query_101

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course MTH 277

7/13/2012 345AM

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_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:

Find the cross product of F(t) and G(t)

F(t) X G(t) = X <1 / (t+2), t-5, sqrt(t)>

= (-t^2 -t^3/2 - 6t - 2sqrt(t) + 5)`i - [t^5/2 - (t-1)/(t+2)]`j + [t^3 - 5t^2 + (t+2) / (t+2)]`k

therefore, the domain is [0,infinity)

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Given Solution:

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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:

0 _< t _< 2pi

x= sin(t)

y = cos(t)

z = 4/3

sin^2(t) + cos^2(t) = (4/3^)2

x^2 + y^2 = 16/9

r = 4/3

Cylinder centered around the z-axis with a radius of 4/3 units

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Given Solution:

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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:

<(t*sin t) + (e^-t)> dot [ X <(t) - 5(e^t) +(t^3)>]

<(t*sin t) + (e^-t)> dot <-5e^t/t, (-t^4 -1), -5te^t>, where t is not equal to zero.

=-5e^tsin(t) + e^-t(-t^4 -1), where t is not equal to zero

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Given Solution:

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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:

x = 0 + 5t

y = 3 + 6t

z = -2 + 4t

F(t) = (5t)`i + (3 + 6t)`j + (-2 + 4t)`k

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Given Solution:

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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:

limit as t -> 2 of r(t) = ((t^4-2)/(t-2))i + ((t^2-4)/(t^2-2t))j + ((t^2 + 3)e^(t-2))k

limit as t -> 2 of r(t) = limit as t->2 of ((t^4-2)/(t-2))i + limit as t->2 of ((t^2-4)/(t^2-2t))j + limit as t->2 of ((t^2 + 3)e^(t-2))k

= 14/0`i + 0/0`j + 7`k

In order to find the correct i

limit as t->2 of ((t^4-2)/(t-2))i

= limit as t -> 2 of (3t^4 - 8t^3 + 2) / (t^2 -4t +4) `i

= -14/0`i

use L'hopitals rule again

limit as t->2 of (12t^3 - 24t^2) / (2t - 4) `i

= 0/0 `i

use L'Hopitals one last time

limit as t -> 2 of (36t^2 - 48t) / 2 = 24`i

In order to find the correct `j component use L'Hopitals rule

limit as t->2 of ((t^2-4)/(t^2-2t))j

= limit as t-> 2 of 2t / (2t - 2)

= 2`j

Therefore the final answer for the limits as t -> 2 of ((t^4-2)/(t-2))i + ((t^2-4)/(t^2-2t))j + ((t^2 + 3)e^(t-2))k

= 24`i + 2`j + 7`k

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Given Solution:

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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:

vertical distance is measured by the z-axis, so if z = 12 =3/4t, then t = 16. 16/2pi = 2.55. So roughly 2.5 revolutions.

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query_101

&#Your work looks very good. Let me know if you have any questions. &#