quiz8Asol - Solutions to Quiz 8A...

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Unformatted text preview: Solutions to Quiz 8A www.math.ufl.edu/˜harringt March 20, 2008 1. A particle moves according to a law of motion s = f (t) = 2t3 − 15t2 + 36t, for t ≥ 0, where t is measured in seconds and s is in feet. (a) Find the velocity at time t. v (t) = ds/st = f (t) = 6t2 − 30t + 36 (b) When is the particle at rest? When v (t) = 0. So, 0 = 6t2 − 30t + 36 = 6(t2 − 5t + 6) = 6(t − 2)(t − 3) Hence, the particle is a rest at 2 seconds and 3 seconds. (c) When is the particle moving in the positive direction? When v (t) > 0. Recall that t ≥ 0. Here I will use a sign chart on the factors found in part (b). 6 t-2 t-3 v(t) + + ++++ 0+++ --0+ 0-0+ 2 3 So the particle is increasing on [0, 2) ∪ (3, ∞). 2. Find the differential dy and evaluate dy for the given values of x and dx 1 (a) y = ex/2 , x = 0, dx = 1 10 1 dy = ex/2 dx 2 After substition we have: 1 11 dy = e0 = 2 10 20 (b) y = sin(x), x = π , dx = 3 1 100 dy = cos(x)dx After substition we have: dy = cos π 1 1 1 1 ∗ =∗ = 3 100 2 100 200 2 ...
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quiz8Asol - Solutions to Quiz 8A...

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