Query Assignment 104

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course Mth 277

10/18 10pm

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Question:

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Your solution:

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Given Solution:

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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)||

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

so find derivative and magnitudes of them to get the normal and tangential vectors of R(t)

OK i found the derivative of R and magnitude to find T(t). this is messy.

I can get the T(t) but to get N(t) you take derivative of a already messy function, I'm only going to type the i component to show you what i dont understand.

T'(t) { ( -e^(2t) sint - 2cost*e^(2t) ) / ( e^(2t) )^2 } (i) / sqrt{ [ ( ( -e^(2t) sint - 2cost*e^(2t) ) / ( e^(2t) )^2 ] ^2 }

how....in the heck do I take the derivative of that? by squaring the inner terms then setting that in parenthasis to ^(1/2) ? then do the quotient rule from there? ( d/dx [u/v] = vdu - udv / v^2 ? this calculation seems very very large and messy derivatives

I've gone through simplifying this equation and separating the components and I see that this is posible, but still obviously messy.

@& There's no question that this process is a mess.

The derivative of R ' (t) / || R(t) || is, by the quotient rule,

((R ''(t) * || R(t) || - R ' (t) * (|| R(t) ||) ' ) / || R(t) || ^ 2.

I recommend first figuring out R ' (t), R ''(t), || R(t) ||, || R(t) || ', then assembling them into a solution.

That's more an exercise in algebra than calculus, but of course taking the derivative isn't trivial either.*@

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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:

equation for curavture of a planar curve is k = | f''(x) | / (1 + [ f'(x) ] ^2 ) ^(3/2)

9sin(3pi/2) / (1 + (-3cos(3pi/2)^2)^(3/2) = 9/1 or 9

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Given Solution:

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Self-critique (if necessary):

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:

OK on the unit tangent vector T(t) [ just find R'(t) / || R'(t) || ]

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

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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:

take the derivative and 2nd to get information for the k= | f""'(x) | / (1+[f'(x)]^2)^(3/2) and you get that

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

take the derivative of this and set it equal to 0 to solve for the maximum curvature and get that e^x = 0 and 1/3^(4/9)

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

ok

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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:

N = A'n

T =

A't = B dot T = 0 means that T is orthogonal to B

B dot B = 0 means it is just a multiple of B. ( i think this was supposed to be B dot N )

This is very difficult and I dont even know where to start, i've tried this on multiple peices of paper and dont need to keep getting bogged down on this one question

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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@& B dot B isn't zero.

(V dot W) ' = V ' dot W + V dot W '. This can be applied to both dot products.

Note that B = T X N is orthogonal to T. What does this tell you about B dot T, and how do you combine this with what you know about the derivative of B dot T?*@

@& Looks like you're doing pretty well here. Check my notes.*@