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course mth 277
12/8 4
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_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)= 8cos(t)i-18sin(t)j+k
F”(t)= -8sin(t)i-18cos(t)j
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):okay
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Self-critique rating:3
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Question: Given the position vector of a particle R(t) = (cos t)i + tj + (4 sin t)k, find the particle's velocity and acceleration vectors and then find the speed and direction of the particle at t = pi/2.
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Your solution:
The velocity equals the derivative of the position vector which is
V(t)= sin(t)i+j-4cos(t)k
The acceleration vector is the derivative of the velocity vector which is
a(t)= -cos(t)i-4sin(t)k
the speed at t=pi/2 is v(pi/2)= sin(t)i+j-4cos(t)k= 2 rad/s im not sure about the units.
And the direction is the unit vector of the velocity which equals (1i+1j+0k)/sqrt(1^2+1^2)= .707i+.707j
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The `i component would be -sin(t).
The speed is || V(t) ||, which is not 2 rad / sec.
At t = pi / 2 we have sin t = 1 and cos t = 0, so V(pi/2) = - `i + `j, the magnitude of which is sqrt(2).
The direction is that of the vector -`i + `j. The unit vector in this direction is sqrt(2) / 2 ( -`i + `j), which except for the sign of the `i component agrees with your result.
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confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):I know I did the derivatives right but the direction im unsure.
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Self-critique rating:3
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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:
Int(df/dt)=Int(sint+cost+t^2)dt= -cost+sint+((t^3)/3)+c
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
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Self-critique (if necessary):okay
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Self-critique rating:3
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Question: Find Integral((e^t)* dt)
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Your solution:
Int(e^t)+Int(ti+4t^2j+sin tk dt)= e^t+ ((t^2)/2)i+((4t^3)/3)j-cos(t)k +c
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
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Self-critique (if necessary):okay
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Self-critique rating:3
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The derivative of your result is not equal to the original integrand. You have not integrated correctly.
You need to integrate
t e^t
4 t^2 e^t and
sin(t) e^t.
Each integral requires the use of integration by parts.
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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:
I took the integral of the acceleration vector to get the velocity vector which is v(t)= ((4t^3)/3)i-((4t^(3/2))/3)j+(5e^3 ((t^2)/2))k and then the position vector is the integral of the velocity vector R(t)= ((4t^4)/12)i-((8t^(5/2))/15)j+(5e^3((t^3)/6))k
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):I am not sure if the integrals are right.
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Self-critique rating:3
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You need integration constants in order to match your results to the initial conditions on `R(0) and `v(0).
For your results R(0) would be 0, not 2 `i + `j - 3 `k.
You have an integration constant with each antiderivative; different antiderivatives have independent integration constants.
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Question: F(t) = e^(-kt)i + e^(kt)k. Show that F and F'' are parallel.
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Your solution:
The F and F” are both equal to e^(-kt)i + e^(kt)k so they are parallel because they lie on top of each other.
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
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Self-critique (if necessary):okay
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Self-critique rating:3
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F '' = e^(-kt)i + e^(kt)k and this is parallel to F for the reason you give.
F ' = -e^(-kt)i + e^(kt)k
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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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Check my notes and let me know if you have questions or revisions.
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