query 102

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course Mth 277

2-27

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) + 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):

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Self-critique rating:

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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:

V= dR/dt = -sin(t)i + j + 4cos(t)k

Accel = dV/dt = -cos(t)i - 4sin(t)k

Speed = ||V|| = `sqrt(-sin(t)^2+ 1^2 + 4cos(t)^2)

= `sqrt(-sin(t)^2+ 1+ 16cos(t)^2) = sqrt(1 - sin(t)^2+ 16cos(t)^2)

= `sqrt(cos(t)^2+ 16cos(t)^2)

= `sqrt(17cos(t)^2)

=`sqrt(17)*|cos(t)|

Speed at t = `pi/2

||V(`pi/2)|| = `sqrt(-sin(`pi/2)^2+ 1^2 + 4cos(`pi/2)^2)

= `sqrt(1+ 1+ 0)

@& This isn't consistent with your preceding result || v || = sqrt(17) cos(t), which would give you || v || = 0.

Took me a minute to see it, but your second step here would be

`sqrt(-1+ 1+ 0) ,

which would be 0.*@

= `sqrt(2)

Direction at t = `pi/2

V(`pi/2)/||V(`pi/2)||

=1/`sqrt(2)[ -sin(`pi/2)i + j + 4cos(`pi/2)k]

= -1.414i + .707j

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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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([sin ti + cos tj + t^2k] dt)

Int(sin t dt)i + Int(cos t dt)j + Int(t^2 dt)k

= -cos(t)i +sin(t)j + (t^3/3)k + C

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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Question: Find Integral((e^t)* dt)

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Your solution:

I ntegral((e^t)* dt)

= Int((e^t)*[ ti +4t^2j + sin(t)k] dt)

= Int(e^t*t dt)i + Int(e^t*4t^2 dt)j + Int(e^t* sin(t) dt)k

Int(e^t*t dt)i = [e^t(t - 1)]i

Int(e^t*4t^2 dt)j =[8e^t((1/2)t^2 - t - 1)]j

Int(e^t* sin(t) dt)k = [(1/2)e^t(sin(t) - cos(t))]k

Integral((e^t)* dt) = [e^t(t - 1)]i + [8e^t((1/2)t^2 - t - 1)]j + [(1/2)e^t(sin(t) - cos(t))]k +C

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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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:

1st we find anti derivative of accel. fn. `A(t) = 4(t^2)i - 2 sqrt(t) j + 5(e^3t)k

Int([4(t^2)i - 2 sqrt(t) j + 5(e^3t)k] dt)

= [(4t^3)/3 + c_1]i - [(4/3)t^(3/2) + c_2]j + [(5/3)e^3t + c_3]k

= [(4t^3)/3]i - [(4/3)t^(3/2)]j + [(5/3)e^3t]k + c_1i + c_2j + c_3k

v(t) = [(4t^3)/3]i - [(4/3)t^(3/2)]j + [(5/3)e^3t]k + [c_1i + c_2j + c_3k]

Where [c_1i + c_2j + c_3k] is our C term, plugging in our init. condition v(0) = 4i + j + 2k.

v(t) = [(4t^3)/3]i - [(4/3)t^(3/2)]j + [(5/3)e^3t]k + C

v(0) = [(4*0^3)/3]i - [(4/3)*0^(3/2) + 1]j + [(5/3)e^3*0]k + C = k + C

Now using init cond

4i + j + 2k = k+C

4i = c_1i

c_1 = 4

j = c_2j

c_2 = 1

2k = k + c_3k

c_3 = 1

So we have solved for init cond and…

v(t) = [(4t^3)/3 + 4]i - [(4/3)t^(3/2) + 1]j + [(5/3)e^3t + 1]k

Now we find position

v(t) = dR/dt = [(4t^3)/3 + 4]i - [(4/3)t^(3/2) + 1]j + [(5/3)e^3t + 1]k, solving differential equ..

dR/dt = [(4t^3)/3 + 4]i - [(4/3)t^(3/2) + 1]j + [(5/3)e^3t + 1]k

Int( dR) = Int([(4t^3)/3 + 4]i - [(4/3)t^(3/2) + 1]j + [(5/3)e^3t + 1]k] dt)

R(t) = Int([(4t^3)/3 + 4]i - [(4/3)t^(3/2) + 1]j + [(5/3)e^3t + 1]k dt)

Int((4/3)t^3 + 4 dt)i = [(1/3)t^4 + 4t)]i

Int((4/3)t^(3/2) + 1 dt)j =[(8/15)t^(5/2) + t]j

Int([(5/3)e^3t + 2 dt)k = [(5/9)e^3t + t]k

So

R(t) = [(1/3)t^4 + 4t + c_1)]i - [(8/15)t^(5/2) + t + c+2]j + [(5/9)e^3t + t + c_3]k

= [(1/3)t^4 + 4t)]i - [(8/15)t^(5/2) + t]j + [(5/9)e^3t + t]k + c_1i + c_2j + c_3k

= [(1/3)t^4 + 4t)]i - [(8/15)t^(5/2) + t]j + [(5/9)e^3t + t]k + [c_1i + c_2j + c_3k]

Where [c_1i + c_2j + c_3k] is our C term, using our init. condition R(0) = 2i + j -3k

R(t) = [(1/3)t^4 + 4t)]i - [(8/15)t^(5/2) + t]j + [(5/9)e^3t + t]k + C

R(0) = [(1/3)0^4 + 4*0)]i - [(8/15)0^(5/2) + 0]j + [(5/9)e^3*0 + 2*0]k + C = k +C

2i + 1j - 3k = k + C

We solve separate equations

2i = C_1i,

c_1 = 2

j = c_2j,

c_2 = 1

-3k = k + c_3k,

c_3 = -4

2i + j - 4k = C

So position at any t is

R(t) = ((1/3)t^4 + 4t)i - ((8/15)t^(5/2) + t)j + ((5/9)e^3t +t)k + [2i + j -4k]

=((1/3)t^4 + 4t + 2) i - ((8/15)t^(5/2) + t + 1)j + ((5/9)e^3t +t - 4)k

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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Question: F(t) = e^(-kt)i + e^(kt)k. Show that F and F'' are parallel.

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Your solution:

For F(t) = e^(-kt)i + e^(kt)k

 F’ = -ke^(-kt)i + ke^(kt)k

 F’’ = -k^2e^(-kt) + k^2e^(kt) = -k^2e^(-kt)i + k^2e^(kt)k = k^2(e^(-kt)i + e^(kt)k)

Components are the same except for F’’(t) has scalar mult. constant of k^2 for both terms

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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Question:

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Self-critique (if necessary):

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Self-critique rating:

&#This looks good. See my notes. Let me know if you have any questions. &#