#$&*
course Mth 277
2/8 2
If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
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Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises.
query_09_5
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Question: Find an explicit relationship between x and y by eliminating the parameter in the following equations: x = e^-t, y = e^t. Sketch the corresponding curve for -inf <= t <= inf. (inf stands for infinity).
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Your solution:
ln(y) = t
x = e^-(ln(y)) = 1/y
For this curve x gets larger very quick as y approaches 0 and y has very large value as x approaches 0. It looks like a fn that runs along y and x axis but stays only in the first Quad of graph.
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
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Question: Find the parametric and symmetric equations for the line passing through the point (-1,-1,0) and parallel to the line (x-3)/4 = (y-1)/3 = (z+3)/2
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Your solution:
L is line containing pt (-1, -1, 0)
v = 4i + 3j + 2k
Parametric
x = 3 + 4t, y = 1 + 3t, z = -3 + 2t
Symmetric
(x -3)/4 = (y - 1)/3 = (z - 3)/2
confidence rating #$&*:
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Given Solution:
@& The equation through the point is characterized by
(x - (-1) ) `i + (y - (-1) ) `j + (z - 0 ) `k parallel to v = 4i + 3j + 2k
so that
(x - (-1) ) `i + (y - (-1) ) `j + (z - 0 ) `k = t * ( 4i + 3j + 2k ) for some t,
which implies that
(x + 1) = 4 t
y + 1 = 3 t
z = 2 t
giving us
t = (x + 1) / 4 = (y + 1) / 3 = z / 2.
and the symmetric equation
(x + 1) / 4 = (y + 1) / 3 = z / 2.*@
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Self-critique (if necessary):
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Self-critique rating:
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Question: Find the intersection of the line represented by the parametric equations x = 3t + 4, y = 1 - 3t, z = 2t - 7 with each of the coordinate planes (if the line doesn't intersect one or more coordinate plane, specify which one).
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Your solution:
Intersects
xy-plane, t = 7/2
x = 14.5, y = -9.5 (14.5, -9.5, 0)
xz-plane, t = 1/3
x = 5, y = 0 (5, 0, 0) ?????Does this satisfy passing through xz-plane???
yz-plane, t = -4/3
y = -3, z = -9.6667 (0, -3, -9.6667)
confidence rating #$&*:
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Given Solution:
@& x = 3t + 4, y = 1 - 3t, z = 2t - 7
so:
If x = 0 we have
0 = 3 t + 4
so that t = -3/4 and
y = 1 - 3 * (-3/4) = 13/4
z = 2 * (-3/4) - 7 = -17/2.
x = 0 describes the y - z plane, so the graph intersects the y z plane at
(0, 13/4, -17/2).
If y = 0 (corresponding to the x z plane) we get
0 = 1 - 3 t so that
t = 1/3,
giving us
x = 5, z = -19/3
and intercept
(5, 0, -19/3)
If z = 0 we get t = 7/2 so that
x = 29/2 and y = -19/2
with xy intercept being (29/2, -19/2, 0).*@
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Self-critique (if necessary):
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Self-critique rating:
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Question: Show whether the line represented by the parametric equations x = 2-t, y = 3t , z = 3 - 2t and the line represented by x = 5-t, y = -1-3t, z = -3 +4t intersect, are parallel, or if they are skew. If they intersect, give the point of intersection.
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Your solution:
I know they are not parallel but have made a mistake somewhere, I’ve worked this problem for an hour now and I’m just going to move on. I got it to
L_1 : x = 2 - s, y = 3s, z =3 - 2s
L_2: x = 5 - t, y = -1 - 3t, z = -3 + 4t
System of Equations
s - t = -3
3s + 3t = -1
2s + 4t = 6
Can’t figure out where I’m making my mistake or maybe this system of equations can’t be solved I just have to move on either way.
Confidence rating: 2
@& You're very close. You have three equations in two unknowns. Generally you can't get a solution, but you might in this case. Solve the first two equations for s and t, and see if your results work in the third.
The lines are respectively parallel to the vectors
-`i + 3 `j - 2 `k
and
-`i - 3 `j + 4 `k
Neither is a multiple of the other, as can be seen either by inspection or by setting up and solving the necessary equations.
The lines intersect if there are two values of t, say t = r and t = s, for which the x, y and z coordinates of the first line at t = r is equal to the x, y and z coordinates of the second line at t = s.
Thus the conditions are:
x coordinates equal:
2 - r = 5 - s
3 r = -1 - 3 s
3 - 2 r = -3 + 4 s.
You will notice that the three equations contain only two unknowns, r and s. A random set of three linear equations in two unknowns is unlikely to have a solution, but the numbers in this equation are all relatively small integers so a solution is conceivable; and of course this is a text problem and the numbers could have been manipulated so that the lines do intersect.
We can solve the first two equations for r and s. Recall that this would give us two values of t, one for each of the respective lines.
If this solution also works in the third equation, we have a solution and the lines intersect. We can then find the point of intersection by plugging our solution into the equation of the line.
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
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Question: Determine whether the vector v = -(7/3)i - (4/3)j - k is orthogonal to the line passing through the points P(-2,2,7) and Q(1/2,-1/2,9/2).
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Your solution:
PQ = 2.5i - 2.5j - 2.5k
v dot PQ = -5.8333 + 3.3333 + 2.5 = 0
Yes these are orthogonal
@& Good.*@
confidence rating #$&*:
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Given Solution:
@& I've inserted some notes. You're on the right track with everything. You'll understand my notes, but just in case, be sure to ask questions if you don't.*@