Assignment 9

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course Mth 277

If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

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

A(t)= -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):

The derivative of sin(t) is cos(t)

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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 unit tangent vector of the position vector is the velocity vector divided by the magnitude of the velocity vector. T(t)= (cos(t)i-sin(t)j+k)/sqrt(cos^2(t)+(-sin(t)^2)+(1)^2)= (cos(t)i-sin(t)j+k)/sqrt(2).

confidence rating #$&*:

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Given Solution:

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

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Self-critique (if necessary):OK

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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:

This is the projection of a(t) onto T(t):

[(<-sin(t), -cos(t)> dot <1/sqrt(2)(cos(t)), -1/sqrt(2)(sin(t)), 1/sqrt(2)>)/sqrt(cos^2(t)+sin^2(t))](<-sin(t), -cos(t)>)=

(-1/sqrt(2)(sin(t))(cos(t)) + 1/sqrt(2)(sin(t))(cos(t))+0)/1 (<-sin(t), -cos(t)>)= 0

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:

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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:

-sin(t)i- cos(t)j

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):

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Self-critique rating:

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

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Your solution:

The normal component vector of the acceleration vector is the acceleration vector minus the unit tangent vector of acceleration. I believe I have made a mistake finding the unit tangent vector of acceleration.

@&

The acceleration has a tangential component (the a component in the direction of motion) and a normal component (the component perpendicular to the direction of motion).

In this case the tangential component is zero, as you have shown, and the normal component is the acceleration vector itself.

*@

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):

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Self-critique rating:

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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 the tangential vector component is 0 and the normal component is <-sin(t), -cos(t)>, and the dot product between them is zero, then they 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):

I still think that I have done something wrong.

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Self-critique rating:1

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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:

If the vectors are in the same direction, then they are parallel. They are parallel if the cos(theta)=(unit tangent vector dot unit normal vector)/(the magnitude of the unit tangent vector times the magnitude of the unit normal vector)=1

confidence rating #$&*:

@&

You have calculated the unit tangent vector, so you can calculate the its derivative.

Then it's easy to show that the angle between the derivative of the unit tangent vector and the acceleration vector is zero.

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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 T’(t)/magnitude of T’(t)

= (d/dt(1/sqrt(2)cos(t)i- 1/sqrt(2)sin(t)j+ 1/sqrt(2)k)/(the magnitude of d/dt(1/sqrt(2)cos(t)i- 1/sqrt(2)sin(t)j+ 1/sqrt(2)k))=

T’(t)= -1/sqrt(2)sin(t)i -1/sqrt(2)cos(t)j

The normal unit vector is : (-1/sqrt(2)sin(t)i -1/sqrt(2)cos(t)j)/sqrt(0.5sin^2(t)+0.5cos^2(t)= (-1/sqrt(t)sin(t)i -1/sqrt(2)cos(t)j)/(1/sqrt(2))= -sin(t)i -cos(t)j

@&

note, related to the preceding, that this is clearly parallel to the acceleration vector

*@

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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Self-critique rating:

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Question: `q009. Find the unit binormal vector.

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Your solution:

B(t)= <1/sqrt(2)cos(t), -1/sqrt(2) sin(t), 1/sqrt(2)>cross<-sin(t), -cos(t)> =<0+1/sqrt(2)cos(t), -1/sqrt(2)sin(t)-0, -1/sqrt(2)cos^2(t)-1/sqrt(2)sin^2(t) >= <1/sqrt(2)cos(t), -1/sqrt(2)sin(t), -1/sqrt(2)>

confidence rating #$&*:

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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):OK

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Self-critique rating:OK

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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:

The velocity and acceleration vectors would be different because the derivative of R(t) would be different. Therefore, changing the unit tangent and unit normal vectors.

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:

The derivative of R(t) would again change, then changing the velocity and acceleration vectors.

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