Keywords 015 100points the curves given

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keywords: 015 10.0points The curves given parametrically by x ( t ) = t, y ( t ) = t 5 , z ( t ) = t 6 ,
meaney (oam527) – Assignment 5 due by 11:59pm on Fri. 10/11 – reyes – (52605) 9 and x ( t ) = sin t, y ( t ) = sin 3 t, z ( t ) = t intersect at the origin. Find the cosine of their angle, θ , of inter- section. 1. cos θ = 2 2 3 2. cos θ = 1 11 correct 3. cos θ = 2 10 4. cos θ = 2 11 5. cos θ = 1 10 6. cos θ = 1 2 3 Explanation: The angle of intersection of the two curves is the angle between the tangent vectors at the point of intersection. Now the tangent vector at the origin to the first curve is r 1 (0) = ( 1 , 5 t 4 , 6 t 5 ) vextendsingle vextendsingle vextendsingle t =0 = ( 1 , 0 , 0 ) , while the tangent vector at the origin to the second curve is r 2 (0) = ( cos t, 3 cos 3 t, 1 ) vextendsingle vextendsingle vextendsingle t =0 = ( 1 , 3 , 1 ) . But then cos θ = r 1 (0) · r 2 (0) | r 1 (0) | | r 2 (0) | = 1 (2 + 9) 1 / 2 . Consequently, cos θ = 1 11 . 016 10.0points When C is parametrized by c ( t ) = (sin 4 t ) i + 3 t j + (cos 4 t ) k , find its arc length between c (0) and c (3). 1. arc length = 12 2. arc length = 15 correct 3. arc length = 6 4. arc length = 9 5. arc length = 18 Explanation: The length of the curve between c ( t 0 ) and c ( t 1 ) is given by the integral L = integraldisplay t 1 t 0 bardbl c ( t ) bardbl dt . Now when c ( t ) = (sin 4 t ) i + 3 t j + (cos 4 t ) k , we see that c ( t ) = (4 cos 4 t ) i + 3 j (4 sin 4 t ) bk . But then by the Pythagorean identity, bardbl c ( t ) bardbl = (16 + 9) 1 / 2 = 5 . Thus L = integraldisplay 3 0 5 dt = bracketleftBig 5 t bracketrightBig 3 0 . Consequently, arc length = L = 15 . 017 10.0points Find the unit tangent vector T ( t ) to the graph of the vector function r ( t ) = 4 sin t i 3 t j + 4 cos t k . 1. T ( t ) = 4 5 cos t i 3 5 j 4 5 sin t kcorrect 2. T ( t ) = 4 5 cos t i 3 5 j + 4 5 sin t k
meaney (oam527) – Assignment 5 due by 11:59pm on Fri. 10/11 – reyes – (52605) 10 3. T ( t ) = 4 5 sin t i + 3 5 j 4 5 cos t k 4. T ( t ) = 4 sin t i + 3 k + 4 5 cos t k 5. T ( t ) = 4 cos t i + 3 j 4 sin t k 6. T ( t ) = 4 sin t i + 3 j 4 cos t k Explanation: The unit tangent vector T ( t ) to r ( t ) is given by T ( t ) = r ( t ) bardbl r ( t ) bardbl . Now when r ( t ) = 4 sin t i 3 t j + 4 cos t k , we see that r ( t ) = 4 cos t i 3 j 4 sin t k , while bardbl r ( t ) bardbl = radicalBig 4 2 (cos 2 t + sin 2 t ) + ( 3) 2 = 5 . Consequently, T ( t ) = 4 5 cos t i 3 5 j 4 5 sin t k . keywords: 018 10.0points Find the unit vector T ( t ) tangent to the graph of the vector function r ( t ) = (big 4 t 2 , 8 t, 4 ln t )big . 1. T ( t ) = (Big t 2 2 t 2 + 1 , t 2 t 2 + 1 , 1 2 t 2 + 1 )Big 2. T ( t ) = (Big t 2 t 2 + 1 , t t 2 + 1 , 1 t 2 + 1 )Big 3. T ( t ) = (Big 2 t 2 2 t 2 + 1 , 2 t 2 t 2 + 1 , 1 2 t 2 + 1 )Big correct 4. T ( t ) = (Big 2 t 2 t 2 + 1 , 2 t t 2 + 1 , 1 t 2 + 1 )Big 5. T ( t ) = (Big 2 t 2 t 2 + 1 , t t 2 + 1 , 1 t 2 + 1 )Big 6. T ( t ) = (Big t 2 2 t 2 + 1 , 2 t 2 t 2 + 1 , 1 2 t 2 + 1 )Big Explanation: The unit tangent vector T ( t ) is given by T ( t ) = r ( t ) | r ( t ) | . Now r ( t ) = (Big 8 t, 8 , 4 t )Big while | r ( t ) | = parenleftBig 64 t 2 + 64 + 16 t 2 parenrightBig 1 / 2 = 4 t (2 t 2 + 1) . Consequently, T ( t ) = (Big 2 t 2 2 t 2 + 1 , 2 t 2 t 2 + 1 , 1 2 t 2 + 1 )Big . keywords: vector function, space curve tan- gent vector, unit tangent vector 019 10.0points Determine the curvature, κ , of the curve r ( t ) = t 2 i + 7 t k .

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