#$&* course Mth 277 7/10 2 pm Question: Find both F' and F'' for F(t) = (4sin^2 t)i + (9cos^2 t)j + tkYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
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Given Solution: Self-critique (if necessary); ------------------------------------------------ Self-critique rating: ********************************************* Question: Given the position vector of a particle R(t) = (cos t)i + tj + (4 sin t)k, find the particle's velocity and acceleration vectors and then find the speed and direction of the particle at t = pi/2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Velocity = (-sin t)i + j + (4 cos t)k Acceleration = (-cos t)i - (4 sin t)k Velocity(pi/2) = -1 + 0 + 4*0 = -1 units/second Acceleration(pi/2) = 0 - 4*1 = -4 units/second^2
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Find Int(
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Find Integral((e^t)*
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Find the velocity and position vectors given the acceleration vector `A(t) = 4(t^2)i - 2 sqrt(t) j + 5(e^3t)k, initial position R(0) = 2i + j -3k and initial velocity v(0) = 4i + j + 2k. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Velocity = 4/3 t^3 + c1 i - 4/3 t^(3/2) + c2 j + 5/2 e^3*t^2 + c3 k Use the initial velocity and substitute 0 for t: 4 i + j + 2k = (0 + c1)i - (4/3*0 + c2) j + (5/2*e^3*t^2 + c3)k 4 i + j + 2k = c1 i - c2 j + c3 k c1 = 4, c2 = -1, c3 = 2 V = (4/3 t^3 + 4) i - (4/3 t^(3/2) - 1) j + (5/2 e^3*t^2 + 2) k Position: 1/3 t^4 + 4t + c1 i - 8/15 t^(5/2) - t + c2 j + 5/6*e^3*t^3 + 2t + c3 Use the initial position, substitute 0 for t: (1/3*0 + 0 + c1)i + (-8/15*0 - 0 + c2)j + (5/6*e^3*0 + 0 + c3)k Set it equal to the initial: c1 i - c2 j + c3 k = 4i + j + 2k c1 = 4, c2 = -1, c3 = 2 S = (1/3 t^4 + 4t + 4) i + (-8/15 t^(5/2) - t - 1) j + (5/6*e^3*t^3 + 2t + 2)k confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: F(t) = e^(-kt)i + e^(kt)k. Show that F and F'' are parallel. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F' = -ke^(-kt)i + ke^(kt)k F'' = k^2*e^(-kt)i + k^2*e^(kt)k Angle between two parallel vectors is 0. theta = acos (F dot F'')/(mag(a)*mag(b)) theta = acos (k^2*e^(-2kt) + k^2*e^(2kt))/(k^2*e^(-2kt) + k^2*e^(2kt)) theta = acos(1) = 0 Therefore, F and F'' are parallel.
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: "