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.)
velocity is
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.
(a) To calculate the average velocity over the time interval 1 ≤ t ≤ 3, we need to find the displacement of the weight during this period and divide it by the time taken.
The displacement can be found by subtracting the height at t = 1 from the height at t = 3:
h(3) = 0.5 * cos(2*3) = 0.5 * cos(6) ≈ 0.5 * (-0.97) ≈ -0.49 ft
h(1) = 0.5 * cos(2*1) = 0.5 * cos(2) ≈ 0.5 * (-0.41) ≈ -0.21 ft
Displacement = h(3) - h(1) = (-0.49) - (-0.21) = -0.28 ft
The time taken is 3 - 1 = 2 seconds.
Average velocity = Displacement / Time taken
Average velocity = -0.28 ft / 2s
Average velocity = -0.14 ft/s
Therefore, the weight's average velocity over the time interval 1 ≤ t ≤ 3 is -0.14 ft/s.
(b) To calculate the average velocity over the time interval 0 ≤ t ≤ π, we again need to find the displacement of the weight during this period and divide it by the time taken.
The displacement can be found by subtracting the height at t = 0 from the height at t = π.
h(π) = 0.5 * cos(2*π) = 0.5 * cos(2π) = 0.5 * 1 = 0.5 ft
h(0) = 0.5 * cos(2*0) = 0.5 * cos(0) = 0.5 * 1 = 0.5 ft
Displacement = h(π) - h(0) = 0.5 - 0.5 = 0 ft
The time taken is π - 0 = π seconds.
Average velocity = Displacement / Time taken
Average velocity = 0 ft / πs
Average velocity = 0 ft/s
When the weight oscillates back and forth between two extremes, its displacement is zero over this time interval. Therefore, the average velocity is zero, even though the weight is moving.
(c) The instantaneous velocity of the weight at t = 2 sec cannot be found using only arithmetic and algebra because it requires knowledge of calculus. More specifically, it requires finding the derivative of the given height function h(t).
(d) We can 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/s.
The average velocity over any time interval is equal to the instantaneous velocity at some point within that interval. Since the average velocity over the time interval 1 ≤ t ≤ 3 is -0.14 ft/s (calculated in part a), the instantaneous velocity at some point within that interval is also -0.14 ft/s.
Therefore, we can conclude that the instantaneous velocity of the weight at t = 2 sec is equal to -0.14 ft/s, which is less than 0.76 ft/s.
(a) To calculate the weight's average velocity over the time interval 1 ≤ t ≤ 3, we need to find the displacement of the weight during this time period and divide it by the time taken.
The displacement is given by the difference in height between the initial and final times, which is h(3) - h(1).
Substituting the values into the equation for h(t), we get:
h(3) = 0.5 cos(2(3)) = 0.5 cos(6)
h(1) = 0.5 cos(2(1)) = 0.5 cos(2)
Therefore, the displacement is 0.5 cos(6) - 0.5 cos(2).
The time taken is 3 - 1 = 2 seconds.
Average velocity = (displacement) / (time taken)
Average velocity = (0.5 cos(6) - 0.5 cos(2)) / 2
Simplifying the expression gives us the average velocity.
(b) To calculate the weight's average velocity over the time interval 0 ≤ t ≤ π, we again need to find the displacement and divide it by the time taken.
The displacement is given by the difference in height between the initial and final times, which is h(π) - h(0).
Substituting the values into the equation for h(t), we get:
h(π) = 0.5 cos(2(π)) = 0.5 cos(2π) = 0.5 cos(0) = 0.5
h(0) = 0.5 cos(2(0)) = 0.5 cos(0) = 0.5
Therefore, the displacement is 0.5 - 0.5 = 0.
The time taken is π - 0 = π seconds.
Average velocity = (displacement) / (time taken)
Average velocity = 0 / π = 0
Even though the weight is obviously moving throughout the time interval, the average velocity over the entire interval is 0 because the displacement is 0. This means that the weight moves back and forth symmetrically around its mean position.
(c) The instantaneous velocity of the weight at the instant t = 2 sec cannot be found using only arithmetic and algebra because it requires the concept of calculus. Instantaneous velocity is the derivative of the displacement function with respect to time.
In this case, the displacement function is h(t) = 0.5 cos(2t). To find the instantaneous velocity at t = 2 sec, we need to take the derivative of this function with respect to t.
(d) The connection between average velocity and instantaneous velocity is given by the Mean Value Theorem for Derivatives, which states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one point c in (a, b) where the instantaneous velocity equals the average velocity.
In this case, the average velocity over the time interval 1 ≤ t ≤ 3 was calculated to be (0.5 cos(6) - 0.5 cos(2)) / 2.
To determine whether the instantaneous velocity at t = 2 sec is greater or less than 0.76 ft/sec, we compare it to the average velocity. If the instantaneous velocity at t = 2 sec is greater than 0.76 ft/sec, then the instantaneous velocity must have exceeded the average velocity at some point in the interval (1, 3). If it is less than 0.76 ft/sec, then the instantaneous velocity must have been lower than the average velocity at some point.
Therefore, by comparing the value of the average velocity (calculated in part a) with the value of 0.76 ft/sec, we can determine whether the instantaneous velocity at t = 2 sec is greater or less than 0.76 ft/sec.