CAL3_ass_02

#$&*

course Mth 277

1230am 9/28/2011

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

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.

At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises.

query_09_2

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Question: Find u + v, u - v, (5/2)u, and 2u + 3v for the following vectors: u = <1,2,-3>, v = < -1,-2,3>.

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Your solution:

u + v = <0,0,0>

u - v = <2,4,-6>

(5/2)u = < 5/2, 5, -15/2>

2u + 3v = <2, 4, -3> + <-3, -6, 9> = <-1, -2, 6>

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating:

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Question: Find the standard form equation of the sphere with center (-1,2,4) and radius 2.

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Your solution:

(x-h)^2 + (y-k)^2 + (z-n)^2 = r^2

(x+1)^2 + (y-2)^2 + (z-4)^2 = 4

confidence rating #$&*:

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Given Solution:

A point (x, y, z) is on the given sphere if its distance from (-1, 2, 4) is 2, so that

sqrt( (x - (-1))^2 + (y - 2)^2 + (z - 4)^2 ) = 2

and

(x + 1)^2 + (y - 2)^2 + (z - 4)^2 = 4.

This is the equation of the sphere in one form.

Expanding the squares we obtain

x^2 + 2 x + 1 + y^2 - 4 y + 4 + z^2 - 8 x + 16 = 4

which we rearrange to the standard form

x^2 + 2 x + y^2 - 4 y + z^2 - 8 z + 13 = 0.

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Self-critique (if necessary):

I didn't put the equation in standard form. I got halfway, but didn't expand. I now understand that standard form is when I expand the equation out.

I don't understand how you got 13?????????????????? I worked out the solution, but I got 17.

(x + 1)^2 + (y - 2)^2 + (z - 4)^2 = 4.

x^2 + 2 x + 1 + y^2 - 4 y + 4 + z^2 - 8 x + 16 = 4

x^2 + 2 x + y^2 - 4 y + z^2 - 8 x + 21 = 4

x^2 + 2 x + y^2 - 4 y + z^2 - 8 z + 17 = 0.

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Self-critique rating:3

@& I believe you're right.

You can generally trust my solutions but my numbers are sometimes off (mental calculations are fast and usually pretty reliable, but not always).*@

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Question: Find the center and radius of the sphere with equation x^2 + y^2 + z^2 - 2x - 6y + 12z - 17 = 0.

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Your solution:

x^2 + y^2 + z^2 - 2x - 6y + 12z - 17 = 0

x^2 -2x +1 + y^2 -6y +9 + z^2 + 12z +36 = 17 +1 + 9 + 36

(x-1)^2 + (y-3)^2 +(z+6)^2 = 63

radius is 3sqrt(7) or approximately 7.94 units

center of the sphere is at (1,3,-6)

confidence rating #$&*:

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Given Solution:

Completing the squares we obtain

(x^2 - 2 x + 1 - 1) + (y^2 - 6 y + 9 - 9) + (z^2 + 12 z + 36 - 36) = 17

which can be written as

(x - 1)^2 - 1 + (y - 3)^2 - 9 + (z + 6)^2 - 36 = 17

and finally as

(x - 1)^2 + (y - 3)^2 + (z + 6)^2 = 63

This sphere is centered at (1, 3, -6) and has radius sqrt(63) = 3 sqrt(7).

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Self-critique (if necessary):

BAM! oh yeh!

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Self-critique rating:3

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Question: Find the standard representation and length of PQ when P = (-3,1,4) and Q = (2,-4,-3).

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Your solution:

PQ= <5, -5, -7>

ll PQ ll = sqrt(25+25+49)= sqrt(99) = 3(sqrt11)

confidence rating #$&*:

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Given Solution:

PQ = (2 - (-3) ) i + (-4 - 1) j + (-3 - 4) k = 5 i - 5 j - 7 k.

|| PQ || = sqrt( 5^2 + 5^2 + 7^2) = sqrt(99).

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Self-critique (if necessary): OK

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Self-critique rating:

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Question: Find a unit vector in the direction of v = <-1, sqrt(3), 4>.

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Your solution:

ll v ll = sqrt(1+3+16) = sqrt(20) = 2sqrt(5)

find unit vector [1/2sqrt(5)] * v = < -1/2sqrt(5), sqrt(3)/2sqrt(5), 2/sqrt(5)>

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

|| v || = sqrt( 1^2 + sqrt(3) ^ 2 + 4^2 ) = sqrt( 26 )

so a unit vector in the direction of v is

v / || v ||= < -1, sqrt(3), 4 > / sqrt(26) = <-sqrt(26) / 26, sqrt(78) / 26, 4 sqrt(26) / 26)> .

4 sqrt(26) / 26 is 2 sqrt(26) / 13.

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Self-critique (if necessary):

the magnitude is sqrt(20). NOT sqrt(26). 1+3+16= 20 NOT 26.

How do I simplify my answer??????????? I don't know how to simplify sqrt(3)/2sqrt(5)

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Self-critique rating: 3

@& I apparently squared the 3.

Standard form doesn't have square roots in the denominator. 1 / sqrt(5) = sqrt(5) / 5, so you would get sqrt(3) sqrt(5) / (2 * 5) = sqrt(15) / 10.

I wouldn't penalize this, but it's a good thing to know.*@

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Question: Sketch and describe the cylindrical surface given by y = cos x.

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Your solution:

I have no earthly idea!????????????? A cosine in 2 dimensions starts at zero and goes from 1 to -1. In 3 dimensions, I assume it does the same thing, but in a circle. So it creates a cosine wrapping around the z axis.

confidence rating #$&*:

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Given Solution:

In the x-y plane y = cos(x) consists of a sinusoidal function oscillating between the lines y = -1 and y = 1, with period 2 pi radians, and containing the point (0, 1).

The surface in 3 dimensions repeats this same curve for every value of z, so that the graph represents a wavy curtain hanging vertically downward, intersecting the xy plane along the sinusoidal curve.

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Self-critique (if necessary):

Where does the value of z come from? I understand in a sphere, we have x^2 + y^2 + z^2 = r^2, but for a cylindrical surface, where is the value of z, when y= cosx?????????????

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Self-critique rating: 3

@& z is not specified in the equation, so for any value of x and y on the graph, all values of z are possible. So each point of the graph in the xy plane extends to an infinite vertical line, resulting in the 'curtain' shape.*@

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Question: Determine if u = 2i + 3j + -4k is parallel to v = <1,-3/2,2>.

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Your solution:

yes, because 1/2u = v.

confidence rating #$&*:

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Given Solution:

Two vectors are parallel if the angle between them is 0 or pi radians (180 degrees), meaning that the cosine of the angle between them is 1 or -1.

u dot v = || u || || v || cos(theta)

so that

cos(theta) = u dot v / (|| u || || v || )

= (2 * 1 + 3 * (-3/2) + (-4 * 2) ) / ( sqrt(2^2 + 3^2 + 4^2) * sqrt( 1^2 + (3/2)^2 + 2^2) )

= (-21/2) / (sqrt( 29) sqrt(29/4).

This is not 1 or -1, so the cosine is neither 0 nor pi rad (i.e., 180 deg).

The vectors are therefore not parallel.

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Self-critique (if necessary):

u dot v = ll u ll * ll v ll cos(theta)

(2-9/2-8) = sqrt(29)*sqrt(s9/4)*cos(theta)

cos(theta) = -10.5/14.5

theta = 136.4° =/ 180°

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Self-critique rating: 3

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Question: Find the lengths of the sides of the triangle and determine if the triangle with vertices A(3,0,0), B(7,1,4) and C(5,4,4) is a right triangle, isosceles triangle, both, or neither.

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Your solution:

AB= < -4, -1, -4>

BC= < -2, -3, 0 >

AC= < -2, -4, -4>

to find BA, BC and CA, just multiply by -1.

ll AB ll = ll BA ll = sqrt(33)

ll BC ll = ll CB ll = sqrt(13)

ll AC ll = ll CA ll = sqrt(24)

to find angles between the vectors

AB dot AC = ll AB ll * ll AC ll * cos(theta)

28/ [sqrt(33)*sqrt(24)] = cos(theta)

theta = 5.7°

BA dot BC = ll BA ll * ll BC ll * cos(phi)

-11/[sqrt(33)*sqrt(13)] = cos(phi)

phi = 122°

Therefore the triangle is obtuse

confidence rating #$&*:

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Given Solution:

The sides can be represented by the vectors

AB = < 4, 1, 4 >,

BC = < -2, 3, 0 > and

AC = < 2, 4, 4 >.

The magnitudes of these vectors are respectively

sqrt(33)

sqrt(13)

sqrt(36).

None of the sides are the same length so the triangle is not isosceles.

The sum of the squares of the shorter two side is 33 + 13 = 46, which is not equal to the sum of the longest, so the triangle is not a right triangle.

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Self-critique (if necessary):

Thats not how I did the problem, but I am confident in my answer. Please verify I did this properly????????????????? Thank you

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Self-critique rating: OK

@& You've seen the dot product, so that's fine. However it hasn't yet been introduced in the text, so the solution based on the Pythagorean Theorem is the one that's given here.

You should understand it both ways.*@

Self-critique (if necessary):

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Self-critique rating:

&#Good work. See my notes and let me know if you have questions. &#

@& Note also the link to videos, recently place on the Assignments page just before the listing of assignments. Let me know if you can't locate that link.*@

CAL3_ass_02

@& I believe you're right. You can generally trust my solutions but my numbers are sometimes off (mental calculations are fast and usually pretty reliable, but not always).*@

@& I apparently squared the 3. Standard form doesn't have square roots in the denominator. 1 / sqrt(5) = sqrt(5) / 5, so you would get sqrt(3) sqrt(5) / (2 * 5) = sqrt(15) / 10. I wouldn't penalize this, but it's a good thing to know.*@

@& z is not specified in the equation, so for any value of x and y on the graph, all values of z are possible. So each point of the graph in the xy plane extends to an infinite vertical line, resulting in the 'curtain' shape.*@

@& You've seen the dot product, so that's fine. However it hasn't yet been introduced in the text, so the solution based on the Pythagorean Theorem is the one that's given here. You should understand it both ways.*@

&#Good work. See my notes and let me know if you have questions. &#

@& Note also the link to videos, recently place on the Assignments page just before the listing of assignments. Let me know if you can't locate that link.*@

@& I believe you're right.

You can generally trust my solutions but my numbers are sometimes off (mental calculations are fast and usually pretty reliable, but not always).*@

@& I apparently squared the 3.

Standard form doesn't have square roots in the denominator. 1 / sqrt(5) = sqrt(5) / 5, so you would get sqrt(3) sqrt(5) / (2 * 5) = sqrt(15) / 10.

I wouldn't penalize this, but it's a good thing to know.*@

@& You've seen the dot product, so that's fine. However it hasn't yet been introduced in the text, so the solution based on the Pythagorean Theorem is the one that's given here.

You should understand it both ways.*@