Query105

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course Mth 277

115pm 10/27/2012

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_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)>

@&

<-3/5sin(t), 45cos(t), -sin(t)> dot <-3/5cos(t), -4/5sin(t), - cos(t)> / sqrt(16 + 18sin^2(t)/25)

*@

= [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))

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:

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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

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:"

&#This looks very good. Let me know if you have any questions. &#