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course Mth 277
If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://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_10_1
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Question: Find the domain of F(t) X G(t) when F(t) = t^2 i - (t+2)j + (t-1)k and G(t) = (1/(t+2))i + (t-5)j + sqrt(t) k.
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Your solution:
F(t) cross g(t)= <(-t^(3/2)-2sqt(t) -t^2 +6t -5), ((t-1)/(t+2) -t^(5/2)), (t^3-5t^2 +(t+2)/(t+2))>= <-t^3/2-t^2+6t-2sqrt(t)-5, (t-1)/(t+2) -t^(5/2), t^3-5t^2+1.
x=-t^(3/2)-t^2+6t-2sqrt(t)-5
y=(t-1)/(t+2) -t^(5/2)
z= t^3-5t^2+1
the domain of x is [0, infinity)
the domain of y is (-infinity, -2) U (-2, infinity)
the domain of z is (-infinity, infinity)
the domain of F X G= [0, infinity).
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):OK
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Self-critique rating:OK
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Question: Describe the graph of G(t) = (sin t)i + (cos t)j + (4/3)k
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Your solution:
This is a circular cylinder since x^2+y^2=1 (sin^2+cos^2=1). t revolves around -1 and 1 in the y direction and 0 and 1 in the x direction but stays fixed at 4/3 on the z axis.
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Good, but a circular cylinder would be
sin(t) `i + cos(t) `j.
With the 4/3 `k, you're confined to the plane z = 4/3, so the graph is just a unit circle in that plane.
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confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):OK
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Self-critique rating:OK
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Question: Given F(t)= (t)i - 5(e^t)j +(t^3)k, G(t) = ti - (1/t)k and H(t) = (t*sin t)i + (e^-t)j, find H(t) dot [G(t) X F(t)]
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Your solution:
G X F= <0-(5e^t)/t, -1-t^4, -5te^t-0>= <-(5e^t)/t, -1-t^4, -5te^t>
(G X F) dot H= <-(5e^t)/t, -1-t^4, -5te^t> dot = ((-5te^tsint)/t)+ (-e^-t -e^-t(t^4))+0= -5e^tsint- e^-t -e^-t (t^4)
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):OK
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Self-critique rating:OK
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Question: Find a vector function F whose graph is the curve given by the equation x/5 = (y-3)/6 = (z+2)/4.
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Your solution:
R(t)= 5x+6(y-3)+4(z+2)=0
X=5t
Y=3t
X=-2t
R(t)=5ti+3tj-2tk
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The point (5, 3, -2) would lie on the graph of your function, but would not satisfy the given equation.
The equation is that of a straight line through (0, 3, -2), parallel to 5 `i + 6 `j + 4 `k.
A point (x, y, z) is on that line if
(x, y, z) - (0, 3, -2) is a multiple of that vector, i.e., if (x, y, z) = (0, 3, -2) + (5 t, 6 t , 4 t), so that
x = 5 t, y = 3 + 6 t and z = -2 + 4 t.
A vector function is therefore
R(t) = 5 t `i + (3 + 6 t) `j + (-2 + 4 t) `k.
For example the point (5, 9, 2) would lie on the graph of R(t), and the coordinates of this point satisfy the given equation.
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confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):OK
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Self-critique rating:OK
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Question: Find the limit as t -> 2 of ((t^4-2)/(t-2))i + ((t^2-4)/(t^2-2t))j + ((t^2 + 3)e^(t-2))k.
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Your solution:
Lim as t approaches 2 of (t^4 -2)/(t-2)i= 14i
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As t -> 2, t - 2 -> 0 while t^4 - 2 -> 14. So this limit would be undefined.
I believe the intent of the problem was that this expression be (t^4 - 16) / (t - 2), in which case the numerator simplifies to (t^2 + 4) ( t + 2) ( t - 2) so that the denominator divides out and the limit is 32.
Alternatively for the intended expression l'Hopital's rule yields 4 t^3 / 1 which approaches 32.
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Lim as t approaches 2 of (t^2 -4)/(t^2 -2t) j= 0
Lim as t approaches 2 of (t^2+3)e^(t-2)k= 7k
So, the limit as t approaches 2 of ((t^4-2)/(t-2))i + ((t^2-4)/(t^2-2t))j + ((t^2 + 3)e^(t-2))k is 14i+7k
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):OK
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Self-critique rating:OK
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Question: How many revolutions are made by the circular helix R(t) = (sin t)i + (cos t)j + (3/4)tk in a vertical distance of 12 units.
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Your solution:
Since x= sin(t) alternates from zero to 1 or -1 three times in 12 units, I assume that the helix makes three revolutions in that time. I’m not really sure.
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Vertical position is 3/4 t so to rise 12 units t would have to change by 4/3 * 12 = 16.
sin(t) `i + cos(t `j) describes a circle, which is completed every time t changes by 2 pi.
So as t changes by 16, it changes by 2 pi 16 / (2 pi) = 8 / pi = 2.6 times, approximately.
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confidence rating #$&*:
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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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Self-critique (if necessary):
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Self-critique rating:
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Not bad but be sure to check my notes. I'll be glad to answer questions if you have them.
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