#$&*
course Mth 277
10/03/2011 @ 11:18 a.m.
If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
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Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises.
query_10_2
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Question: Find both F' and F'' for F(t) = (4sin^2 t)i + (9cos^2 t)j + tk
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Your solution: F’(t) = [8sin(t)cos(t)]i - [18cos(t)sin(t)]j + k, F’’(t) = [8sin^2(t) - 8cos^2(t)]i + [18cos^2(t) + 18sin^2(t)]j
confidence rating #$&*:y
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Given Solution:
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Question: Given the position vector of a particle R(t) = (cos t)i + tj + (4 sin t)k, find the particle's velocity and acceleration vectors and then find the speed and direction of the particle at t = pi/2.
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Your solution: R’(t) = -sin(t)I + j + (4cos(t))k which is the velocity function and R’’(t) = -cos(t)i - (4sin(t))k. It is 5.57m from start at this time, with a velocity of -1unit/sec and an acceleration of -4unit/sec^2.
confidence rating #$&*:erately
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Given Solution:
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Question: Find Int( dt) (Where Int( f(t) dt) is the integral of f with respect to t)
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Your solution: I don’t quite follow your notation, but I’m going to assume you just want us to integrate the three functions within the brackets. Int.sin(t) = -cos(t) + c. Int.cos(t) = sin(t) + c. Int.t^2 = (t^3)/3 + c.
confidence rating #$&*:I understood the notation, very.
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Given Solution:
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@& So int( dt ) = <-cos t + c1, sin t + c2, t^3 / 3 + c3 >
Note that the different integrals willl have different constants.*@
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Question: Find Integral((e^t)* dt)
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Your solution: I don’t think you’ll mind if I leave out the steps and all of the u and v values for these integrations. I used integration by parts to solve all of these. Int.e^t * t = te^t + C. int.e^t * 4t^2 = 4t^2e^t - 8te^t +8e^t + C. The next one we had to integrate by parts twice: int.e^t * sin(t) = ½(e^tsin(t) - e^tcos(t)) + C.
confidence rating #$&*:y on the first two and just ok on the third one.
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Given Solution:
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@& No problem.*@
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Question: Find the velocity and position vectors given the acceleration vector `A(t) = 4(t^2)i - 2 sqrt(t) j + 5(e^3t)k, initial position R(0) = 2i + j -3k and initial velocity v(0) = 4i + j + 2k.
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Your solution: The first integral of the acceleration is the velocity, which I found to be 4/3*t^3 - 4/3*t^(3/2) + 5/2*e^3*t^2 + C. The second integral is the position, which is 1/3*t^4 - 8/15*t^(5/2) + (5/6)*e^3*t^3 + C.
confidence rating #$&*: too confident, the 2nd half of your question giving a position and velocity vector threw me. I didn’t use this in my answer at all and that doesn’t seem right.
@& You left off `i, `j and `k.
v(t) = 4/3 t^3 `i - 2/3 t^(3/2) `j + 5/3 e^(3 t) `k + `c1, where `c1 is an arbitrary vector.
Since v(0) = 4 `i + `j + 2 `k, we have
`v(0) = 0 `i - 0 `j + 5/3 `k + `c1 = 4 `i + `j + 2 `k.
Thus `c1 = 4 `i + `j + 1/3 `k, and
v(t) = (4/3 t^3 + 4) `i + (- 2/3 t^(3/2) + 1) `j + (5/3 e^(3 t) + 1/3) `k .
When you integrate v(t) you again get a constant of integration, which is a vector we could denote by `c2. The initial condition on `R(0) is used to evaluate `c2.
*@
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Given Solution:
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Question: F(t) = e^(-kt)i + e^(kt)k. Show that F and F'' are parallel.
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Your solution: F’(t) = -ke^(-kt) + ke^(kt) F’’(t) = k^2*e^(-kt) + k^2*e^(kt). They are parallel because they are multiples of each other by a factor of k^2.
confidence rating #$&*:y
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Given Solution:
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Question:
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#*&!
&#Good work. See my notes and let me know if you have questions. &#