#$&* course Mth 277 query_10_4*********************************************
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Find T(t) and N(t) when R(t) = (e^-2t cost)i + (e^-2t sint )j + e^-2t. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: R'(t)=(e^-2t cost)i+(e^-2t sint)j+e^-2tk R'(t)=2*e^-2t*sin(t)i-2e^-2t*cos(t)j-2e^-2tk T(t)=R'/||R'|| = [2*e^-2t*sin(t) i - 2e^-2t*cos(t) j - 2e^-2t k] / || (4e^-4t)*sin^2(t) + (4e^-4t)*cos^2(t) + (4e^-4t) || = [2*e^-2t*sin(t) i - 2e^-2t*cos(t) j - 2e^-2t k] / sqrt(8e^-4t) = [4*e^-4t*sin^2(t) i + 4e^-2t*cos^2(t) j + 4e^-4t k] / (8e^-4t) = [0.5*sin^2(t) i + 0.5*cos^2(t) j + 0.5 k] N(t)= T' / ||T'|| = 0.5*sin^2(t) i + 0.5*cos^2(t) j + 0.5k / ||0.5*sin^2(t)+0.5*cos^2(t)+ 0.5|| = 0.5*sin^2(t) i + 0.5*cos^2(t) j + 0.5k / sqrt(.25sin^4(t)+0.5cos^4(t)+0.25) = 0.5*sin^2(t) i + 0.5*cos^2(t) j + 0.5k / sqrt(.25sin^4(t)+0.255cos^4(t)+0.25)
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: Find the curvature of the plane curve y = sin (-3x) at x = pi/2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = sin (-3x) k = y""(x)/(1+[y'(x)]^2)^(3/2) y'=-3cos(-3x) y'(pi/2)=0 y""=sin(-3x) y""(pi/2)=9 k=9/(1+0)^(3/2) k=9 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* 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) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 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 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: Find the maximum curvature for the curve y = e^3x. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = e^3x y’ = 3e^3x y’’ = 9e^3x k=9e^(3x)/(1+27e^(9x))
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: NO question confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 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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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: