course Mth 164
I am still waiting to hear from you about some of the orientation steps but I wanted to get this in since it is already a day late.
The last two postings on your page included Orientation part iii and a submission entitled 'problems with orientation'. The latter included the following, part of which is a request for more information from you. Can you provide that information?
(your question) I have been working with the q a initial problems form for several hours and I do not understand how this form works. I have discovered that these forms did not open using Vista on my computer at home so I am having to redo everything at school. I am trying to finish everything for the orientation so I can submit the 6.1 assignment, which I have finished.
(my posted response) Every computer has a different setup, but as far as I know there has been no problem at all related to these programs working with Vista. It is possible that a firewall or some other software blocks the running of .exe files.
Can you give me the keystroke-by-keystroke details of what happens when you try to open these files on your home computer?
Math 164Assignment #1
Textbook problems Chapter 6, assigned problems and odd multiples of 3
3. An angle is in standard position if its vertex is at the origin of a rectangular coordinate system and its initial side coincides with the positive x-axis.
9. False - The area A of the sector of a circle of radius r formed by a central angle of theta radians is
A=(?r^2theta.
11.30 degree angle 15. 450 degree angle
21. 16 pi/3 radians = (16 pi/3)*(180/pi)= 960 degrees
960 degree angle
23. 40 degrees 10?5?expressed as a decimal:
= 40*10(1/60) + 25((1/60)*(1/60))
= 40 degrees + 17 degrees + .0069 degrees
= 40.1769 degrees
( the calculator gave me 40.1736 but I did it several times manually and got 40.1769 - not sure why the difference)
27. 9 degrees 9?9?expressed as a deciimal:
= 9 degrees + 9*(1/60) + 9*((1/60)*(1/60))
= 9 degrees + .15 + .0025
= 9.1525 degrees
29. Express 40.32 degrees in DMS notation:
= 40 degrees + .32 degrees
= 40 degrees + .32(60?
= 40 deg + 19.2? 40 deg + 19?+ 2? 40 deg+ 19? 2?60?= 40 deg+ 19?2?
33. Express 19.99 deg in DMS:
19 deg + .99(60?= 59.4 .4(60? = 24? = 19 deg 59?4?
35. 30 degrees = 30(pi/180) = pi/6 rad. = .524 rad.
39. -60 degrees (pi/180) = -pi/3 rad.
45. -90 deg. (pi/180) = -pi/2 rad.
47. Pi/3(180/pi)=60 deg.
51. Pi/2 (180/pi) = 90 deg.
57. -pi/6(180/pi) = -30 deg.
61. -40 deg.(pi/180) - (-2pi/9) rad. = -.698 rad.
63. 125 deg. (pi/180) = 2.18 rad.
69. 6.32(180/pi) = 362.11 deg.
75. Radius = 5 miles, s = 3 miles (arc) , theta = ? 3 = 5(theta) theta = .6 rad.
79. R - 10 meters, theta = ?rad , A = ? A = ?10)squared(?=25m squared
81. Theta = 1/3 rad. A = 2 sq. ft. , r = ? A = 1/2rsquared(theta) 2 = 1/2r sq. (1/3) 2 = 1/6r sq. r=3.46
87. Theta = pi/3, r = 2 ft, arc length s = ? A= (?r sq. theta A= (?(4)(pi/3) = 2.09
S = r(theta) = 2(pi/3) = 2.09
93. R = 4, theta = 45 deg. = pi/4, A = ? A = (?r sq. theta A= (? (16) (pi/4) = 12.57
99. Linear speed 35 mph, radius = 13 in., rpm = ? Linear speed = radius * angular speed(w)
35 mi/hr(5280 ft/mi) = (184800 ft/hr * 12 in/ft)=(hr/60 min) = 36960 in/min/(26 pi) = 452. 49 rpm
105. Radius = 2.39 * 10^5 miles, 1 rev./27.3 days linear speed ? Mph
2 pi (2.39*10^5) = 478000pi /27.3 days (24 hrs/day) = 478000 pi miles/655.2 hrs = app.2292 mph
111.linear speed 9.55 mph, diameter 8.5 ft, radius 4.25 ft How fast is the wheel rotating ?
9.55miles/hr(5280 ft/mile) = 50424 ft/hr(hr/60 min)(min/8.5pi) = 50424 ft/510 pi min = app. 31.47 rpm
117. ?
123. A nautical mile corresponds to measures of longitude.
Chapter 5 Problems(online)
6. Draw 540 deg. Angle (straight line)
12. Draw 21 pi/4 = 1890 deg. Same as 90 deg. 
Convert from degrees to radians or radians to degrees:
18. -30 deg. = -30(pi/180) = -pi/6 rad.
24. -180 deg. = -180(pi/180) = -pi rad.
30. 4 pi = 4 pi(180/pi) = 720 deg.
36. -3 pi/4 = -3 pi/4(180/pi) = -135 deg.
48. 51 deg. = 51(pi/180) = .283333...pi rad.
50. 200 deg. = 200(pi/180) = 10pi/9 = 1.1... Pi rad.
54 pi = pi(180/pi) = 180 deg.
If r is the radius and s is the length of the arc, find the missing quantity.
40. Theta = ?rad., s = 6 cm, r= ? s = r(theta) 6 = .25r r = 24
42. R = 6m, s = 8m, theta = ? 8 = 6 (theta) theta = 8/6 = 4/3
78. Radius = 15 in. wheels turning @ 3 rev/sec. How fast is the car moving ?
2(pi)r = 30 pi(3 rev/sec) = 90 pi in/sec
90 pi in/sec(ft/12 in) (60 sec/min)(60 min/hr) = 27000 ft/hr(mile/5280 ft) = 5.1 mph
80. Windshield wiper = 18 in. If it takes 1 sec. to trace out 1/3 rev. how fast is the tip of the wiper moving ?
C= 2(pi)r = 2(pi)18 = 36 pi 1/3 rev/sec = 36 pi in/sec
36 pi in/sec(ft/12 in) = 3 ft/sec(mile/5280 ft.)(60 sec/min)(60 min/hr) = 3*60*60/(5280)mph =
10800/5280 = app. 2.05 mph
84. Ferris wheel radius 30 ft., one rev. in 70 sec. What is the speed of a pt. on the outer part of the wheel ?
C= 2(pi)(30) = 60 pi ft. , 1 rev. 70 sec. , 1 rev. 7/6 min. 60/(7/6) = 51.42pi/min.
51.42pi ft/min(mile/5280 ft)(min/60)(60/hr) = .0306 mph
90. If the radius of the earth is 3960 miles express 1 nautical mile in terms of statute mile:
R = 3960 mi. , nautical mile = arc length = central angle of great circle
60 angles @ 6 degrees = 360 deg. Theta = 6 deg. = (pi/180) rad. = .1047 rad *3960 = 414.69
1 nautical mile = 414.69 statute miles(?)
Sketching Exercise:
1. When the paper moves to the left, a straight line is formed.
2. T = 4.5 seconds
3. Distance between peaks is cut in half when you double the angular velocity. When you increase the angular velocity to 9 rad/ sec. the distance will be halved again.
I f you decrease the angular velocity to 1 rad/sec., the distance between the peaks will be = to the radius.
Summary: As angular velocity increases, the distance between peaks decreases.
Modeling Periodic Phenomena with Graphs based on Circles
A graph of the y coordinate of a point moving around a circle at 3 rad/s/ vs. clock time t.
How far is it between the peaks of the graph ? 3 rad/sec (?)
The peaks of the graph are separated by a distance that corresponds to the horizontal coordinate, which represents clock time.
At 3 rad / s it takes a little more than 2 seconds to complete a cycles. So the peaks will be separated by a little more than 2 seconds.
The accurate separation will be 2 pi rad / ( 3 rad / sec) = 2 pi / 3 rad * sec / rad = 2 pi / 3 sec. To three significant figures this is 2.09 seconds.
Maximum pt. = 5, minimum pt. -5 (radius = 5)
If the center of the circle lies on y=12 and the radius =5, maximum = 17 and minimum = 7.
Where would you position the circle, and what would be its radius, if you wished to model a quantity y which varies cyclically from a value of 6 to a value of 14 ? Y= 10 , radius = 4
Find the difference between the 2 numbers(14-6=8). Half of that is the radius. Subtract the radius from maximum and/or add to the minimum to determine where to position the circle.(y=)
At a certain latitude near here the length of a day varies from app. 9 hrs. to app. 15 hrs. over a period of 1 yr. , or 12 months. 15 - 9 = 6 r= 3, y = 12
Angular velocity(don? know how to compute this)
A cycle lasts a year, or about 365.25 days.
A cycle corresponds to 2 pi radians.
So the angular velocity (which is the rate at which the angle changes with respect to clock time) is
2 pi radians / (1 year) = 2 pi rad / yr or alternatively
2 pi radians / (365.25 days), which since the .25 isn't exactly correct can be approximated by a decimal with 5 significant figures (i.e., divide 2 pi by 365.25 and express as a decimal).
Avg daily mean temp. varies from 75 deg. To 35 deg over a 52 wk. period
75-35 = 40 r= 20 y= 55 angular velocity ? That would be 2 pi radians / (52 weeks) = pi /26 radians / week.
Water level varies from 2 ft. below to 2 ft above a walkway, over a 10 hr. period
From -12 to +2 = 14 r= 7 y = -5 angular velocity ? 2 pi rad / (10 hr) = pi/5 rad / hr
Not sure how to figure time into the graph. Does it have something to do with the computation of angular velocity ?
Pendulum Exercise
Angular velocity was 4.700. Measurement was app. 8.5 in. from end of string to head of screw. I had to use a screw instead of a washer and had a hard time making it swing.
I assume you mean 4.7 rad / s?
You did very good work on this assignment. See my notes.
Future assignments can be submitted using the q_a_ and query programs, as indicated on your Assignments page.
I'm also very close to posting 'open qa's' for your course. Email me for a estimated time of posting and instructions.
I was hoping to get to it tonight, but might not be able to. Should get it by mid-week if not sooner.