#$&*
course Mth 277
135pm 11/14/2011
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_5
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Question: Find the tangential and normal components of an object's acceleration which has the position vector R(t) = <3/5 cos t, 4/5(1+sin t), cos t>.
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Your solution:
R(t) = <3/5 cos t, 4/5(1+sin t), cos t>
V(t) = <-3/5sin(t), 45cos(t), -sin(t)>
a(t) = <-3/5cos(t), -4/5sin(t), - cos(t)>
||v(t)|| = sqrt(9/25sin^2(t) + 16/25cos^2(t) + sin^2(t)
= sqrt(16/25 + 18/25sin^2(t))
a_t = <-3/5sin(t), 45cos(t), -sin(t)> dot <-3/5cos(t), -4/5sin(t), - cos(t)>
@&
a_t = <-3/5sin(t), 45cos(t), -sin(t)> dot <-3/5cos(t), -4/5sin(t), - cos(t)> /sqrt(9/25sin^2(t) + 16/25cos^2(t) + sin^2(t).
You accounted for the denominator in subsequent steps.
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= [9/25sin(t)cos(t) - 16/25sin(t)cos(t) + sin(t)cos(t)] / sqrt(16 + 18sin^2(t)/25)
= [9/25sin(t)cos(t) - 16/25sin(t)cos(t) + sin(t)cos(t)]^2 / [(16 + 18sin^2(t))/25]
= 8100sin^2(t)cos^2(t)/(16 + 18sin^2(t)
a_n = sqrt([-4/5cos^2(t) - 4/5sin^2(t)]^2 + [3/5sin(t)cos(t) + 3/5sin(t)cos(t)]^2 + [12/25 sin^2(t) + 12/25sin^2(t)]^2
= sqrt(649/625) / sqrt(16/25 + 18/25sin^2(t))
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a_n = a - a_t.
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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: If V(0) = <5,-2,4> and A(0) = <1,3,-9>, what is A_T and A_N at t = 0?
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Your solution:
a_t = V dot A / sqrt(V)
= 5 -6 -36) / sqrt(25 + 4 + 16)
= -37 / sqrt(45)
= -37sqrt(5) / 15
a_n = ||(v X a)|| / ||v||
= ||<5,-2,4> X <1,3,-9>|| / sqrt(45)
= ||<6, -49, 17>|| / 3sqrt(5)
= sqrt(36 +2401 + 289) / 3sqrt(5)
= sqrt(2726) / 3sqrt(5)
= 7.78 units
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
@&
You got || a_t || and || a_n ||, which might be all that was asked.
To get the vector a_t you would then multiply by T = V / || V ||.
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Question: An object moves with a constant angular velocity omega around the circle x^2 + y^2 = r^2 in the xy-plane.
Find a parameterization for the circle.
Compute the tangential and normal acceleration for the object.
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Your solution:
x=rcos(theta)
y=rsin(theta)
r(t) = rcos(theta)i + rsin(theta)j
v(t) = -rsin(theta)i + rcos(theta)j
a(t) = -rcos(theta) - rsin(theta)
||v(t)|| = sqrt[r^2sin(theta) + r^2cos(theta)]
= r
a_t=
v(t) dot a(t) / ||v(t)|| = <-rsin(theta)i + rcos(theta)j> dot <-rcos(theta)i - rsin(theta)j> / r
= r^2sin(theta)cos(theta) -r^2sin(theta)cos(theta) / r
= 0/r = 0
Which makes since because there is constant velocity, acceleration would not be in the tangential direction, but rather the normal direction
a_n = ||v X a|| / ||v||
= ||r^2sin(theta) + r^2cos(theta)|| / r
= sqrt(r^4) / r
= r^2 / r
= r
Which makes since because for a circle with constant acceleration, r(t) = -a(t)
confidence rating #$&*:
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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:
Consider the vector function R(t) = <3 sin t, 4t, 3 cos t>.
Evaluate V(t) = R'(t), N(t), and A(t) = R''(t) when t = 1.
Find the vector projection of A(1) onto V(1). Denote this proj_V(1) (A(1)).
Find the vector projection of A(1) onto N(1). Denote this proj_N(1) (A(1)).
What is the sum of proj_V(1) (A(1)) and proj_N(1) (A(1)).
How does proj_V(1) (A(1)) relate to A_T when t = 1.
How does proj_N(1) (A(1)) relate to A_N when t = 1.
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Your solution:
R(t) = <3 sin t, 4t, 3 cos t>
R'(t) = <3cos(t), 4, -3sin(t)>
R'(1) = <3cos(1), 4, -3sin(1)>
A(t) = <-3sin(t), 0, -3cos(t)>
A(1) = <-3sin(1), 0, -3cos(1)>
T = R'/||R'||
= <3cos(t), 4, -3sin(t)> / sqrt[9cos^2(t) + 16 +9sin^2(t)]
= <3cos(t)/5, 4/5, -3sin(t)/5>
T' = <(-3/5)sin(t),0,(-3/5)cos(t)>
||T'|| =
= 3/5
N(t) = T'/||T'||
N(t) = <(-3/5)sin(t),0,(-3/5)cos(t)> / (3/5)
= <-sin(t), 0, -cos(t)>
N(1) = <-sin(1), 0, -cos(1)>
proj_V(1) (A(1)) = V(1) dot A(1) / ||V(1)||
= <3cos(1), 4, -3sin(1)> dot <-3sin(1), 0, -3cos(1)> / sqrt[9cos^2(1) + 16 + 9sin^2(1)]
= 0/3
proj_N(1) (A(1)) = N(1) dot A(1) / ||N(1)||
= <-sin(1), 0, -cos(1)> dot <-3sin(1), 0, -3cos(1)> / sqrt[
= [3sin^2(1) + 0 + 3cos^2(1)] / 1
= 3
sum of proj_V(1) (A(1)) and proj_N(1) (A(1)) = 3
@&
These are specified to be vector projections. So you need to base this on the vector projections, which are easily enough obtained from the scalar projections.
When you add the two vector projections you will obtain the original velocity vector.
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How does proj_V(1) (A(1)) relate to A_T when t = 1.
Well, they are the same. Just like the last problem, in order to find A_t, we used the dot product of v(t) and a(t) / ||v(t)||
How does proj_N(1) (A(1)) relate to A_N when t = 1
We used the cross product in the last problem, same answer, just different way of looking at it..
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
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Question: Let B = T X N when T and N are the unit tangent and normal vectors to a curve C with position vector R. Show that dB/ds = T X (dN/ds).
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Your solution:
unit tangent vector is perpendicular to the normal vector. B is orthgonal to these two vectors. To understand use the right-hand rule, where your thumb is B and normal vector is your wrist and your fingers are the tangent vector
a_t = v(t) dot a(t) / ||v(t)||
Tangent is the scalar
a_n = ||v X a|| / ||v||
Normal is the vector.
dB/ds = T X (dN/ds) says
to find the rate of change of B with respect to s, you have to find the rate of change of N with respect to s. becuase T is a scalar.
confidence rating #$&*:
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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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dB/ds = dT/ds X N + T X dN/ds.
The result would be proven if you could show that dT/ds X N = 0. This is so, and the reason is pretty straightforward.
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Self-critique (if necessary):
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Self-critique rating:
Self-critique (if necessary):
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#*&!
&#Good responses. See my notes and let me know if you have questions. &#