Assignment 9 QA

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course Mth 277

10/02/2011 @ 3:33p.m. I have had this done for several days but have been neglecting to take the time to submit this and our other assignments due last Wednesday, so I'm sorry for the pretty large number of submissions going out today.

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.

qa 10_02

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Question:

Section 10.2

The velocity vector corresponding to position vector `R(t) = x(t) `i + y(t) `j + z(t) `k is the derivative `v(t) = `R ' (t) = x ' (t) `i + y ' (t) `j + z ' (t) `k, and the acceleration vector is `a(t) = `v ' (t) = `R '' (t) = x '' (t) `i + y ''(t) `j + z '' (t) `k.

The unit tangent vector is the vector function `T(t), equal at every instant to the unit vector in the direction of the velocity `v(t).

The acceleration vector has components `a_T(t) in the direction of the unit tangent vector, and `a_N(t) = `a(t) - `a_T(t) in the direction perpendicular to the unit tangent vector.

The unit normal vector is the unit vector in the direction of `a_N(t), and is perpendicular to the unit tangent vector.

The direction of the derivative `T ' (t) of the unit tangent vector is the same as that of the unit normal vector.

The unit binormal vector `B(t) is the cross product of the unit tangent and unit normal vectors.

Note that ` in front of a symbol indicates that the symbol is a vector. The only exception: `d means 'Delta'. I will eventually search-replace the document to convert the notation to boldface.

If `R(t) = sin(t) `i + cos(`t) j + t `k then:

`q001. What are the associated velocity and acceleration vectors?

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Your solution:

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

A = sin(t)I + cos(t)j

confidence rating #$&*:

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Given Solution:

`v(t) = `R ' (t) = -cos(t) `i + sin(t) `j + `k

`a(t) = `v ' (t) = `r '' (t) = sin(t) `i + cos(t) `j

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Self-critique (if necessary):

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Self-critique rating:

@& No problem on this end, where it all evens out in the long run. If it works for you, fine.*@

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Question: `q002. What is the function describing the unit tangent vector?

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Your solution: v(t)/||v(t)||

@& You have the expression for the velocity vector (which is the derivative from the first question), so you can write out the expression for this vector.*@

confidence rating #$&*:

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Given Solution:

Divide `v(t) by || `v(t) || and simplify

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Self-critique (if necessary):

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Question: `q003. What is the component of the acceleration vector in the direction of the unit tangent vector?

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Your solution: acceleration vector projected on a tangent unit vector.

@& You need to calculate this projection, given the information you obtained from the first two questions.*@

confidence rating #$&*:

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Given Solution:

The component is denoted `a_T (t) . The desired component is the projection of `a(t) on `T(t).

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Self-critique (if necessary):

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Self-critique rating: moderately

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Question: `q004. What is the component of the acceleration vector in the direction perpendicular to the unit tangent vector?

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Your solution: ‘a(t) - ‘a_T(t)

@& This needs to actually be computed.*@

confidence rating #$&*:

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Given Solution:

Subtract the component `a_T(t) from `a(t).

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Question: `q005. What is the normal component of the acceleration?

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Your solution: The component perpendicular to ‘T(t)

@& You need to compute this result and give the expression.*@

confidence rating #$&*:y

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Given Solution:

This is the component perpendicular to the unit tangent vector.

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Self-critique (if necessary):

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Question: `q006. Show that the normal component of the acceleration is perpendicular to the tangential component.

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Your solution: If a_’N(t) dot ‘T(t) = 0, then they are perpendicular

confidence rating #$&*:y

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Given Solution:

Two vectors are perpendicular if their dot product is zero.

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Self-critique (if necessary):

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Question: `q007. Show that the direction of the derivative of the unit tangent vector is the same as that of the unit normal vector.

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Your solution: cos(’theta) = 0 means they are in the same direction (parallel or the same vector entirely)

confidence rating #$&*:y

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Given Solution:

Two vectors are parallel if the cosine of the angle between them is zero.

How therefore can to test to see if the vectors are parallel?

What further test allows us to determine if they are in the same direction, vs. in the opposite directions.

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Self-critique (if necessary): Won’t they be multiples with different signs if they are parallel in different directions or is this going to just make the vectors different entirely?

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Question: `q008. Find the unit normal vector.

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Your solution: To make any vector a unit vector, we simply divide it by its magnitude.

confidence rating #$&*:

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Given Solution:

You have at least one vector in the normal direction (in fact in the preceding questions you have found two). Use either to find the unit normal.

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Self-critique (if necessary):

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Question: `q009. Find the unit binormal vector.

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Your solution: We first f ind a unit vector in the same direction as the normal acceleration then cross it with ‘T(t) .

confidence rating #$&*: very confident

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Given Solution:

You should have the unit normal and unit tangent. Use them to easily find the unit binormal. How do you know that your result is a unit vector?

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Self-critique (if necessary): I essentially approached this question by using your equations at the top, but I want to divide by the magnitude of some vector somewhere to find the “unit” part of the binormal vector…but this doesn’t seem to be the case here.

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Question: `q009. What difference would it make in the above results if the function was `R(t) = sin(t^2) `i + cos(t^2) `j + t `k?

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Your solution: When we took the derivatives of this function to find velocity and acceleration, the derivate of the “inside” would be 2t instead of just 1 and our 2nd derivative used to find acceleration would have the derivate of our inside times the derivative of 2t*-cos(t^2)i instead of just -cos(t)i. It appears that the acceleration and velocity are just probably going to just be quite a bit larger in this t^2 equation.

confidence rating #$&*:y

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Given Solution:

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Self-critique (if necessary):

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Question: `q010. What difference would it make in the above results if the function was `R(t) = sin(t^2) `i + cos(t^2) `j + t^2 `k?

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Your solution: k wouldn’t disappear by the time we took the 2nd derivative, so we would accelerate in the k direction.

confidence rating #$&*:y

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Given Solution:

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#*&!

@& Throughout, you need to compute your results and give the resulting expressions.

&#Please see my notes and, unless my notes indicate that revision is optional, submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).

Be sure to include the entire document, including my notes.

If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you.

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Self-critique (if necessary):

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Self-critique rating:

#*&!

@& Throughout, you need to compute your results and give the resulting expressions.

&#Please see my notes and, unless my notes indicate that revision is optional, submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).

Be sure to include the entire document, including my notes.

If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you.

&#

*@

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