Assignment 11b

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course mth 164

11/12 6pm

SOLUTIONS/COMMENTARY ON QUERY 11Query problem 8.4.24 (was 8.4.26?) ai+bj form of vector from P (-1, 4) to Q

(6,2).

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Your solution:

6-(-1) = 7

2-4 = -2

The vector is v = 7i - 2j

confidence rating #$&*: 3

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Given Solution:

** If we have the initial point at (-1,4) and terminal point at (6,2).

The x displacement is from -1 to 6, a displacement of 6 - (-1) = 7

The y displacement is from 4 to 2, a displacement of 2 - 4 = -2

The i and j unit vectors are in the x and y directions, respectively, so our vector is

v = 7 i - 2 j. **

** If we have the initial point at (1,4) and terminal point at (6,2).

The x displacement is from 1 to 6, a displacement of 6 - (1) = 5

The y displacement is from 4 to 2, a displacement of 2 - 4 = -2

The i and j unit vectors are in the x and y directions, respectively, so our vector is

v = 5 i - 2 j. **

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Self-critique (if necessary):

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Self-critique Rating:

Query problem 8.4.36 (was 8.4.38?) ||v|| + ||w|| if v = 3i-5j, w = -2i+3j.

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Your solution:\

||v|| = sqrt(3^2 + 5^2) = sqrt(34)

||w|| = sqrt(2^2 + 3^2) = sqrt(13)

So the answer is sqrt(34) + sqrt(13) = 9.437

confidence rating #$&*: 3

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Given Solution:

** to find ||v + w|| if v = 3i-5j, w = -2i+3j, we find that

||v||= sqrt(3^2 + (-5)^2) = sqrt 34

and

||w||= sqrt (-2^2 + 3^2 ) = sqrt 13

so

||w+v|| = sqrt34+ sqrt 13

or in decimal form 9.44.

To get || w + v || you have to first find w + v, then take the magnitude of this resultant.

|| w + v || = || (3 i - 5 j) + (-2 i + 3 j) || = || i - 2 j || = sqrt(1^2 + 2^2) = sqrt(5).

This, with your work, demonstrates that || w + v || is not generally equal to || w || + || v ||.

The two expressions are in fact equal if, and only if, the two vectors are parallel and in the same direction. **

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Self-critique (if necessary):

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Self-critique Rating:

Query problem 8.4.42 unit vector having same direction as v = -5 i + 12 j.

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Your solution:

First we need to find ||v|| which is sqrt(5^2 + 12^2) = sqrt(169) = 13

u = -5i + 12j / 13, -5/13i + 12/13j

To check to make sure this is a unit vector we can do this:

(-5/13)^2 + (12/13)^2 = 1

confidence rating #$&*: 3

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Given Solution:

GOOD STUDENT SOLUTION WITH AMPLIFICATION BY INSTRUCTOR

We need to find a vector having magnitude 1 and the same direction as v = -5i + 12j.

We first need to find ||v||:

||v||=sqrt(25 + 144)=sqrt 169 or 13.

||v||=13

v/||v|| = (-5i + 12j) / 13= -5/13 i + 12/13 j.

This vector would have the same direction.

** Correct solution.

The vector has the same direction because it is a multiple of the vector by a positive constant.

The vector is a unit vector because when you divide a vector by a constant its new magnitude is also divided by that constant; you have divided the original vector by its original magnitude so its new magnitude is equal to its original magnitude divided by its original magnitude; the result is new magnitude 1. **

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Self-critique (if necessary):

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**** Query problem 8.4.50 airplane ends up due South 200 miles with 30

mph wind from Northwest.

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Your solution:

I do not understand the setup up of this question?

confidence rating #$&*: 0

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Given Solution:

** The airplane moves as some velocity with respect to the wind, which since it is being carried along with the wind is different that its velocity with respect to the ground.

If w is the direction of the wind and v the velocity of the airplane with respect to the wind.

The velocity of the wind is added to the velocity of the airplane with respect to the wind.

Then the actual velocity of the airplane is v + w.

Since the airplane ends up 200 miles due South after 1 hour, its actual velocity is 200 mph to the South. Thus we know that

v + w = -200 j,

understanding that the -200 is in mph.

If w is from the Northwest at 30 mph then, placing North and East in the positive y and x directions respectively we see that the direction of w is toward the Southeast, at angle 315 deg with respect to the positive x axis. Thus

w = 30 cos(315 deg) i + 30 sin(315 deg) j

=21.2 i -21.2 j, approx..

Now if v + w = -200 j it follows that

v = -200 j - w = 200 j - (21.2 i -21.2 j) = 200 j -21.2 i +21.2 j = 21.2 i -221.2 j.

The magnitude of this vector is sqrt( (21.2)^2 + (-221.2)^2) = 222.2, approx.. **

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Self-critique (if necessary): How am I supposed to know that the angle is 315deg when we are given no indication of any angle in the question. The questions in the book give what angle we are working with. This question is not in the 8th edition of the textbook and the numbers hardly ever match with the ones we are assigned.

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Self-critique Rating:2

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The positive x axis is taken to be to the east, with the positive y axis north. This is the standard orientation of axes for compass directions.

However there is information in the text that isn't represented in the abbreviated statement given here.

Unfortunately I can't double-check the correspondences with the text from this location.

Email me and I'll be able to take a look at it when I'm on campus.

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**** Query problem 8.5.10 (was 8.5.8?) dot product of v = 3i - 4j and w = 4i - 3j.

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Your solution: This is question 14 of section 9.5 of the 8th edition.

v*w: 3(4) + (-4) -3 = 24

||v|| : sqrt(3^2 +4^2) = sqrt(25) = 5

||w|| : sqrt(4^2 + 3^2) = sqrt(25) = 5

Cos(theta) = 24/25

Theta = cos^-1(24/25) = 16.26 deg.

Neither.

confidence rating #$&*: 3

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Given Solution:

** Using the definition of the dot product:

v dot w = ||v|| * ||w|| cos(`theta) so we have

cos(`theta) = v dot w / ( ||v|| ||w|| ) = 24 / 25 so

`theta = cos^-1(24 / 25) = 16 deg.

**

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Self-critique (if necessary):

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Self-critique Rating:

**** Query problem 8.5.24 resultant displacement 200 miles west then

150 miles 60 degrees north of west.

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Your solution:

?

confidence rating #$&*:

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Given Solution:

** 200 miles west is represented by the displacement vector v = -200 i + 0 j.

150 miles north of west is 150 miles at an angle of 120 deg so the displacement vector here is

w = 150 cos(120 deg) * i + 150 sin(120 deg) * j = -75 i + 130 j, approx.

So net displacement is

v + w = -200 i + 0 j + -75 i + 130 j = -275 i + 130 j.

Magnitude is `sqrt( 275^2 + 130^2) = 300 approx..

Angle is tan^-1(130/(-275)) + 180 deg = -30 deg + 180 deg = 150 deg (approximately). **

INSTRUCTIVE STUDENT ERROR:

we can draw the points on the graph and we see that it travels 200 units in

the negative on the x-axis and then at an angle of 60 deg it travels 150

miles north of west. we thus have two sets of points on the graph so we can

say P sub 1(200,0) and P sub 2(150, 60 deg)

** correct if both of these representations are in polar coordinates **

so by using the distance formula

we can say sq.rt.((200-150)^2+(0-60)^2 = 78.10 mi

** the distance formula for rectangular coordinates doesn't work with polar coordinates. **

these sections were a lot of fun to work on however there were some

difficult problems.i thought it was nice that we were allowed to make use of

some of the basic properties we learned in algebra one such as the

commutative, associative and distributive properties to solve problems

involving vectors.

** vector algebra is an entire subject in itself. Take a linear algebra course sometime if you get the chance. **

POSSIBLE ERROR IN STATEMENT OF PROBLEM:

POSSIBLE CORRECTION BY RELIABLE STUDENT:

The problem gives the angle 60 deg and the leg adjacent to this angle so you use the tangent to find the opposite leg, the displacement.

tan 'theta = opp/adj

tan 60 deg = x/150

x = 150 tan 60 deg

x = about 259.81 mi.

confidence rating #$&*: 2

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Self-critique (if necessary): Confused by these types of problems.

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Self-critique Rating:1

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The negative x axis is to the west, the positive y axis to the north.

So 60 degrees north of west is at 120 degrees on the standard coordinate system.

Thus

200 miles west is represented by the displacement vector v = -200 i + 0 j.

150 miles north of west is 150 miles at an angle of 120 deg so the displacement vector here is

w = 150 cos(120 deg) * i + 150 sin(120 deg) * j = -75 i + 130 j, approx.

The rest follows pretty easily from this, but let me know what you do and do not understand up to this point.

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Self-critique (if necessary):

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Self-critique rating:

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Self-critique (if necessary):

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Be sure to check my notes and respond as indicated.

Note also that your test results are posted at the Supervised Study Fall 12 site.

Be sure to email me as indicated both in this document and in the gradebook.

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