95.
What is the average rate between t = 10 and t = .1, and interval of length .1?
Using the same process we get slope -1.995
What do you conjecture the average rates will be for intervals of length .01, .001 and .0001, each interval starting at t = 10?
We get -1.9995, -1.99995, -1.999995
What do you think will happen as the interval length continues to decrease?
Is there a limiting value to the average rate as the interval length approaches zero?
How would we interpret such a limiting value?
Write down the full statement of the quadratic formula.
What three equations would you get if you plugged the points (2, 88), (5, 37) and (9, 11) into the model y = a t^2 + b t + c of a quadratic?
We get
4 a + 2 b + c = 88
25 a + 5 b + c = 37
81 a + 9 b + c = 11.
If you obtained the model y = .05 t^2 - 3 t + 72 what are the first two steps you would take to determine the clock time when depth is 15?
We first substitute 15 for y to get the equation 15 = .05 t^2 - 3 t + 72.
Then we subtract 15 from both sides to get into the standard form of the quadratic:
.05 t^2 - 3 t + 57 = 0.
From here we can factor, use the quadratic formula, or whatever to solve for t.
What are the forms of the basic functions we have looked at so far?
We've looked at least briefly at the following:
quadratic, linear, polynomial, power and exponential functions.
The basic forms are
y = a t^2 + b t + c (quadratic)
y = m x + b (linear)
y = an x^n + a(n-1) x^(n-1) + a(n-2) x^(n-2) + ... + a1 x + a0. (polynomial)
y = A x^n (power)
y = A b^t (exponential)
We want to look at the how quickly depth is changing over different time intervals for the function y(t) = .05 t^2 - 3 t + 72:
t = 0 gives us y(0) = 72 and
t = 10 gives us y(10) = .05 * 10^2 - 3 * 10 + 72 = 47
so the graph goes from (0, 72) to (10, 47) between t = 0 and t = 10.
The average rate of change is the slope between the points, which is
slope = rise / run = (47 - 72) / (10 - 0) = -25 / 10 = -2.5.
We follow the same process and get ave rate = -1.5.
We again follow the same process and get ave rate = -.5.
What do you think would be the average rate on each interval?
What was the average rate between t = 10 and t = 20, and interval of length 10?
Slope was -1.5.
What is the average rate between t = 10 and t = 11, and interval of length 1?
t = 11 point is (11, 45.05), found by substituting t = 11.
Slope is thus rise / run = (45.05 -47) / (11-10) = -1.95.
What is the average rate between t = 10 and t = .1, and interval of length .1?
Using the same process we get slope -1.995
What do you conjecture the average rates will be for intervals of length .01, .001 and .0001, each interval starting at t = 10?
We get -1.9995, -1.99995, -1.999995
What do you think will happen as the interval length continues to decrease?
Is there a limiting value to the average rate as the interval length approaches zero?
How would we interpret such a limiting value?
The temperature of that Coke increased from 2 Celsius to 4 Celsius in the first 10 minutes.
(now trap graphs)
(then analysis of a quadratic)
y(t) := .05 t^2 - 3 t + 72
VECTOR(y(t), t, 0, 30, 10)
[72, 47, 32, 27]
vector(y(t+10)-y(t),t,0,20,10)
[-25, -15, -5]
vector((y(t+10)-y(t))/10,t,0,20,10)
[-2.5, -1.5, -0.5]
dt :=
aveRate(t, dt) := (y(t + dt) - y(t))/dt
aveRate(0, 10) -2.5
aveRate(10, 10)
-1.5
aveRate(10, 1)
-1.95
aveRate(10, 0.1)
-1.995
VECTOR(aveRate(10, 10^(-z)), z, 0, 4)
[-1.95, -1.995, -1.9995, -1.999950001, -1.999994936]
VECTOR(aveRate(20, 10^(-z)), z, 0, 4)
[-0.95, -0.995, -0.9995, -0.9999500024, -0.9999947368]
VECTOR(aveRate(0, 10^(-z)), z, 0, 4)
VECTOR(aveRate(30, 10^(-z)), z, 0, 4)
[0.05, 0.005, 0.0005, 0, 0]
smooth curve thru all data points ...
Ask specific questions, get specific answers, email first and you need to do the qa; also ask questions at the beginning of class, not stick around after and ask, general phil being that I need to address your questions in a way that benefits everyone and not just you. You ask me a question, you write out a transcript and share with everyone, anonymously or otherwise, your choice.
another precalc question: why is the quad model not good for the freq vs. length data; make sure they graph data for all the first three situations.
note how grainy our depth data was but how smooth the model was and how well it approximated when we consider that any of the data points could have been half a unit to the right or left
take a look at the major quiz and see where we're heading
gotta move into trap graphs and interp etc.
challenge problem: if we know at each instant how quickly temp is changing how can be figure out the temp change over a given time interval?
maybe first bring up the concept of rate in terms of how fast temperature changes; hate to get into numerical examples but
algebra review topics need to be brought up on a regular and scheduled basic
we're not going to bother with lin eqns but gonna need quad form and applications, techniques for solving power fn models, right from the beginning and we should state in review examples of the kinds of things we can do to solve equations: adding, mult, taking powers of both sides etc.
if y = at^2 + bt + c represents depth vs. clock time (and incidentally note that I need to find the curve fit before the next class) then to find the clock time at which a certain depth occurs
plug depth in for t (maybe say x instead) and evaluate
plug in depth for y and rearrange the equation so the quad form applies then use the quadratic formula
For the function y(t) = .05 t^2 - 3 t + 72, recall that the average rate of change from t = 0 to t = 10 was -2.5, while the average rate from t = 10 to t = 20 was -1.5.
y(0) = 72 and y(5) = 58.25 so the rise of the graph is 58.25 - 72 = -13.75. Between t = 0 and t = 5 the run is 5 so the average rate is represented by the average slope `dy / `dt = -13.75 / 5 = -2.75.
When we worked with this function previously we found that the average slope increased by +1 for each new interval (e.g., from -2.5 to -1.5, as given in the statement of this question). We therefore expect that the average slope for an interval of 5 (half as long as the length-10 interval) will change by +1/2.
The average rate between t = 0 and t = 5 is -2.75, so we conjecture that the average rate between t = 5 and t = 10 is
-2.75 + 1/2 = -2.25.
If the patterns we've observed for slopes of a quadratic are indeed valid the average rates will -2.25 + .5 = -1.75 and -1.75 + .5 = -1.25.
According to the results we just obtained the average slope between t = 10 and t = 15 is -1.75. So between t = 10 and t = 15 we have rise / run = -1.75. We know that the run is 5. So we conclude that the rise is
rise = -1.75 * run = -1.75 * 5 = -8.75.
The rise is the change in y, so to find the value of y at t = 10 we add this rise to the t = 10 value y = 47. We get y = 47 + (-.875) = 38.25.
The Coke temperature approaches room temperature T = 22 Celsius, with data points (0, 2), (10, 4), (60, 10.2) and (70, 11.2).
Between t = 0 min and t = 10 min the average rate of change is rise / run = `dT / `dt = (4 Celsius - 2 Celsius) / (10 min - 0 min) = .2 Celsius / minute.
Between t = 60 min and t = 70 min the average rate of change is rise / run = `dT / `dt = (4 Celsius - 2 Celsius) / (10 min - 0 min) = .2 Celsius / minute.
The temperature was further from room temperature during the first 10 minutes than during the 10-minute period starting at t = 60.
We expect the temperature to change less and less quickly as we approach room temperature.
Room temperature is around 22 degrees. At 11 degrees the temperature is 11 degrees below room temperature. At 3 degrees the temperature is 19 degrees below room temperature, which is nearly twice as far as the 11 degree difference. We might expect half the temperature difference to result in half the rate.
In mathematical terms we would say that the rate of change of the temperature is proportional to the difference in temperature between the Coke and the room.
At t = 0 we have Coke temperature 2 Celsius, with the room temperature at its constant 22 Celsius. We therefore obtain Tdiff = 22 Celsius - 2 Celsius = 20 Celsius.
At t = 10 min we have Coke temperature 4 Celsius, with the room temperature at its constant 22 Celsius. We therefore obtain Tdiff = 22 Celsius - 4 Celsius = 18 Celsius.
At t = 20 min we have Coke temperature 10.2 Celsius, with the room temperature at its constant 22 Celsius. We therefore obtain Tdiff = 22 Celsius - 10.2 Celsius = 11.8 Celsius.
At t = 30 min we have Coke temperature 11.2 Celsius, with the room temperature at its constant 22 Celsius. We therefore obtain Tdiff = 22 Celsius - 11.2 Celsius = 10.8 Celsius.
The graph decreases at a decreasing rate from (0, 20) through (10, 18) then (60, 11.8) and (70, 10.8). We conjecture that the graph will be asymptotic to the horizontal axis and sketch a smooth curve with these characteristics.
We will now analyze the graph of the function y(t) := 0.05·t^2 - 3·t + 72.
We solve the equation 0.05·t^2 - 3·t + 72 = 50.
The discriminant is (-3)^2 - 4 * .05 * 72 = -5.4 so we don't have real solutions.
Complex solutions are t = 30 - 23.23790007·î OR t = 30 + 23.23790007·î
We solve the equation 0.05·t^2 - 3·t + 72 = 50. We get
t = 51.44761058 OR t = 8.552389410
The axis of symmetry is at t = -b / (2a) = -(-3) / (2 * .05) = 30.
The vertex lies on the axis of symmetry x = 30 so at the vertex we have y(t) = y(30) = 27.
stretching and shifting basic y = x^2 parabolas
meaningful names
basic pts graphs, fundamental properties