#$&*
course mth 277
10/5/201112:36 am
Your access code is 42454. Be sure you provide the correct code when submitting work. An incorrect code throws a monkeywrench into the system.
I had no clue how to solve the derivative in the last problem, I will probably ask about it in class." "Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises.
query_10_2
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Question: Find both F' and F'' for F(t) = (4sin^2 t)i + (9cos^2 t)j + tk
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Your solution:
F’ = (8cos t)i + (-18sin t) + k
@& The derivative of sin^2(t) is 2 sin(t) cos(t), not cos(t).
You need to apply the chain rule to calculate this derivative. You'll need the product rule to get the second derivatives.
*@
F’’ = (-8sin t)I + (-18cos t) + 0
confidence rating #$&*:232; 2
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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:
Velocity = (-sin t)i + j + (4cos t)k
Acceleration = (-cos t)i + 0 + (-4sin t)k
Speed = sqrt((-sin t)^2 + 1^2 + (4cos t)^2) = sqrt(1 + 4t)
@& sin^2(t) + 4 cos^2(t) = (sin^2(t) + cos^2(t)) + 3 cos^2(t) = 1 + 3 cos^2(t).
It's not equal to 4 t.*@
Speed = sqrt(1 + 4(pi/2)) = sqrt(2pi + 1) =roughly 2.699
Direction = ((-sin t)i + j + (4cos t)k)/2.699
@& The direction would be given by an angle relative to the positive x axis.*@
@& Alternatively the direction could be given by a unit vector.*@
confidence rating #$&*:232; 2
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Question: Find Int( dt) (Where Int( f(t) dt) is the integral of f with respect to t)
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Your solution:
(sin t)i + (cos t)j + t^2k
(-cos t)i + (sin t)j + (t^3/3)k
confidence rating #$&*:232; 3
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@& You should include the integration constant, but otherwise OK.*@
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Question: Find Integral((e^t)* dt)
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Your solution:
I think this is a definite integral (or at least intended to be) but I have no idea how to read it. I only see one value. As such, I will solve for the indefinite integral.
Ti + (4t^2)j + (sin t)k
(t^2/2)i + ((4t^3)/3)j + (-cos t)k
@& (e^t)* = <(e^t)*t,(e^t)*4t^2,(e^t)*sin t>
Each component is then integrated separately.
You didn't multiply the vector by e^t. You did correctly integrate .*@
confidence rating #$&*:232; 2
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Question: Find the velocity and position vectors given the acceleration vector `A(t) = 4(t^2)i - 2 sqrt(t) j + 5(e^3t)k, initial position R(0) = 2i + j -3k and initial velocity v(0) = 4i + j + 2k.
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Your solution:
F’’ = acceleration vector
F’ = velocity vector
F = position vector
4(t^2)i - 2sqrt(t)j + 5(e^3t)k
velocity vector = ((4t^3)/3)i + (4(t^(3/2))/3)j + ((5 e^3)/3)k
confidence rating #$&*:232; 2
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Given Solution:
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@& You also need an integration constant with each component.
((4t^3)/3 + c1)i + (4(t^(3/2))/3 + c2)j + ((5 e^3)/3 + c3)k,
or
((4t^3)/3)i + (4(t^(3/2))/3)j + ((5 e^3)/3)k + c1 i + c2 j + c3 k, would do it.*@
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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:
I don’t know how to take the derivative of this vector. Without that, I can’t find F’’. I understand that after finding F’’, I would see if there is a scalar value to pull out, that makes the vectors the same (IE: do they share a unit vector?).
@& Your strategy is good.*@
@& You need to review differentiation and integration.
For example the derivative of e^(k t) = k e^(k t).*@
confidence rating #$&*:232; 1
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#*&!
@& You need to brush up on integration and differentiation.
You appear to understand the vector operations OK, but you also have to do the calculus.*@
Your access code is 42454. Be sure you provide the correct code when submitting work. An incorrect code throws a monkeywrench into the system.