query 104

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course Mth 277

2-27

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.

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Question:

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Your solution:

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Given Solution:

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Self-critique (if necessary):

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Question: Find T(t) and N(t) when R(t) = (e^-2t cost)i + (e^-2t sint )j + e^-2t.

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Your solution:

T(t) = R’(t)/||R’(t)||

R’(t) = (-2e^(-2t)cos(t) - e^(-2t)sin(t))i + (-2e^-2t sint + e^(-2t)cos(t) )j - 2e^-2t

=-2e^(-2t)[( cos(t) -.5sin(t))i + (sint + .5cos(t) )j + 1]

||R’(t)|| = `sqrt((-2e^(-2t)cos(t) - e^(-2t)sin(t))^2 + (-2e^-2t sint + e^(-2t)cos(t))^2 +

(-2e^(-2t))^2)

=`sqrt((-4e^(-4t)cos^2(t) - e^(-4t)sin^2(t)) + (-4e^-4t sin^2t + e^(-4t)cos^2(t)) +(-4e^(-

4t)))

= `sqrt(-4e^(-4t)cos^2(t) - 4e^-4t sin^2t-e^(-4t)sin^2(t)+e^(-4t)cos^2(t)-4e^(-4t))

=`sqrt(-4e^(-4t)[cos^2(t) + sin^2(t)] - e^(-4t) [sin^2(t) - cos^2(t)]- 4e^(-4t))

= `sqrt(-8e^(-4t) - e^(-4t) [sin^2(t) - cos^2(t)])

=`sqrt (-e^(-4t)( 8 + (sin^2(t) - cos^2(t)))

So that R’(t)/||R’(t)||

[1/`sqrt (-e^(-4t)( 8 + (sin^2(t) - cos^2(t)))]*[ -2e^(-2t)[( cos(t) -.5sin(t))i + (sint + .5cos(t) )j + 1]

=< -2e^(-2t)( cos(t) -.5sin(t))/ `sqrt (-e^(-4t)( 8 + (sin^2(t) - cos^2(t))), -2e^(-2t )(sint + .5cos(t))/

`sqrt (-e^(-4t)( 8 + (sin^2(t) - cos^2(t)))

???I’m not sure about this problem, is the e^-2t term in R(t) supposed to be the k component. I also believe I have messed up my calculations somewhere?????????????????

@& The derivative of the unit tangent vector is perpendicular to the unit tangent vector, and therefore to the velocity.

T ' (t) = sqrt(6) / 6 ( ( -cos(t) + 2 sin(t) ) `i + (-sin(t) - 2 cos(t) ) `j).

Note how the components compare to those of T(t), the `i component of one being equal in magnitude to the `j component of the other, and with the sign of one being switched. Be sure you see how this ensures that the dot product of T and T ' is zero.

The unit normal vector is found by dividing T ' (t) by || T ' (t) || .

The acceleration vector is

a(t) = v ' (t) = ((-sin(t) - 2 cos(t)) `i + (cos(t) - 2 sin(t)) `j + `k) '

= ( -cos(t) + 2 sin(t) ) `i + (-sin(t) - 2 cos(t)) `j

The component of the acceleration vector in the direction of the unit tangent vector is

a_parallel(t) = a(t) dot T(t) = sqrt(6) / 6 * ( (sin(t) cos(t) + 2 cos^2(t) - 2 sin^2(t) - 4 sin(t) cos(t)) - cos(t) sin(t) - 2 cos^2(t) + 2 sin^2(t) + 4 sin(t) cos(t) )

= 0.

The acceleration vector is therefore perpendicular to the unit tangent vector, so the unit normal vector is just equal to the acceleration vector divided by its magnitude.

You can verify that the magnitude of the acceleration vector is sqrt(6), so that the unit normal vector is

N(t) = ( sqrt(6) / 6 * ( -cos(t) + 2 sin(t) ) `i + (-sin(t) - 2 cos(t)) `j )

*@

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Given Solution:

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Question: Find the curvature of the plane curve y = sin (-3x) at x = pi/2.

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Your solution:

k = ( |f’’(x)|/((1+ [f’(x)]^2)^(3/2)))

Where…

f’(x) = -3cos(-3x)

f’(`pi/2) = 0

f’’(x) = -9sin(-3x)

f’’(`pi/2) = -9

k = ( |f’’(x)|/((1+ [f’(x)]^2)^(3/2)))

k = 9/(1+0)^(3/2) = 9

So curve of plane is equal to 9.

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Given Solution:

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Question: Let C be the curve given by R(t) = (1-cos t)i + (t-sin t)j + (4sin(t/2))k

• Find the unit tangent vector T(t) to C

• Find dT/ds and the curvature k(t)

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Your solution:

To find T(t) we find the following

R’(t) = (sin(t))i + (1-cos(t))j + (2cos(t/2))k

||R’(t)|| = `sqrt((sin(t))^2 + (1-cos(t))^2 + (2cos(t/2))^2)

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

== `sqrt(sin(t)^2 + sin(t)^2 + 4(1+cos(t)/2))

=`sqrt(2 sin(t)^2 + 2+2cos(t)^2)

= `sqrt(2(sin(t)^2 + 1 + cos(t)^2))

= `sqrt(2(1 + 1 )) = `sqrt(4) = 2

T(t) = R’(t)/||R’(t)|| = (sin(t))i + (1-cos(t))j + (2cos(t/2))k/2

Now we find k(t)

T’(t) = (cos(t)/2)i + (sin(t)/2)j + (cos(t/2)/2)k

||T’(t)|| = `sqrt( (cos(t)/2)^2 + (sin(t)/2)^2 + (cos(t/2)/2)^2)

=`sqrt((1/2)*(cos(t)^2 + sin(t)^2 + cos(t/2))

=`sqrt((1/2)*(1 + cos(t/2)^2))

=`sqrt((1/2)*(1 +(+-`sqrt((1 + cos(t))/2))^2))

= `sqrt((1/2)*(1 +(1/2) + cos(t)/2))

= `sqrt((1/2)*(1.5+ cos(t)/2))

= `sqrt(.75 + cos(t)/4)

@& Easier:

1/2 sqrt( cos^2(t) + sin^2(t) + cos^2(t/2)) = 1/2 sqrt(1 + cos^2(t/2) ).

*@

??? I think I made a simply algebria mistake when calculating ||R’(t)||, which would throw off everything else. But I can’t see it right off??????

k = ||T’(t)||/||R’(t)|| = `sqrt(.75 + cos(t)/4)/2

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Given Solution:

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Question: Find the maximum curvature for the curve y = e^3x.

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Your solution:

For f(x) = e^(3x)

f’(x) = 3e^(3x) and f’’(x) = 9e^(3x)

So curvature is

k = 9e^(3x)/((1+(3e^(3x))^2)^(3/2))

=9e^(3x)/(1+3e^(6x))^(3/2)

@& Need to square the 3 in 3 e^(3 x); you get

=9e^(3x)/(1+9e^(6x))^(3/2)*@

Now we find k’(x)

k’(x) = ((-27e^(3x)(6e^(6x) - 1))/(3e^(6x) + 1)^(5/2)

= -162e^(9x) + 27e^(3x)/ (3e^(6x) + 1)^(5/2) = (27e^(3x) - 162e^(9x))/ (3e^(6x) + 1)^(5/2)

We need to find k’(x) = 0 and that will be when numerator is = 0 so…

(27e^(3x) - 162e^(9x)) = 0

27 = 162e^(6x)

e^(6x) = .167

6x = ln(.167)

x = ln(.167)/6 = approx. -.2983

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Given Solution:

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Question:

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Your solution:

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Given Solution:

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Self-critique (if necessary):

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Question: If T and N are the unit tangent and normal vectors on the trajectory of a moving body, we can define B = T X N to be the unit binormal vector. A coordinate system with three planes can be made at each point with these vectors since they are mutually orthogonal.

• Show that T is orthogonal to dB/ds (Hint: Differentiate B dot T)

• Show that B is orthogonal to dB/ds (Hint: Differentiate B dot B)

• Show that dB/ds = -(tau)N for some constant tau. (We call the constant tau the torsion of the trajectory)

• **Prove the Frenet-Serret formulas: (dT/ds = kN, dN/ds = -kT + tauB, dB/ds = -tauN). Where k is the curvature and tau = tau(s) is a scalar function.

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Your solution:

(B dot T)’(s) = (B’dot T)(s) + (B dot T’)(s), we know B’ is orthogonal to B and T’ is orthogonal to T

= (dB/ds Dot T(s)) + (B(s) Dot dT/ds)

????I’m having a hard time understanding how to approach this problem???????

@& For example:

The derivative if B dot T is

B ' dot T + B dot T '

= (T X N ) ' dot T + B dot T '.

B = T X N is perpendicular to T and to T '; since T ' = N, then, B is perpendicular to T ' so B dot T ' = 0.

This holds whether the derivative is with respect to s or with respect to t.

So what can be made of (T X N ) ' dot T?

*@

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