Query Assignment 102

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course Mth 277

10/7 12

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) = (8sin t cos t) i + (-18sin t cos t )j + k + C

F''(t) = ( -8sin^2 t - 8sin t cos t)i + (18sin^2 t - 18cos^2 t)j + 0k + C

@& Good instinct. There are things you can do with that expression.

However none of them really simplifies it, so the form in which you gave it is good.

The `j component could be written 18 ( 1 - 2 cos^2(t)), which would be a little easier if you needed to take another derivative. However the form in which you wrote it is OK.*@

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Given Solution:

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Self-critique (if necessary):

???????????????????????

I cant help but think that the i component in the F '' (t) can be reduced

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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:

to find velocity, take first derivitave.

R'(t) = -sin(pi/2) i + j + (4cos(pi/2) ) = -1 i + 1 j + 0

to get acceleration, take second derivative

R''(t) = -cos(pi/2)i + 0j - 4sin(pi/2)k = 0i + 0j - 4k

which implies it is decelerating in the negative k direction of 4 units / s^2

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Given Solution:

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Self-critique (if necessary):

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@& -1 `i + 1 `j is the velocity; it has magnitude sqrt(2), which is the speed at this instant. *@

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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:

take deriv; ln t(cos t, -sin t, t^3/3) dt

now do ln rule. u'/u

(-sin t/cos t , cos t/sin t , 3t^2/t^3/3 ) dt

reduce; (-tan t, cot t, 9t^2/t^3) dt

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Given Solution:

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Self-critique (if necessary):

??????????? are my order of opperations correct finding ln t and taking derivative? or do I even take the derivative in the first place? if not, I can easly do this

@& 'Int' is the integral.

The integral would be

<-cos(t) + c1, sin(t) + c2, t^3 / 3 + c3> =

<-cos(t), sin(t), t^3 / 3> + *@

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

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Your solution:

take integral of all components and get e^t(t^2/2, 4t^3/3, cos t)

now distribute; (e^t*t^2) / 2 , (e^t * 4t^3) / 3 , e^t*cos t

@& If you take the derivative of your result you don't get the original expression.

You need to multiply through by e^t before integrating, then you need to use appropriate integration techniques.*@

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Given Solution:

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Self-critique (if necessary):

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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:

Long problem, not going to write out all algebra because it took whole paper.

to find vel vector, take the first integral of the A(t) vector, to get ( (4t^3)/3 +c'1)i - ( (4t^3/2)/3 +c'2)j + ( 5e^3(t) / 3 + c'3)k. set the v(0) vector = to this vector to solve for the c1, c2, and c3

solve for constants and find that the v vector is (4i + j + 11/3k)

now to get the position vector, take the 2nd integral of the A(t) vector and set that = to the R(0) vector and solve for those constants, now we have a lot of constants

2nd integral for A(t) vector is; (1/3 t^4 + c'1t)i + (8/15t^(5/2) +c'2t)j + (5e^(3t)/9 + c'3t)k + c'1i + c'2j + c'3k

set R(0) = d^2/dt A(t) and seperate i j and k components and set the t = 0 to solve for the new constants, these constants are the position vector,

once I solved for the constants in the new position vector, i now have R(t) = (2i + j - 27/5k)

@& v(t) = (4t^3)/3 +c'1)i - ( (4t^3/2)/3 +c'2)j + ( 5e^3(t) / 3 + c'3)k, with c1 = 4, c2 = 1 and c3 = 1/3

so

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

When you integrate you get brand new constants. We've evaluated c1, c2 and c3 for the preceding, so we can start over with brand new c1, c2, c3 and evaluate them.

An antiderivative of the `k component is 5 e^(3 t) / 9 + 1/3 t + c3.

When t = 0 this component has value 5/9 + c3 = 3, so that c3 = -32/9, assuming my arithmetic is correct.*@

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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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:

if they are parallel, they will be multiples of eachother.

when you integrate e^(-kt) with respect to t, you get the same thing, so yes, they are multiples of eachother, and they are parallel

@& You wouldn't integrate, you would differentiate, and you would do so twice.

You will find that F '' (t) = k^2 F(t), so the two vectors are parallel.

An antiderivative of F(t), incidentally, would give you a -1/k multiply of the `i component and a +1/k multiple of the `j component. This would make this particular antiderivative perpendicular to F(t).*@

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Question: "

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Question: "

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@& Pretty good, but there are some errors so check my notes. Let me know if you have questions.*@