assignment 11

course mth 164

mr. SmithHow long of an extention can you and will you grant me if necessary? I am working as diligently as possible but i am worried that I may not finish in time

Thanks Tony" "`ÍMØÒéÄäœÒÉìRíË¥Ž¸¶·ãÕ~åBø øðxnðñ

assignment #011

011.

Precalculus II

12-10-2008

......!!!!!!!!...................................

19:47:58

Query problem 8.4.24 ai+bj form of vector from P (1,4) to Q (6,2).

......!!!!!!!!...................................

RESPONSE -->

1-6=-5

4-2=2

sqrt-5^2+2^2=-4.58

confidence assessment: 1

.................................................

......!!!!!!!!...................................

19:48:56

What is the a i + b j form of the vector from P to Q?

......!!!!!!!!...................................

RESPONSE -->

sqrt a^2+b^2)

confidence assessment: 2

** 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. **

.................................................

......!!!!!!!!...................................

19:51:53

Query problem 8.4.36 ||v + w|| if v = 3i-5j, w = -2i+3j.

......!!!!!!!!...................................

RESPONSE -->

3i-2i=i

3j-5j=-2j

i-2j result

confidence assessment: 2

.................................................

......!!!!!!!!...................................

19:52:50

What is ||v+w|| and how did you obtain it?

......!!!!!!!!...................................

RESPONSE -->

it is the sum of the magnitude of v and w

confidence assessment: 2

** 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. **

.................................................

......!!!!!!!!...................................

19:54:23

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

......!!!!!!!!...................................

RESPONSE -->

dont understand quiestion

confidence assessment: 0

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 as the original, but would have magnitude 1 unit.

** 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. **

.................................................

......!!!!!!!!...................................

19:55:38

What is the unit vector having the same direction as v = -5i + 12j and how did you obtain it?

......!!!!!!!!...................................

RESPONSE -->

it is 5,12

confidence assessment: 2

.................................................

......!!!!!!!!...................................

19:56:28

What is the magnitude ||v|| of v and how was it used in the process of finding the unit vector?

......!!!!!!!!...................................

RESPONSE -->

it is the sqrt of a and b squared

confidence assessment: 2

.................................................

......!!!!!!!!...................................

19:57:23

Query problem 8.4.50 airplane ends up due South 200 miles with 30 mph wind from Northwest.

......!!!!!!!!...................................

RESPONSE -->

ok

confidence assessment: 1

** 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.. **

.................................................

......!!!!!!!!...................................

19:58:22

What is the average air speed of the airplane?

......!!!!!!!!...................................

RESPONSE -->

the airspeed is 230 due to the tailwind effect

confidence assessment: 2

.................................................

......!!!!!!!!...................................

20:01:24

Explain how you obtained your result using vectors.

......!!!!!!!!...................................

RESPONSE -->

actaully the travel speed is 202.23 and i used the formula sqrt30^2+200^2

This would work if the vectors were at right angles, but that's not the case here.

confidence assessment: 2

.................................................

......!!!!!!!!...................................

20:04:30

Query problem 8.5.10 dot product of v = 3i - 4j and w = 4i - 3j; angle between.

......!!!!!!!!...................................

RESPONSE -->

dot product is 3,4+4,-3

angle is cos theta=vtimesw/absolute value of v times w so 3,4+4,4/-3,4+4,3

confidence assessment: 2

** 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.

**

.................................................

......!!!!!!!!...................................

20:04:52

What is the dot product between the two vectors?

......!!!!!!!!...................................

RESPONSE -->

v times w

confidence assessment: 2

.................................................

......!!!!!!!!...................................

20:05:23

What is the angle between the vectors and how did you obtain it?

......!!!!!!!!...................................

RESPONSE -->

vtimesw divided by absolute value of v w

confidence assessment: 2

.................................................

......!!!!!!!!...................................

20:06:37

Query problem 8.5.24 resultant displacement 200 miles west then 150 miles 60 degrees north of west.

......!!!!!!!!...................................

RESPONSE -->

ok

confidence assessment: 2

** 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). **

.................................................

......!!!!!!!!...................................

20:09:52

What is the resultant displacement of the airplane and how did you obtain your result?

......!!!!!!!!...................................

RESPONSE -->

by using v=ai+bj+c=IIwII

confidence assessment: 2

.................................................

......!!!!!!!!...................................

20:10:33

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

......!!!!!!!!...................................

RESPONSE -->

this assignment was challenging and used real world applications that interest me

confidence assessment: 3

.................................................

Be sure you understand the solutions I've inserted. To get maximum benefit from this feedback you should insert self-critiques, marked with &&&&, into a copy of this document.