Query101

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course Mth 277

215pm 11/7/2011

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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Self-critique (if necessary):

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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

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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@& The circle is described by the `i and `j components, which ensure that x^2 + y^2 = 1.

So the circle is in the x-y plane.

4/3 `k means that the z component of the vector `G(t) is always 4/3.

So the points described by `G(t) are the points lying 4/3 of a unit above the circle x^2 + y^2 = 1 in the xy plane.

Thus `G(t) defines a circle of radius 1 in the plane z = 4/3.

Note that there is nothing in the given function that implies x^2 + y^2 = z^2.*@

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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

@& G X H is t e^t `k - sin(t) `j + e^(-t) / t `i, or

This can be calculated as

det [ `i, `j, `k; t, 0, -1/t; t sin(t), e^-t, 0 ]

i.e., the determinant of the matrix with the three rows as listed.*@

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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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

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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@& Good.*@

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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

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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@& The derivatives of the `i component are 4 t^3 and 1. Neither is zero when t = 2 so the limit is 4 * 2^3 = 32.

For the `j component the derivative of t^2 - 4 is 2 t, while the derivaive of t^2 - 2 t is 2 t - 2. Evaluated at t = 2 you get 4 and 2, so the limit of the `j component is 4 / 2 = 2.

The last term is as you say.*@

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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.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

????????????????/The QA10.1 on the MTH277 homepage gave me a link to QA9.7 So this assignment was very time consuming, but I feel confident in most of my answers.????????????????

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&#Good work. See my notes and let me know if you have questions. &#