#$&*
course Mth 277
2/8 2
If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
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Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises.
query_09_2
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Question: Find u + v, u - v, (5/2)u, and 2u + 3v for the following vectors: u = <1,2,-3>, v = < -1,-2,3>.
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Your solution:
u + v = <1+(-1), 2+(-2), -3+3> = <0, 0, 0>
u - v = <1-(-1), 2-(-2), -3- 3> = <2, 4, -9>
(5/2)u =(5/2)<1, 2, -3> = <1*(5/2), 2*(5/2), -3*(5/2)> = <5/2, 5, -15/2>
2u + 3v = …= <-1, -2, 3>
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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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+1)^2 + (y-2)^2 + (z-4)^2 = 4, where standard form is (x-a)^2 + (y-b)^2 + (z-c)^2 = r^2
confidence rating #$&*:
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Given Solution:
@& Good. For reference:
A sphere of radius r centered at (x0, y0, z0) consists of all points whose distance from a center point (x0, y0, x0) is equal to r.
Thus a point (x, y, z) is on the sphere if
sqrt( (x - x0)^2 + (y - y0)^2 + (z - z0)^2 ) = r.
We can square both sides to get the standard form of the equation:
(x - x0)^2 + (y - y0)^2 + (z - z0)^2 = r^2.
For the given center and radius we therefore obtain standard form
(x+1)^2 + (y-2)^2 + (z-4)^2 = 2^2 or
(x+1)^2 + (y-2)^2 + (z-4)^2 = 4.
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Self-critique (if necessary):
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Question: Find the center and radius of the sphere with equation x^2 + y^2 + z^2 - 2x - 6y + 12z - 17 = 0.
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Your solution:
Completing the square
(x^2-2x) + (y^2-6y) + (z^2-12z) = 17
(x^2-2x + 1) + (y^2-6y+9) + (z^2-12z+36) = 17+1+9+36
(x-1)^2 + (y - 3)^2 + (z - 6)^2 = 63
Center = (1, 3, 6)
radius = `sqrt(63)
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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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 = 5i - 5j - 7k
||PQ|| = `sqrt(5^2 + (-5)^2 + (-7)^2) = `sqrt(99) = approx. 9.9499
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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Question: Find a unit vector in the direction of v = <-1, sqrt(3), 4>.
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Your solution:
Let u be unit vector of v
u = v/||v|| = <-i+`sqrt(3)+4k>/`sqrt(-1^2 + `sqrt(3)^2 + 4^2) = <-i+`sqrt(3)+4k>/`sqrt(20)
= -i/`sqrt(20) + `sqrt(3)/`sqrt(20) + 4/`sqrt(20)
confidence rating #$&*:
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Given Solution:
@& Good. For reference (note the rationalized denominators near the end, which is the standard form):
A vector divided by a positive scalar c yields a vector in the same direction as the original, but with 1/c times the magnitude of the original. If the scalar is negative the same is true, except that the direction is reversed.
A unit vector has magnitude 1.
It follows that a unit vector in the direction of a given vector is equal to that vector, divided by its magnitude.
In this case we obtain the unit vector `u as follows:
`u = `v/||`v||
= <-1,`sqrt(3),4>/`sqrt((-1)^2 + `sqrt(3)^2 + 4^2)
= <1,`sqrt(3),4>/`sqrt(20)
= -1 /`sqrt(20) `i + `sqrt(3)/`sqrt(20) `j + 4/`sqrt(20) `k
= -sqrt(5) / 10 `i + sqrt(15) / 10 `j + 2 sqrt(5) / 5 `k
A very rough approximation to this vector is
`u = -.2 `i + .4 `j + .9 `k (very approx.)
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Self-critique (if necessary):
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Question: Sketch and describe the cylindrical surface given by y = cos x.
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Your solution:
It’s the wave looking fn which could be seen as half elliptic’s shape with alt of z coord.
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
@& In the x-y plane you get the graph of the cosine function, where y(0) = 1, y (pi/2) = 0, y(pi) = -1, y(3 pi/2) = 0 and y (2 pi) = 1. The curve is periodic with period 2 pi.
Since z is unspecified, for any point (x, y) on the curve z can take any value. Therefore if from every point of the cosine curve you extend a line in the vertical z direction, you will have the surface.
If you take the y-z plane and deform it in the x direction so that it follows the curve y = cos(x) in the x-y plane, much like the folds of a hanging curtain (but much more smooth and regular), you will get the surface. You can imagine hanging a curtain from a wavy track having the shape of the cosine curve.
A figure depicting part of this curve is at
http://vhcc2.vhcc.edu/dsmith/geninfo/qa_query_etc/multivar/qy02_q6.gif*@
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Question: Determine if u = 2i + 3j + -4k is parallel to v = <1,-3/2,2>.
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Your solution:
We could mult v by scalar s= -(1/2) and all terms but i term would match and because scalar must be mult through entire vector then I say no I do not believe v is or can be parallel to u.
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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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:
Length
AB = `sqrt((7-3)^2 + 1^2 + 4^2) = `sqrt(33)
AC = `sqrt((5-3)^2 + 4^2 + 4^2) = `sqrt(2^2 + 4^2 + 4^2) = `sqrt(36) =6
BC = `sqrt((5-7)^2 +(4-1)^2 + (4-4)^2) = `sqrt(13)
With these sides I believe triangle is none of the above, but I think I miss calculated somewhere.
Confidence rating: 3
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Given Solution:
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Self-critique (if necessary):
@& Your statement is correct but would be better if stated in a way closer to the definitions.
Consider the following:
In the x-y plane you get the graph of the cosine function, where y(0) = 1, y (pi/2) = 0, y(pi) = -1, y(3 pi/2) = 0 and y (2 pi) = 1. The curve is periodic with period 2 pi.
Since z is unspecified, for any point (x, y) on the curve z can take any value. Therefore if from every point of the cosine curve you extend a line in the vertical z direction, you will have the surface.
If you take the y-z plane and deform it in the x direction so that it follows the curve y = cos(x) in the x-y plane, much like the folds of a hanging curtain (but much more smooth and regular), you will get the surface. You can imagine hanging a curtain from a wavy track having the shape of the cosine curve.
A figure depicting part of this curve is at
http://vhcc2.vhcc.edu/dsmith/geninfo/qa_query_etc/multivar/qy02_q6.gif*@
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Self-critique rating:
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Self-critique (if necessary):
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Self-critique rating:
@& For the problem regarding the sides of the triangle, your solution is good, but consider the following:
It is straightforward to find the lengths.
The Pythagorean Theorem can be applied to test whether the triangle is a right triangle.
If two of the lengths are equal the triangle is isosceles.
It is possible for the triangle to be both, in which case you should easily see that the angles of the triangle are 45 deg, 45 deg and 90 deg.
Specifically:
AB = `sqrt((7-3)^2 + 1^2 + 4^2) = `sqrt(33)
AC = `sqrt((5-3)^2 + 4^2 + 4^2) = `sqrt(2^2 + 4^2 + 4^2) = `sqrt(36) =6
BC = `sqrt((5-7)^2 +(4-1)^2 + (4-4)^2) = `sqrt(13)
None of these are equal, so the triangle is not isosceles.
The sum of the squares of the two shorter sides is equal to the square of the longest and we have
sqrt(13)^2 + sqrt(33)^2 = 6^2.
This gives us
13 + 33 = 36.
This is clearly not true, and we conclude that the triangle is not a right triangle.*@
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