assignment 9

course mth164

Mr. Smith when you recieve this if you have not already emailed me my grade please do so for test 1.thanks Tony

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assignment #009

009. Vectors

Precalculus II

11-13-2008

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10:21:01

Previous Assignments: Be sure you have completed Assignment 8 as instructed under the Assts link on the homepage at 164.106.222.236 and submitted the result of the Query and q_a_ from that Assignment.

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RESPONSE -->

ok

self critique assessment: 3

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10:22:52

`q001. Sketch the points (2,3) and (9,14) on a set of coordinate axes. Give the x and the y displacements from (2,3) to (9,14).

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RESPONSE -->

2-9/3-14=approx .6363

confidence assessment: 2

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10:23:46

As shown in Figure 75, the x displacement is from 2 to 9, a displacement of 9 - 2 = 7, while the y displacement is from 3 to 14, a displacement of 14 - 3 = 11. The arrows represent the direction of the displacements, from the initial point (2, 3) to the terminal point (9, 14).

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RESPONSE -->

used the reverse order on the x coordinates and this skewed the results.

self critique assessment: 2

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10:26:48

`q002. In the preceding example we saw that the x and y displacement from from (2,3) to (9,14) are 9-2 = 7 and 14-3 = 11. Sketch an arrow which originates at (2,3) and terminates at (9,14), with the point of the arrow at the terminating end. If we were to sketch a geometrically similar arrow, having the same slope, orientation and length as the preceding, but starting at the point (-2, 5) at what point would the arrow terminate? Note that we can and should really incorporate information from the physics introductory problems.

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RESPONSE -->

-2+7

11+5

yields 5/15

confidence assessment: 2

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10:27:32

Originating at (-2,5) the arrow will displace 7 units in the x direction and 11 units in the y direction. Starting at x = -2 the arrow will displace 7 units in the x direction to end up at x = -2 + 7 = 5. Starting at y = 5 the arrow will displace 11 units in the y direction and end up at y = 5 + 11 = 16. The arrow therefore originates at (2,-5) and terminates at (5, 16).

If we sketch the same arrow starting from the point (-2, 5) then it will again displace 7 units in the x direction, ending up at x = -2 + 7 = 5, and 11 units in the y direction, ending up at y = 5 + 11 = 16. As shown in Figure 58 the arrow terminates at the point (5, 16).

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RESPONSE -->

incorrectly added the y values and gave 15 instead of 16 overall correct formula

self critique assessment: 3

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10:28:14

`q003. We say that the displacement of 7 units in the x direction and 11 units in the y direction is a vector, represented by the arrows used in the preceding problems and denoted using 'pointy braces' as < 7, 11 >. If we apply this vector, starting this time at the origin, at what point do we end up?

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RESPONSE -->

7,11

confidence assessment: 2

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10:28:21

As shown in Figure 40 this vector takes us from the origin (0,0) to the point (7, 11).

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RESPONSE -->

ok

self critique assessment: 3

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10:29:21

`q004. If we start from the terminal point of the vector in the preceding exercise and sketch a new vector having x displacement 3 and y displacement -8, at what point do we end up? Sketch the arrows representing these two vectors, the first running from (0,0) to (7,11) and the second from that point to its terminal point.

Now sketch a vector from directly from (0,0) to the terminal point of the second vector.

How can the x displacement of this new vector be calculated from the x displacements of the first two vectors? Answer the same question for the y coordinates.

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RESPONSE -->

10,3

confidence assessment: 2

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10:29:27

As shown in figure 11, the second vector runs from (7, 11) to (7 + 3, 11 + (-8) ) = (10, 3).

The vector from the initial point of the first vector to the terminal point of the second therefore runs from (0, 0) to (10, 3), as shown in Figure 72.

It should be clear from the calculations done above and from the sketches that the x displacement of the new vector is calculated by adding the x displacements of the original two vectors, and that the same strategy works for the y displacements.

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RESPONSE -->

ok

self critique assessment: 3

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10:29:54

`q005. In what sense can we say that the vector <10,3> is the sum of the two vectors <7, 11> and <3, -8>?

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RESPONSE -->

yes because the x and the y values have been added

confidence assessment: 2

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10:30:10

The x coordinate of the new vector to is 10, which is the sum 7 + 3 of the x coordinates of the two vectors. The y coordinate of the new vector is 3, the sum 11 + (-8) of the y coordinate of the two vectors. In this respect it is the sum.

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RESPONSE -->

ok

self critique assessment: 2

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10:30:35

`q006. http://164.106.222.236/ph1introsets/default.htm and choose Set 5, Vectors. Click in turn on Problems 1 - 9 and see if you can solve these problems. Solutions are given and are generalized and many are accompanied by figures. If you can't immediately solve them, study the solutions and learn to solve them. Explain the solution to the first problem.

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RESPONSE -->

ok

confidence assessment: 3

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10:30:51

In case clarification is needed, displacement is just movement through a distance and in a certain direction. The vector (3, -8) of the preceding problem (and figur 72) corresponds to a displacement of 3 units in the x direction and -8 units in the y direction.

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RESPONSE -->

ok

self critique assessment: 3

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&#Good work. Let me know if you have questions. &#

assignment 9

course mth164

WŁאyNassignment #009

009.

Precalculus II

11-13-2008

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10:33:59

Query problem 7.3.14 b = 4, c = 1, `alpha = 120 deg

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RESPONSE -->

i will use the law of sines to complete.

confidence assessment: 3

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10:36:45

specify the unknown sides and angles of your triangle.

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RESPONSE -->

a=4.123

b=4

c=1

a=120

b=58

c=2

confidence assessment: 2

You don't have an angle and the side opposite that angle, so you can't use the law of sines to start out. You need to use the law of cosines.

a= sqrt(4^2 + 1^2 -2(4)(1)cos120)

a = sqrt(16+1-8cos120)

a = sqrt(17-8cos120)

a = 4.58

b^2 = a^2 + c^2 -2accos`beta

4^2 = 4.58^2 + 1^2 - 2(4.58)(1)cos`beta

9.16cos`beta = 4.58^2 + 1^2 - 4^2

9.16cos`beta = 5.98

beta = cos^-1(5.98/9.16)

beta = 49.2deg

c^2 = a^2 + b^2 -2abcos`gamma

1^2 = 4.58^2 + 4^2-2(4.58)(4)cos`gamma

36.64cos`gamma = 35.98

gamma = cos^-1(35.98/36.64)

gamma = 10.89

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10:37:03

Explain how you obtained your results.

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RESPONSE -->

by using the pythagorean theory and the law of sines

confidence assessment: 3

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10:53:45

Query problem 7.3.32 500 ft tower 15 deg slope 2 guy wires 100 ft on either side of base on slope

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RESPONSE -->

ok

confidence assessment: 3

** GOOD STUDENT SOLUTION: A triangle containing the tower side and the side of the slope up the hill will have angle 75 deg between tower and hill. The triangle with the side down the slope has angle 105 deg between tower and hill.

The guy wire will form the hypotenuse c on each side.

If

c1 = length of guy wire on the right hand side of the tower and

c2 = length of guy wire on the left hand side of the tower

then the Law of Cosines gives us

(c1)^2 = a^2 + b^2 - 2ab cos `gamma

= 500ft^2 + 100ft^2 - 2(500)(100)(cos75 deg)

= 260,000 - 25,881.90451

(c1)^2 = 234118.0955

c = approx. 483.8575 feet

(c2)^2 = a^2 + b^2 - 2ab cos`gamma

= 500ft^2 + 100ft^2 -2(500)(100)(cos105 deg)

= 260,000 - (-25881.90451)

(c2)^2 = 285,881.9045

c = approx. 534.68 feet **

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10:59:21

how long should the two guy wires be?

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RESPONSE -->

386.37 is the length of the guy wires

confidence assessment: 2

There are two guy wires; both would be longer than 387 ft.

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10:59:37

Explain how you obtained your result.

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RESPONSE -->

by using the law of sines

confidence assessment: 2

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11:01:00

query problem 7.5.6 coiled spring 10 cm displ period 3 sec moving downward at equil when t = 0.

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RESPONSE -->

ok

confidence assessment: 2

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11:03:31

What is the equation of motion of the object?

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RESPONSE -->

the equation of motion is v=vi+adelta t

This equation would apply if acceleration was uniform, provided you have the initial velocity and acceleration.

However acceleration in this case is not uniform.

confidence assessment: 2

** Modeling with reference-circle point moving counterclockwise at angular velocity omega on a circle of radius A, and using the sine function, displacement will be

y(t) = A sin (`omega * t + theta0), where theta0 is the initial angular position on the reference circle.

We know that the spring is at equilibrium when t = 0, so theta0 is either 0 or pi.

Since the spring is initially moving downward, we conclude that theta0 is pi.

The amplitude of motion will be 10 cm so A = 10 cm.

The period of motion is 2`pi/`omega = 3 sec, so `omega = 2`pi/(3 sec), so

the equation of motion is y = 10 cm sin( 2 pi / (3 sec) * t + pi).

Alternatively we could model using the cosine function, though this is less intuitive for an object moving up and down:

x(t) = A cos(omega * t + theta0),

in which case since the spring is at equilibrium when t = 0, theta0 is pi/2 (motion at t = 0 therefore being in the negative direction)

and we get

x(t) = 10 cm cos(2 pi / (3 sec) * t + pi/2).

**

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11:04:21

What is the amplitude of the motion and how does this amplitude affect the equation of motion?

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RESPONSE -->

the amplitude is 10

this effects the equaiton by changing the range that the object travles

confidence assessment: 2

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11:04:46

What is the period of the motion, how does the period affect the angular frequence of the motion and how does the angular frequency appear in the equation of motion?

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RESPONSE -->

the period of motion is the length that the object travels

confidence assessment: 2

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11:05:21

What give condition determines the phase shift for this situation and how does the phase shift affect the equation of motion?

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RESPONSE -->

the phase shift is a condition that is effected by the t function

confidence assessment: 2

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11:05:42

query problem 7.5.12 d=5 cos(`pi /2)t

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RESPONSE -->

ok

confidence assessment: 3

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11:06:41

Describe the motion of the object

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RESPONSE -->

the motion of the object is moving is up and down with a range of 5 and no lateral movement

confidence assessment: 2

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11:06:48

What is the maximum displacement for its resting position?

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RESPONSE -->

5

confidence assessment: 2

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11:06:59

What is the time required for the oscillation?

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RESPONSE -->

1

confidence assessment: 2

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11:07:19

What is the frequency?

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RESPONSE -->

pi,2

confidence assessment: 2

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11:08:25

what is the area of the cone?

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RESPONSE -->

the cone has an area of 1x5xpi

confidence assessment: 2

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11:08:59

Explain how you obtained eac result

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RESPONSE -->

by using the formula in the book for a cone and movements of an object

confidence assessment: 3

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11:09:16

Comm on any surprises or insights you experienced as a result of this assignment.

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RESPONSE -->

need to reaserach the moving of object formulas

confidence assessment: 3

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You aren't showing me enough detail for me to give you feedback specific to your work. Try to include more detail in your solutions.

&#Please see my notes and submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end). &#