posted by Dede on .
The height (in feet) of a small weight oscillating at the end of a spring is h(t) = 0.5 cos(2t) where t is in seconds, 0 ≤ t ≤ 5. (This is calculus. Use radians!)
(a)Calculate the weight’s average velocity over the time interval 1 ≤ t ≤ 3. Show clearly how you calculated the average velocity.
(b)Calculate the weight’s average velocity over the time interval 0 ≤ t ≤ π. Show clearly how you calculated the average velocity. (Exact values required!) Explain how this result can be right when the weight is obviously moving all the time!
(c)Explain why the instantaneous velocity of the weight at the instant t = 2 sec cannot be found using only arithmetic and algebra.
(d)Use the connection between average velocity and instantaneous velocity to determine whether the instantaneous velocity of the weight at time t = 2 sec is greater than or less than 0.76 ft/sec. Explain clearly how you decided. (Don’t evaluate a limit or differentiate the cosine to answer this question.)
v = dh/dt = -sin(2t)
average v over an interval is
∫[1,3] dv/dt / (3-1)
= 1/2 ∫[1,3] -sin(2t) dt
= 1/2 (1/2 cos 2t) [1,3]
= 1/4 (cos6-cos1)
= 1/4 (.4198) = 0.105
Note that this is just the total displacement / total time
h(pi)-h(0) = .5(1) - .5(1) = 0
after pi seconds, the weight is back where it started. Hence, average velocity = 0, since it has not moved (in effect).
instantaneous velocity is defined as a limit, using cos(2). Trig functions of most integers cannot be evaluated without more advanced techniques.