Query102

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course Mth 277

745pm 9/12/2012

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) = (4sin^2 t)i + (9cos^2 t)j + tk

F'(t) = [4 * (1-cos2t)/2] dt `i + [9 * (1+cos2t)/2] dt `j + 1`k

F'(t) = (2 - 2cos2t) dt `i + [9/2 + (9cos2t)/2] dt `j + 1`k

F'(t) = 4sin2t `i - 9sin2t `j + `k

F""(t) = 8cos2t `i - 18cos2t `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:

R(t) = (cos t)i + tj + (4 sin t)k

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

A(t) = -cos(t)i - 4sin(t)k

speed = sqrt(V(t)^2)

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

= sqrt{(1-cos2t)/2 + 1 + 16[(1+cos2t)/2]}

= sqrt[1/2 - cos2(pi/2) + 9 + 8cos2(pi/2)]

= sqrt[9.5 +7cos(pi)]

= sqrt(9.5 - 7)

=sqrt(2.5) appr= 1.58 units

direction = A dot B = ||A|| * ||B|| cos(theta)

cos^-1 = theta = [V(t) dot A(t)] / {sqrt[V(t)^2] * sqrt[A(t)^2]}

= [(-sinti + 1j + 4costk) * (-costi - 4sintk)] / [sqrt(2.5) * sqrt(cos^2t + 16sin^2t)]

= [sin(t)cos(t) - 16sin(t)cos(t)] / (sqrt(2.5) * {sqrt[(1+cos2t)/2 + 16*(1-cos2t)/2]})

= -15sin(t)cos(t) / {sqrt2.5 * sqrt[.5 + cos(pi)/2 + 8 - 8cos(pi)]}

since sin2u = 2sinucosu, then 7.5sin2u = 15sinucosu

therefore

= 7.5sin(pi) / sqrt(2.5) * sqrt(.5 -.5 + 8 + 8)

= 0/8.5

therefore cos(theta) = cos^-1 which means that theta equals 90°

The velocity is 1.58 units in the direction of 90 degrees.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I am fairly confident in my answer. idk.

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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:

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

Integral((e^t)* dt)

= Integral of t*e^t i + 4t^2*e^t j + e^t*sint k

Use integration by parts to find each component

intergal of u*dv = u*v - intergral of v*dv

u=t,dv= e^t

du=dt, v=e^t

Intergral t*e^t = t*e^t - integrale e^t*dt

= e^t(t-1)i + C

integral 4t^2 * e^t

u=4t^2, dv = e^t

du=8tdt, v = e^t

=4t^2*e^t - integral 8t*e^tdt

u=8t, dv=e^t

du=8dt, v=e^t

= 4t^2*e^t - 8t*e^t - 8e^t

= 4e^t *(t^2 - 2t -2)j + C

integral e^t*sin(t)

u=sin(t), dv=e^t

du=cos(t), v=e^t

integralu*dv = u*v - integral vdu

integral e^t*sint = sin(t)*e^t - integral e^t*cos(t)

u=sin(t), dv=e^t

du=cos(t), v=e^t

intergral e^tsin(t) = sin(t) - cos(t)*e^t - integral sin(t)*e^t

2integral e^t*sin(t) = sint*e^t - cos(t)*e^t

integral e^t*sin(t) = {e^t*[sin(t)^e^t - cos(t)*e^t]/2} + C

Therefore the final solution for the integral of ((e^t)* dt)

is

= e^t(t-1)i + 4e^t *(t^2 - 2t -2)j + {e^t*[sin(t)^e^t - cos(t)*e^t]/2}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:

A(t) = 4(t^2)i - 2 sqrt(t) j + 5(e^3t)k

V(t) = (4/3)t^3 i - (4/3)t^(3/2) j + (5/3)e^3t k + C_1

V(0) = 0 i - 0 j + 5/3 k + C_1 = 4i + j + 2k

C_1 = 4i + j + (1/3)k

The velocity vector is:

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

R(t) = [((t^4)/3 + 4t]i - [(8/15)t^(5/2) + t]j + [(5/9)e^3t + (t/3)] k + C_2

R(0) = 0i - 0j + 5/9k + C_2 = 2i + j -3k

C_2 = 2i + j - (32/9)k

The position vector is:

R(t) = [((t^4)/3 + 4t + 2]i - [(8/15)t^(5/2) + t +1]j + [(5/9)e^3t + (t/3) - (32/9)] 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:

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

F'(t) = -[e^(-kt) / k]i + [e^(kt) / k] 'k

for clerification, k is constant and 'k is the z component

F""(t) = [e^(-kt)]/(k^2)i + e^(kt)/(k^2)k

F and F"" are vectors. The F'' is a scalar multiple of F. The two are parrallel

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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&#Very good responses. Let me know if you have questions. &#