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
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).
Your solution:
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Given Solution:
We could solve the first equation for t, taking natural log of both sides to get -t = ln(x). Substituting this into the second equation we would get y = e^t = e^(- ln(x) ) = 1 / e^(ln(x) ) = 1 / x.
Or we could obseved that since e^-t is 1/ e^t, we have x = 1 / e^t and y = e^t, implying that x = 1 / y, which is equivalent to y = 1 / x.
You should be able to easily sketch this curve. If necessary substitute +- 1/2, +- 1 and +- 2 for x and think about where the horizontal and vertical asymptotes should be (think also about where the function is undefined).
Self-critique (if necessary):
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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
The given equations describe a line through (3, 1, -3) parallel to the vector 4 `i + 3 `j + 2 `k.
Our line will not be through (3, 1, -3), but will be parallel to the same vector 4 `i + 3 `j + 2 `k.
Your solution:
Confidence rating:
Given Solution:
Our line is through (-1, -1, 0) so its symmetric equations are
(x + 1) / 4 = (y + 1) / 3 = z / 2 .
Let t be the parameter, and set t equal to each of these expressions, so that
(x + 1) / 4 = (y + 1) / 3 = z / 2 = t.
The parametric equations are thus
x = 4 t - 1
y = 3 t - 1
z = 2 t.
Self-critique (if necessary):
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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).
Your solution:
Confidence rating:
Given Solution:
The xy plane is the z = 0 plane, so our parametric equation for z yields
2 t - 7 = 0, with solution t = 7/2.
For this value of t we get
x = 3 * 7/2 + 4 = 29/2 = 14.5
y = 1 - 3 * 7/2 = -19/2 = -19.5.
So the intersection with the xy plane is (14.5, -19.5, 0).
The xz plane is the y = 0 plane, giving use 1 - 3 t = 0 so that t = 1/3. Our resulting point is (5, 0, -19/3), approximately (5, 0, -6.33).
The intersection with the y z plane is found similarly.
Self-critique (if necessary):
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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.
Your solution:
Confidence rating:
Given Solution:
We don't want to use the same parameter for both lines, so let's express the second line as x = 5 - s, y = -1 - 3 s, z = -3 + 4 s.
The lines then intersect provided there are values of s and t such that all three coordinates are the same for both lines. That condition is
2 - t = 5 - s
3 t = -1 - 3 s
3 - 2 t = -3 + 4 t
Eliminating t between the first two equations we get
6 = 14 - 6 s
so that s = 4/3 and t = - 5 / 3.
So if there is an intersection, it must be for these values of s and t.
Plugging these values into the third equation does not lead to an identity, so no simultaneous solution for s and t exists and the lines do not intersect.
Self-critique (if necessary):
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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).
Your solution:
Confidence rating:
Given Solution:
The vector PQ is 5/2 i -5/2 j - 5/2 k.
The two vectors are orthogonal if and only if their dot product is zero.
(-(7/3)i - (4/3)j - k ) dot (5/2 i -5/2 j - 5/2 k ) = -35/6 + 20/6 + 5/2 = -15/6 + 5/2 = -5/2 + 5/2 = 0
so the vectors are orthogonal.
Self-critique (if necessary):
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