Query10_4mth

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course Mth 277

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Question:

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Your solution:

No question

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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:

R'(t)=(e^-2t cost)i+(e^-2t sint)j+e^-2tk

R'(t)=2*e^-2t*sin(t)i-2e^-2t*cos(t)j-2e^-2tk

T(t)=R'/||R'||

= [2*e^-2t*sin(t) i - 2e^-2t*cos(t) j - 2e^-2t k] / || (4e^-4t)*sin^2(t) + (4e^-4t)*cos^2(t) + (4e^-4t) ||

= [2*e^-2t*sin(t) i - 2e^-2t*cos(t) j - 2e^-2t k] / sqrt(8e^-4t)

= [4*e^-4t*sin^2(t) i + 4e^-2t*cos^2(t) j + 4e^-4t k] / (8e^-4t)

= [0.5*sin^2(t) i + 0.5*cos^2(t) j + 0.5 k]

N(t)= T' / ||T'||

= 0.5*sin^2(t) i + 0.5*cos^2(t) j + 0.5k / ||0.5*sin^2(t)+0.5*cos^2(t)+ 0.5||

= 0.5*sin^2(t) i + 0.5*cos^2(t) j + 0.5k / sqrt(.25sin^4(t)+0.5cos^4(t)+0.25)

= 0.5*sin^2(t) i + 0.5*cos^2(t) j + 0.5k / sqrt(.25sin^4(t)+0.255cos^4(t)+0.25)

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Your sequence of steps is correct, but you are missing some details in differentiation.

To start with:

R ' (t) = e^(-2 t) * ( (-sin(t) - 2 cos(t)) `i + (cos(t) - 2 sin(t)) `j - 2 `k)

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You will find that

T ' (t) = 1/3 ( ( -cos(t) + 2 sin(t) ) `i + (-sin(t) - 2 cos(t) ) `j)

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

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Your solution:

y = sin (-3x)

k = y""(x)/(1+[y'(x)]^2)^(3/2)

y'=-3cos(-3x)

y'(pi/2)=0

y""=sin(-3x)

y""(pi/2)=9

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

k=9

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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: 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:

R(t)=(1-cost)i+(t-sint)j+(4sin(t/2))k

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

||R'||=sqrt[sin^2(t) + (cos^2(t) -2cos(t) + 1) + 4cos^2(t/2)]

=sqrt(1+1-2cos(t)+2+2cos(t))

=2

T=R'(t)/||R'||

=[sin(t)i + (1 - sin(t)) j + 2cos(t/2) k] / 2

=[sin(t) / 2] i + [(1-cos(t)) / 2] j + [cos(t/2)] k

dT/ds = K = ||T'||/||R'||

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

||T'|| = sqrt[cos^2(t)/4 + sin^2(t)/4 + sin^2(t/2)/4]

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

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

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

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Your solution:

y = e^3x

y’ = 3e^3x

y’’ = 9e^3x

k=9e^(3x)/(1+27e^(9x))

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You're not going to get e^(9x) by squaring e^(3x)

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max curve k'=0

k'=((1+27e^(9x)(27e^(3x)-((9e^(3x))(243e^(9x))))/((1+27e^(9x))^2

Since k'=0 we set the numerator equal to 0

27e^3x+729e^27x-2187e^27x=0

27e^3x-1458e^27x=0

27e^3x(1-54e^6x)=0

We see here that x could equal 0 or the other expression can be set to 0.

SO,

54e^6x=1

6x=ln(1/54)

x=-.66

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Right sequence of steps, but see my note on at least one detail.

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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:

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Your solution:

NO question

confidence rating #$&*:

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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 is always 0 and since the derivative of the dot product is (B'dot T)+(B dot T'). Therefore, T would have to be orthogonal to dB/ds.

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You also need to show that B dot T ' = 0.

The derivative of B dot T with respect to s is dB/ds dot T + B dot dT/ds. Since T is a unit vector its derivative is orthogonal to T, as is B. You can conclude that dT/ds is orthogonal to B, which since B is orthogonal to T (making B dot T zero) leaves you with dB/ds dot T = 0.

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Im not sure how to prove the rest of it.

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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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Self-critique (if necessary):

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Self-critique rating:

@&

You're missing some details in some of your differentiation, but overall you are using the right sequence of reasoning.

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

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