Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 4 - Integration - 4.1 Exercises - Page 253: 61

Answer

(a) $x'(t)$ is velocity and $x''(t)$ is acceleration. $x'(t) = 3t^2 - 12t + 9$ $x''(t) = 6t - 12$ (b) $(0, 1)$ U $(3, 5)$ (c) $-3$

Work Step by Step

(a) $x(t) = t^3 - 6t^2 + 9t - 2$ To find the velocity function, differentiate the position function. $x'(t) = 3t^2 - 12t + 9$ To find the acceleration function, differentiate the velocity function. $x''(t) = 6t - 12$ (b) The particle moves to the right when the velocity is positive. Therefore, we must find the intervals on which $x'(t) > 0$. Start by finding where the velocity is zero. $0 = 3t^2 - 12t + 9$ $0 = 3(t^2 - 4t + 3)$ $0 = 3(t - 3)(t - 1)$ $v(t) = 0$ at $t = 1$ and $t = 3$ Find if the velocity is positive or negative on each interval by substituting a time from each interval. $t < 1$: $x'(0) = 3(0 - 3)(0 - 1) > 0$ $1 < t < 3$: $x'(2) = 3(2 - 3)(2 - 1) < 0$ $t > 3$: $x'(4) = 3(4 - 3)(4 - 1) > 0$ So, velocity is positive when $t < 1$ and when $t > 3$. Since the position function only applies to the time interval $0 < t < 5$, the intervals on which the particle is moving right are: $(0, 1)$ and $(3, 5)$ (c) Find the time at which acceleration is zero by setting $x''(t) = 0$. $0 = 6t - 12$ $6t = 12$ $t = 2$ Now, find the velocity at this time. $x''(2) = 3(2)^2 - 12(2) + 9$ $x''(2) = 3(4) - 24 + 9$ $x''(2) = 12 - 24 + 9$ $x''(2) = -3$
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