qa10_02

#$&*

course Mth 277

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:

****I was confused if we were suppossed to use this position vector for all the problems or if we were merely discussing how to achieve an answer if a problem was given. Since the first one gave a solution with the initial vector and the rest did not I did it similarly. Also, since our velocity vectors are different none of the answers are simplifying and therefore I would be left with expressions that are lines long. I know how to do all of these questions when the opportunity presents itself.****

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

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Your solution:

The velocity vecotr associate is the derivative of the position vector.

v(t)=cos(t)i-sin(t)j+(1)k

a(t)=-sin(t)i-cos(t)j+0k

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):

Is there a reason why you did the integral instead of the derivative for the velocity? Or did I mess something up?

@&

I think I changed the problem and missed an edit.

However I will tell you that there are errors in these qa's and even in the Queries. Somehow a set of old uncorrected versions got posted. I had intended to correct them, but an unexpected surge of enrollment has kept me behind on responding to student work and until I get caught up, I can't properly do the corrections.

I hope to get to this within the next couple of days.

So don't take the given solutions too seriously. As long as you know there are errors, you'll know that you're free to disagree and point them out.

*@

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Self-critique rating:3

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

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Your solution:

The function describing the unit tangent vector is found by doing the velocity vector divide by the magnitude of the velocity vector.

(cos(t)-sin(t)+1)/(sqrt(cos^2(t)-sin^2(t)+1^2))

@&

That would be

(cos(t) `i - sin(t) `j + `k ) / sqrt( cos^2(t) - sin^2(t) + 1 ).

*@

confidence rating #$&*:

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Given Solution:

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

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Self-critique (if necessary):

I tried to simplify and I could not seem to simplify it. I believe the reason why is because our velocity vectors are different.

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Self-critique rating:OK

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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:

The component of the acceleration vector in the direction of the unit tangent vector is the projection of the acceleration vector onto the unit tangent vector.

(-sin(t)i-cos(t)j dot (cos(t)-sin(t)+1)/(sqrt(cos^2(t)-sin^2(t)+1^2)))/(-sin(t)i-cos(t)j)

@&

(-sin(t)i-cos(t)j dot (cos(t) `i-sin(t) `j+`k)/(sqrt(cos^2(t)-sin^2(t)+1^2)))

multiplied by the unit vector in the direction of `v. (Note that you can't divide by a vector).

The dot product can be multiplied out.

*@

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):OK

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Self-critique rating:OK

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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:

To find this, we need to do `a(t) - `a_T(t). The result of that gives us the vector a_N(t).

@&

Right. Be sure you can calculate this (see previous note).

*@

confidence rating #$&*:

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Given Solution:

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

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Self-critique (if necessary):OK

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Self-critique rating:OK

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

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Your solution:

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

confidence rating #$&*:

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Given Solution:

This is the component perpendicular to the unit tangent vector.

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Self-critique (if necessary):OK

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Self-critique rating:OK

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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:

To do this, we would take the normal componenet of the acceleration and dot it by the tangential component. If there result is 0 then the vectors are perpendicular.

confidence rating #$&*:

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Given Solution:

Two vectors are perpendicular if their dot product is zero.

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Self-critique (if necessary):OK

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Self-critique rating:OK

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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:

The direction of the derivative of the unit tangent vector is the same as the the unit normal vector if cos(theta)=0.

confidence rating #$&*:

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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):

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Self-critique rating:

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Question: `q008. Find the unit normal vector.

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Your solution:

The unit normal vector is the derivative of the unit tangent vector divided by the derivative of 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):OK

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Self-critique rating:OK

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@&

... divided by the magnitude of the derivative

*@

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:

If the position function was changed,many of the above results would be different considering the chain rule will be used to differentiate. However, that is why it is good that I explained how to do things. Even though the position functions are different, it is easily tackled with the steps given above.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:OK

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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:

It would change the values quite a bit. This is because the velocity and acelleration vectors would change drastically since the chain rule would be used for the derivative and a domino effect would cause the other values to change as well. However, the same concepts will be used as established above. Also, there would be a k component to the velocity vector of this problem since the k component of the position vector is t^2.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:OK"

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#*&!

&#Good responses. See my notes and let me know if you have questions. &#