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course Mth 164
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
&&&& query modeling exercise
&&&& What is the distance between the peaks of the graph for which the angular velocity of the reference point is 3 rad / sec, and what are the maximum and minimum vertical coordinates if the circle is centered at vertical coordinate 12?
Your solution:
2pi/3sec 12-5= 7 min
12+5= 17 max
Confidence Assessment: 3
Given Solution:
** At 3 rad/sec a complete trip around the reference circle takes 2 pi / 3 seconds, close to but not exactly 2 seconds. 2 pi / 3 seconds is the distance between the peaks on the graph of y vs. t.
If the circle has radius 5 the max and min will be 5 units above and below the center of the circle, at 12 - 5 = 7 and 12 + 5 = 17. **
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15:10:20
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&&&& Given the values between which a cyclical quantity varies, how you determine where to position the circle that models the quantity, and how the determine the radius of the circle?
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15:22:13
Your solution:
Take the highest value and the lowest value and take the average of the two. This will give you the center of the circle.
Find the diameter by taking the difference between the max and min then the radius is half of the diameter.
Confidence Assessment: 3
Given Solution:
** The center of the circle will be halfway between the max and min values, which can be found by averaging the two values (i.e., add and divide by 2).
The diameter will be the difference between the max and min values and the radius will be half of the diameter. **
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15:22:16
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&&&& What is the vertical coordinate of the center of the circle, and what is the angular velocity of the reference point, for the daylight model?
Your solution:
2pi is a complete circle
Using 12 months:
2pi/12 cycles/month
Pi/6 cycles/month
using the max value of 15 hours and the min of 9 hours and take the average
15+9= 24/2= 12hours
Confidence Assessment:3
Given Solution:
** If the period is 52 weeks then you have 2 pi / 52 cycles in a week or pi/26 cycles per week.
If the period is in months then you have 2 pi / 12 cycles per month, or pi/6 cycles per month.
The vertical coordinate of the center will be the day length midway between the min and max day lengths, which is 12 hours.**
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&&&& What is the vertical coordinate of the center of the circle, and what is the angular velocity of the reference point, for the temperature model?......!!!!!!!!...................................
15:55:04
Your solution:
2pi/52= pi/26 cycles/week
95+35= 110/2= 55 degrees F
Confidence Assessment:2
Given Solution:
** If the period is 52 weeks then you have 2 pi / 52 cycles in a week or pi/26 cycles per week.
The vertical coordinate of the center will be the temperature midway between the min and max temperatures.**
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15:58:33
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&&&& What is the vertical coordinate of the center of the circle, and what is the angular velocity of the reference point, for the tide model?
Your solution:
2pi/10= pi/5 ft/hr
12+2= 14/2= 7ft
Confidence Assessment: 3
Given Solution:
** If you have a cycle in 10 hours then you have 2 pi rad in 10 hours, or 2 pi / 10 = pi/5 rad / hour.
The vertical coordinate of the center will be the water level midway between the min and max water levels. **
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15:58:35
Self-critique (if necessary):
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&&&& What is the vertical coordinate of the center of the circle, and what is the angular velocity of the reference point, for the ocean wave model?
Your solution:
5 cycles/min * 2pi= 10picycles/min
40+36= 76/2= 38ft
Confidence Assessment:2
Given Solution:
** The center will lie halfway between the highest and lowest levels.
At 5 waves per minute the angular frequency would be 5 periods / minute * 2 pi rad / period = 10 pi rad / min. **
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&&&& query ch. 5 # 78 15 in wheels at 3 rev/sec. Speed in in/s and mph: rpm?
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Your solution:
radius= 15in
Circumference= 2pi*r
C=95.24 * 3= 282.74 in/sec
282.74 in/sec * (1ft/12in)= 23.56 ft/sec * (1mi/5280ft)= .00446 mi/sec
.00446 mi/sec * (60sec/1min)= .26776 mi/min * (60min/1hr)= 16.4 mi/hr
3rev/sec * (60sec/1min)= 180 rpm (revolutions per min)
Confidence Assessment:3
Given Solution:
** If 15 inches is the diameter of the wheel then the radius is 15 inches.
The angular velocity is 3 rev / sec * 2 pi rad / rev = 6 pi rad / sec.
905 pi rad / sec.
If you approximate this you get around 280 in/sec.
This is 280 in / sec * 1 ft / 12 in = 23 ft / sec approx.
A mile is 5280 ft and an hour is 3600 sec so this is
23 ft/sec * 1 mile / 5280 ft * 3600 sec / 1 hr. = 16 miles / hr approx.. **
Note that 3 revolutions / second is 180 revolutions / minute, since there are 60 seconds in a minute.
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