#$&*
course Mth 277
745pm 11/8/11
If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
.
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises.
query_10_2
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Question: Find both F' and F'' for F(t) = (4sin^2 t)i + (9cos^2 t)j + tk
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Your solution:
F(t) = (4sin^2 t)i + (9cos^2 t)j + tk
F'(t) = [4 * (1-cos2t)/2] dt `i + [9 * (1+cos2t)/2] dt `j + 1`k
F'(t) = (2 - 2cos2t) dt `i + [9/2 + (9cos2t)/2] dt `j + 1`k
@& The above two expressions are for F(t), not F ' (t).
The expression in the next line is then correct for F ' (t).*@
F'(t) = 4sin2t `i - 9sin2t `j + `k
F""(t) = 8cos2t `i - 18cos2t `j
@& Be sure you also aware that the derivative of sin^2(t) can be calculated from the 'chain' consisting of the squaring function applied to the sine function, giving you the result (cos t) * 2 sin(t) = 2 sin(t) cos(t).*@
confidence rating #$&*:
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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
@& At t = pi/2 we have sin(t) = 1 and cos(t) = 0, so speed = sqrt( 1^2 + 1 + 16 * 0) = sqrt(2).
The double- and half-angle formulas can be handy, but probably complicate things on a problem like this.*@
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°
@& This would give you the angle between the velocity and acceleration vectors.
The direction of motion, however, would be specified by the unit vector in the direction of v(t). That would be just v(t) / || v(t) ||, easily enough calculated.
At t = pi/2 you have v(t) = - `i + `j, and the unit vector in the direction of motion would be (-`i + `j) / sqrt(2) = etc..
When t = pi / 2 the v and a vectors would indeed be perpendicular.*@
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(t) and F""(t) are scalars. The only difference is F(t) is multiplied by (1/k^2), therefore the two are parrallel
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
@& F(t) and F '' (t) are both vectors, not scalars.
However F '' ( t) is a scalar multiple of F(t), by factor 1 / k^2, which shows that the vectors are parallel.
Right thinking but you didn't quite get the words down correctly.*@
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Question: F(t) = e^(-kt)i + e^(kt)k. Show that F and F'' are parallel.
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Your solution:
F(t) = e^(-kt)i + e^(kt)k
F'(t) = -[e^(-kt) / k]i + [e^(kt) / k] 'k
for clerification, k is constant and 'k is the z component
F""(t) = [e^(-kt)]/(k^2)i + e^(kt)/(k^2)k
F(t) and F""(t) are scalars. The only difference is F(t) is multiplied by (1/k^2), therefore the two are parrallel
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
@& F(t) and F '' (t) are both vectors, not scalars.
However F '' ( t) is a scalar multiple of F(t), by factor 1 / k^2, which shows that the vectors are parallel.
Right thinking but you didn't quite get the words down correctly.*@
#*&!
@& Good work overall. Check my notes.*@