http://www.vhcc.edu/pc1fall9/gov03/

For explanations on Trapezoidal Graph quizzes:  http://www.vhcc.edu/cal1fall/classes_2002/notes_from_fall_02_classes.htm

 09:40:36
Go to the address http://www.vhcc.edu/ph2spring99/experiments/kinetic_model.htm (copy and paste this address into the Address box of your browser). Click on the link KINMODEL and choose the default settings with medium speed. You'll know what to do when you see those prompts. Observe the red ball to a couple of minutes and get a feel for the fastest speed at which you've seen the ball travel, and for the slowest. (The model may be paused by pressing the pause-break key (top right on the keyboard) or by striking the letter s). Assuming the fastest speed is 12 and the slowest is 1 decide what percent of its time the ball appears to spend in each of the six speed ranges 1-2, 3-4, 5-6, 7-8, 9-10, and 11-12. Give your percents. Sketch a bar graph representing your estimates.
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Speed range: Percent:
1-2 7%
3-4 15%
5-6 34%
7-8 31%
9-10 8%
11-12 5%

** these percents add up to more than 100%.  That just cain't be. **

Since I didn't understand the question or how to find the answers, I don't think I can answer this question. I understand what to do in this question, but I don't have the information needed to answer this.
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09:45:12
Using your estimates from the preceding question give your estimate of the average speed of the red ball, which you will calculate as follows: Multiply the middle of each speed range by the percent, except express the percent as a decimal. For example if you estimated that ball was in the speed range 1-2 about 80% of the time then you would multiply the midpoint of the 1-2 range by .80. The midpoint of that range is halfway between 1 and 2, or 1.5, so you would multiply .80 * 1.5. Add up your results. This is your average speed. Give your results and explain why you think this should represent the average speed.
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RESPONSE -->

The avgerage speed is calculated using this equation: (1.5*.07)+(3.5*.15)+(5.5*.34)+(7.5*.31)+(9.5*.08)+(11.5*.05) = 6.16

** since the percents add up to more than 100% this is gonna be a little more than the correct average. **

I think this should give me avg. speed because it calculates how long the ball was in each speed range.
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09:50:17
You will now find the root-mean-square of your results. This time you will multiply the percent by the square of the middle of each speed range, add up these results, and then take the square root of this sum. What do you get? Why do we call this the root-mean-square? Any idea why this should be a significant quantity?
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I obtained 4.4763 as my root-mean-square. If I had to guess, I would say the name comes from squaring the avg. (or mean) and once all of them are added together, taking the root. However, I have no idea why this quantity is significant.

(1.5^2*.07)+(3.5^2*.15)+(5.5^2*.34)+(7.5^2*.31)+(9.5^2*.08)+(11.5^2*.05)

** Find the sum of all these products and take the square root to get the root-mean-square **

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09:57:43
Pick a number between 0 and 1. Don't pick anything obvious like 0, .5 or 1. Call this number x. Figure out the number 3 x * ( 1 - x). Call this number x. Figure out the number 3 x * ( 1 - x). Call this number x. Figure out the number 3 x * ( 1 - x). Continue this process until you have ten numbers and give the results. Is there any pattern to your results?
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RESPONSE -->

x = .3
3x * (1 - x)
3 (.3) * (1-.3)
.9 * (.7)
.63

x = .63
3 (.63) * (1 - .63)
1.89 * (.37)
.6993

x = .6993
3 (.6993) * (1 - .6993)
2.0979 * .3007
x = .63083853 (x = .63)

x = .63083853
3 (.63083853) * (1 - .63083853)
1.89 * .369
.6986

x =.6986
3 (.6986) * (1 - .6986)
2.09 * .301
.63

x = .63
3 (.63) * (1 - .63)
1.89 * .3684
.698

x = .698
3 (.698) * (1 - .698)
2.09 * .301
.63

x = .63
3 (.63 ) * (1 - .63)
1. 89 * .3676
.69

x = .69
3 (.69) * (1 - .69)
2.09 * (.3025
.63

x= .63
3 ( .63) * (1 - .63)
1.89 * .3677
.69

Final results: .63, .69, .63, .69, .63, .69, .63, .69, .63, .69

The pattern is obvious. .63, .69, .63, .69 etc.


There is a slight pattern, if you would call it that, where-in after the pattern really gets going, the numbers increase, then decrease, then increase, then decrease. This pattern continues until the end after the 2nd number.
 

** The numbers bounce up and down but gradually get closer and closer together.

This map, where x is repeatedly replace by 3 x ( 1 - x), is called a logistic map. 

The general form of a logistic map is m x ( 1 - x), where m is a number between 0 and 4.

Repeat this exercise for m = 1 (i.e., using 1 * x * (1-x))  and for m = 2.

Then try m = 3.99.

The closer m is to 4 the more unpredictable our results will be.  **

ok, for my fisrt number I chose 2.

So, 3*2(1-2)=-6
Now my x number is -6

So, 3*(-6)*(1-(-6))=-12
Now my x number is -12

So, 3*(-12)*(1-(-12))=-24
Now my x number is -24

So, 3*(-24)*(1-(-24))=-48
Now my x number is -48

So, 3*(-48)*(1-(-24))=-96
Now my x number is -96

So, 3*(-96)*(1-(-96))=-192
Now my x number is -192

So, 3*(-192)*(1-(-192))=-384
Now my x number is -384

So, 3*(-384)*(1-(-384))=-768
Now my x number is -768

So, 3*(-768)*(1-(-768))=-1536
Now my x number is -1536

So, 3*(-1536)*(1-(-1536))=-3072

** note that if x isn't between 0 and 1 the result 'blows up'. **

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10:01:47
Find the mean of the numbers in the n = 5 row of Pascal's Triangle. The numbers in that row are 1, 5, 10, 10, 5 and 1, standing for the numbers of ways to get 0, 1, 2, 3, 4 and 5 Heads. To find the mean first add up all the possible outcomes (i.e., add up the row). Then multiply each number of Heads by the number of ways in which it can happen. Add up your results and divide by what you got when you added up the row. What did you get, why does it make sense and why does this work?
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RESPONSE -->
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

1+5+10+10+5+10=32

1*0=0
5*1=5
10*2=20
10*3=30
5*4=20
1*5=5

0+5+20+30+20+5=80

80/32=2.5 Avg mean

I do not know why this works and does not make much sense to me.

When I divided, I received an answer of 2.5, whcih makes sense because that is the avg. of the first and last number of heads (i.e. 0 and 5).

** Note that if you make a histogram of the number of ways you can get different numbers of Heads the graph is symmetric about n = 2.5. 

In the figure the numbers of Heads n = 0, 1, 2, 3, 4, 5; there is a bar for each number.

**
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10:03:04
Suppose that you have water flowing into a container at 100 cm^3 / second. If the cross-sectional area of the container is 20 cm^2 then at what rate will water be rising in the container?
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** To get the rate at which water rises we divide the number of cm^3 per second by the cross-sectional area.

At 100 cm^3 / second water will rise in a 20 cm^2 tube at 100 cm^3/second / (20 cm^2) = 5 cm/s. **

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10:12:18
Suppose you have water flowing into a container at 100 cm^3 / second. If the cross-sectional area of the container at depth y, in cm, is 10 cm^2 * y then at what rate will the water be rising when y = 5 cm? At what rate will the water be rising when y = 10 cm? At what rate will the water be rising when y = 15 cm? Estimate the time required for the water to rise from 5 cm to 10 cm, and from 10 cm to 15 cm.
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RESPONSE -->

At y = 5 cm the area is 10 cm^2 * 5 = 50 cm^2.

So water is rising at the rate 100 cm^3 / (50 cm^2) = 2 cm / s

At y = 10 cm the area is 10 cm^2 * 10 = 100 cm^2.

So water is rising at the rate 100 cm^3 / (100 cm^2) = 1 cm / s

At y = 15 cm the area is 10 cm^2 * 15 = 150 cm^2.

So water is rising at the rate 100 cm^3 / (50 cm^2) = .66...  cm / s

Between y = 5 cm and y = 10 cm the rate of change of depth changes from 2 cm/s to 1 cm/s so the average rate of change of depth is about 1.5 cm/s. 

At this average rate the water will require

change in depth / rate = 5 cm / (1.5 cm/s) = 3.33... sec to rise the 5 cm from y = 5 cm to y = 10 cm.

Between y = 10 cm and y = 15 cm the rate of change of depth changes from 1 cm/s to .666.. cm/s so the average rate of change of depth is about .833..  cm/s. 

At this average rate the water will require

change in depth / rate = 5 cm / (.833.. cm/s) = 6 sec to rise the 5 cm from y = 10 cm to y = 15 cm.

ANOTHER SOLUTION:

To get this, you must divide your answer after sub. in "y" into 100cm^3/sec.
y=5----->2cm/sec
y=10--->1cm/sec
y=15--->0.666......cm/sec
To do the last part, I may be wrong, but I believe you avg. the two rates for the depths desired and divide that into the overall "flow-in" rate.

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10:17:15
Suppose you have water flowing into a container at 100 cm^3 / second. If the cross-sectional area of the container at depth y, in cm, is 10 cm^2 * y then at what rate will the water be rising when y = 5 cm? If this rate continued for 2 seconds what would then be the water level at the end of this time? At what rate would water be rising at this new level, and if this rate continued for another 2 seconds what would then be the water level at the end of this time? At what rate would water be rising at this new level, and if this rate continued for another 2 seconds what would then be the water level at the end of this time?
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WELL ORGANIZED SOLUTION; RATE IS CALCULATED UPSIDE DOWN.

y = 5: 10cm^2(5) = 50 cm^2

rate = 50cm^2/100cm^3/second

** You need to divide the rate of volume flow by the cross-sectional area, not the other way around **

rate = 0.5 cm/second

y = 10: 10 cm^2(10) = 100 cm^2

rate = 100 cm^2/100 cm^3/second
rate = 1 cm/second

y= 15: 10 cm^2(15) = 150 cm^2

rate = 150 cm^2/100 cm^3/second
rate = 1.5 cm/second

5 - 10: 10 seconds
10 - 15: 5 seconds

** good work except for the reversal in calculating rate of depth change **

GOOD SOLUTION WITH NARRATIVE DETAILS

The rate at a depth of 5cm would be 100cm^3 / s divided by 50cm^2, or 2cm^3/s. If this continued for 2 secs, then the new level would be 9cm.

The rate at a depth of 9cm would be 100cm^3 / s divided by 90cm^2, or 1.1cm^3/s. If this continued for 2 secs, then the new level would be 11.2cm.

The rate at this depth would be 100cm^3/s.  The area would be

area = 10 cm^2 * y = 10 cm^2 * 11.2 = 112 cm

so the rate of depth change would be the 100 cm^3 / sec flow rate divided by 112cm^2, or 0.89cm^3/s. If this continued for 2 secs, then the new level would be 12.98cm.

ANOTHER SOLUTION

Water would rise at 2cm/sec at y=5. If this rate continues for 2sec, the water lvl would be 9cm. The new rate would be 1.1111111cm/sec. After 2sec, the water lvl would be 11.222cm. The new rate would then be 0.891cm/sec. After 2sec, the new lvl would be 13.004cm.
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10:22:54
Trace out the first 5 steps of the Christmas Tree problem whose rule is that a tree becomes lit provided exactly one of the three trees (the tree itself or one of its neighbors) is lit or the tree and its neighbor to the right but not the neighor to the left are both lit. Start with a row of 11 trees (not in a circle) with all trees off except the one in the middle. Use 0's and 1's for 'on' and 'off' trees.
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100 or 010 or 001 or 011 light the tree in the next step
 

0 0 0 0 0 1 0 0 0 0 0
0 0 0 0 1 1 1 0 0 0 0
0 0 0 1 1 0 0 1 0 0 0
0 0 1 1 0 0 1 0 1 0 0
0 1 1 0 1 1 1 0 1 1 0
1 1 0 0 1 0 0 0 1 0 1



1 1 1 1 1 0 1 1 1 1 1
1 1 1 1 0 1 0 1 1 1 1
1 1 1 0 1 0 1 0 1 1 1
1 1 0 1 1 1 1 1 0 1 1
1 0 1 1 1 1 1 1 1 0 1

I'm not sure if this is exactly correct.
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10:48:35
Find on the Governor's School homepage the link to Trapezoidal Graph Interpretation and click on that link. Answer the questions in Quizzes 1, 2 and 3.
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CHECK TO SEE IF THESE RESPONSES ARE CORRECT

Quiz 1--->3.625gal/hr; the number of gallons in the bucket after 1.5hrs; 1.5 is the number of hrs. since the start of the rain; 3cm/sec; 0.75cm; 3.75cm; the rate at which the rate is changing between 1 and 1.5min.

Quiz 2--->alt-$ in bank, slope-rate of increase, width-time in bank, area is the money made during the interval, accum area is the money over the entire time; alt-$ change=2.625, slope-change of the rate= 1.5, width- time in bank= .5 - 1 months, area - ??, accum area -??; alt- cs area, slope- change of cs area over height above floor measured, width- height above floor that container measured at, area- total change of the container over the width, accum area- total change over the entire measuring

Quiz 3--->alt- avg. height= vel. of 5.125cm/sec, slope- avg. change of vel.= 3.5cm/sec/sec, width- time interval= 1.5-2sec, area is the total distance traveled in the interval, accum area is the total distance during the entire time
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Rule:  Cell turns on if exactly one neighbor is on.

Rule:  exactly one of the three trees (the tree itself or one of its neighbors) is lit or the tree and its neighbor to the right but not the neighbor to the left are both lit.

Rule number is 30.

Rule number is 55