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Hw13-1

# Hw13-1 - Math 1920 Solutions for 13.1 4 We know x = cos 2t...

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Math 1920 Solutions for 13.1 4. We know x = cos 2 t y = 3 sin 2 t By trigonometry, we obtain: x 2 + 1 9 y 2 = 1 which is the equation of an ellipse centered at zero. By differentiating each component, we obtain the velocity and acceleration vectors: ~v = d~ r dt = ( - 2 sin 2 t ) ~ i + (6 cos 2 t ) ~ j ~a = d~v dt = ( - 4 cos 2 t ) ~ i + ( - 12 sin 2 t ) ~ j Thus at t = 0, we get ~v t =0 = 6 ~ j ~a t =0 = - 4 ~ i 8. By differentiating each component, we obtain the following velocity and acceleration vectors: ~v = d~ r dt = ~ i + 2 t ~ j ~a = d~v dt = 2 ~ j Hence at t = - 1 , 0 and 1, we have: 1

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~v t = - 1 = ~ i - 2 ~ j ~v t =0 = ~ i ~v t =1 = ~ i + 2 ~ j ~a t = - 1 = 2 ~ j ~a t =0 = 2 ~ j ~a t =1 = 2 ~ j 17. By differentiating each component, we obtain the following velocity and acceleration vectors: ~v = d~ r dt = 2 t t 2 + 1 ~ i + 1 t 2 + 1 ~ j + t t 2 + 1 ~ k ~a = d~v dt = - 2 t 2 + 2 ( t 2 + 1) 2 ~ i - 2 t ( t 2 + 1) 2 ~ j + 1 ( t 2 + 1) 3 / 2 ~ k At t = 0, the vectors and norms are: ~v t =0 = ~ j | ~v t =0 | = 1 ~a t =0 = 2 ~ i + ~ k | ~a t =0 | = 2 2 + 1 2 = 5 The angle between the two is therefore equal to: θ = cos - 1 ( ~v t =0 · ~a t =0 | ~v t =0 || ~a t =0 | ) = cos - 1 ( 0 1 · 5 ) = π 2 or 90 degrees. (They are orthogonal!) 2
22.

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Hw13-1 - Math 1920 Solutions for 13.1 4 We know x = cos 2t...

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