#$&* course Mth 277 935pm 11/13/2011 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Find T(t) and N(t) when R(t) = (e^-2t cost)i + (e^-2t sint )j + e^-2t. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I am going to assume that e^-2t is the k component. R(t) = (e^-2t cost)i + (e^-2t sint )j + e^-2tk R'(t) = 2*e^-2t*sin(t) i - 2e^-2t*cos(t) j - 2e^-2t k T= R' / ||R'|| = [2*e^-2t*sin(t) i - 2e^-2t*cos(t) j - 2e^-2t k] / || (4e^-4t)*sin^2(t) + (4e^-4t)*cos^2(t) + (4e^-4t) || = [2*e^-2t*sin(t) i - 2e^-2t*cos(t) j - 2e^-2t k] / sqrt(8e^-4t) = [4*e^-4t*sin^2(t) i + 4e^-2t*cos^2(t) j + 4e^-4t k] / (8e^-4t) = [0.5*sin^2(t) i + 0.5*cos^2(t) j + 0.5 k] N = T' / ||T'|| = sin(t)cos(t) i - sin(t)cos(t) j / || [sin^2(t)cos^2(t) + sin^2(t)cos^2(t)] || = 1/sqrt(2) i - 1/sqrt(2) j confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Find the curvature of the plane curve y = sin (-3x) at x = pi/2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y= sin (-3x) y' = -3cos(-3x) y"" = -sin(-3x) K = y"" / [1+(y')^2]^(3/2) = -9sin(-3x) / [1 + 9cos^2(-3x)]^(3/2) = -9sin(-3 * pi/2) / [1 + 9cos^2(-3pi/2)]^(3/2) = -9 / [1 + 9/2 + 9/2cos(-3pi)]^(3/2)] = -9 / (11/2 - 9/2)^(3/2) = -9sqrt(2) / 4 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Let C be the curve given by R(t) = (1-cos t)i + (t-sin t)j + (4sin(t/2))k Find the unit tangent vector T(t) to C Find dT/ds and the curvature k(t) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: R(t) = (1-cos t)i + (t-sin t)j + (4sin(t/2))k R'(t) = sin(t)i + (1 - cos(t)) j + 2cos(t/2) k ||R'|| = sqrt[sin^2(t) + (cos^2(t) -2cos(t) + 1) + 4cos^2(t/2)] = sqrt(1+ 1 - 2cos(t) + 2 + 2cos(t)) = sqrt(4) = 2 T = R'(t) / ||R'|| = [sin(t)i + (1 - sin(t)) j + 2cos(t/2) k] / 2 = [sin(t) / 2] i + [(1-cos(t)) / 2] j + [cos(t/2)] k dT/ds = K = ||T'||/||R'|| T' = cos(t)/2 i + sin(t)/2 j - sin(t/2) / 2 k ||T'|| = sqrt[cos^2(t)/4 + sin^2(t)/4 + sin^2(t/2)/4] = sqrt(1/4 + sin^2(t/2)/4) K = sqrt(1/4 + sin^2(t/2)/4) / 2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Couldn't figure out how to break down the equation [1/4 + sin^2(t/2)/4} any further ------------------------------------------------ Self-critique rating: ********************************************* Question: Find the maximum curvature for the curve y = e^3x. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = 3^3x y' = 3e^3x (y')^2 = 9e^6x y"" = 9e^3x K = y"" / [1 + (y')^2]^(3/2) K = (9e^3x) / [1 + 9e^6x]^(3/2) = (9e^3x) * [1 + 9e^6x]^(-3/2) k'= (9e^3x)*(54e^6x)*(-2/5)*(1+9e^6x)^(-5/2) + (1+9e^6x)^(-3/2)*27e^3x 0 = (972e^9x) / -5*(1+9e^6x)^(5/2) + (27e^3x) / (1+9e^6x)^(3/2) (972e^9x) / [5*(1+9e^6x)] = 27e^3x 972*e^9x = (135e^3x)*(1+9e^6x) 7.2*e^6x = 1 + 9e^6x I tried to work the solution out, but I can't find x. I tried to do the limit as x->-infinity and found x = 0 So either there is no maximum(which doesn't make since) since an exponential function has a curve. The only time you would find where there is no maximum is when you have an absolute value function or possibly a linear function So I take x = 0 and plug back into K and I get K = 9/[10sqrt(10)] =approx. .285 units confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):
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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: I don't understand how you want me to show you how they are orthoganl, but I believe that I can explain it in words. B = T X N T = R'/||R'|| N = T'/||T'|| B = R'/||R'|| X T'/||T'|| When you do the right-hand rule, B is your thumb. So the vector R is coming from the x-y axis from point (x_0,y_0) along some curve. At the point where R(t) meets the curve, there is a tangent line at that point. The dirivative of that tangent is normal to the tangent line. Similar to a position vector from (x_0,y_0), a velocity vector is tangent to the curve. So the unit tangent vector is the derivitive of the position vector divided by the maginitude. thats why they call it a unit vector, it shows direction. The Normal unit vector is perpendicular to the Tangent unit vector, similar to the way an acceleration vector is to a velocity vector, when there is constant acceleration like the video of the planet around the sun. So, now we have a normal vector that is perpendicular to the tangent vector which is perpendicular to the position vector. So since B= T X N, then it can be easily assumed that B dot T and B dot B are orthoganl and equal zero. If the normal vector is perpendicular to the tangent which is perpendicular to the position, that makes the normal vecotr opposite the position vecotr and dB/ds = -(tau)N is understood to be true. (dT/ds = kN k is curvature or the reciprocal of the radius. So kN is N/r which is the rate of change of the tangent unit vector with respect to position. Once again, normal is perpendicular to tangent. dN/ds = -kT + tauB, The rate of change of the Normal vector with respect to position is equal to the negative value of curvature + tau * B vector. The normal vector is orthoganl to B and goes back towards the center of the curve. The more that normal vector changes, the greater the angle of tau multiplies the B vector. B vector is your thumb on the right hand rule. dB/ds = -tauN And as far as the last equation is the B vector(right hand thumb) changes with respect to position eqauls the negative normal vector times the angle between those two vectors. I'm not certain at all I got the answer to this question right, I just wanted to be sure that I gave a good answer to my understanding of this chapter. Its all interlinked, through phyics and normal vectors, accelerations and gravity, radians and calculus. I am connecting the dots and the picture is still hazy, but becoming much clearer. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: