Find an overestimate and underestimate that the bug crawls during an hour v= 1 / [2+t]?
For time, t, in hours, 0 <= t <= 1, a bug is crawling at a velocity, v, in meters/hours given by v = 1 / [2+t].
Use delta t = 0.2 to estimate the distance that the bug crawls during this hour. Find an over and underestimate.
To estimate the distance that the bug crawls during the hour using delta t = 0.2, we can use the midpoint rule to calculate the average velocity over each interval and then sum them up.
First, let's calculate the number of intervals using delta t.
Number of intervals = (1 - 0) / 0.2 = 5
Next, we will calculate the average velocity over each interval.
Interval 1: t = 0.2, v = 1 / (2 + 0.2) = 0.333 m/h
Interval 2: t = 0.4, v = 1 / (2 + 0.4) = 0.25 m/h
Interval 3: t = 0.6, v = 1 / (2 + 0.6) = 0.2 m/h
Interval 4: t = 0.8, v = 1 / (2 + 0.8) = 0.1667 m/h
Interval 5: t = 1, v = 1 / (2 + 1) = 0.143 m/h
Now, let's calculate the distance increment for each interval using the average velocity.
Interval 1: distance increment = 0.333 * 0.2 = 0.0666 m
Interval 2: distance increment = 0.25 * 0.2 = 0.05 m
Interval 3: distance increment = 0.2 * 0.2 = 0.04 m
Interval 4: distance increment = 0.1667 * 0.2 = 0.0333 m
Interval 5: distance increment = 0.143 * 0.2 = 0.0286 m
Now, let's add up all the distance increments to get the estimated distance.
Estimated distance = 0.0666 + 0.05 + 0.04 + 0.0333 + 0.0286 = 0.2185 m
To find the overestimate and underestimate, we need to consider the maximum and minimum values for the velocity in each interval.
Overestimate:
We will use the maximum values for velocity in each interval.
Interval 1: v = 0.333 m/h
Interval 2: v = 0.25 m/h
Interval 3: v = 0.2 m/h
Interval 4: v = 0.1667 m/h
Interval 5: v = 0.143 m/h
Now, let's calculate the distance increment for each interval using the maximum velocity.
Interval 1: distance increment = 0.333 * 0.2 = 0.0666 m
Interval 2: distance increment = 0.25 * 0.2 = 0.05 m
Interval 3: distance increment = 0.2 * 0.2 = 0.04 m
Interval 4: distance increment = 0.1667 * 0.2 = 0.0333 m
Interval 5: distance increment = 0.143 * 0.2 = 0.0286 m
Overestimate = 0.0666 + 0.05 + 0.04 + 0.0333 + 0.0286 = 0.2185 m (same as the estimated distance)
Underestimate:
We will use the minimum values for velocity in each interval.
Interval 1: v = 0.143 m/h
Interval 2: v = 0.1667 m/h
Interval 3: v = 0.2 m/h
Interval 4: v = 0.25 m/h
Interval 5: v = 0.333 m/h
Now, let's calculate the distance increment for each interval using the minimum velocity.
Interval 1: distance increment = 0.143 * 0.2 = 0.0286 m
Interval 2: distance increment = 0.1667 * 0.2 = 0.0333 m
Interval 3: distance increment = 0.2 * 0.2 = 0.04 m
Interval 4: distance increment = 0.25 * 0.2 = 0.05 m
Interval 5: distance increment = 0.333 * 0.2 = 0.0666 m
Underestimate = 0.0286 + 0.0333 + 0.04 + 0.05 + 0.0666 = 0.2185 m (same as the estimated distance)
Therefore, the overestimate and underestimate for the distance that the bug crawls during the hour are both 0.2185 meters.
To estimate the distance that the bug crawls during the hour, we will use the concept of Riemann sums. We can break the interval [0, 1] into smaller subintervals by using a small interval size, Δt.
Given that Δt = 0.2, we will divide the interval [0, 1] into five subintervals: [0, 0.2], [0.2, 0.4], [0.4, 0.6], [0.6, 0.8], and [0.8, 1].
Now, let's calculate the velocity for each subinterval:
For the subinterval [0, 0.2]:
v(0) = 1 / (2 + 0) = 1/2
v(0.2) = 1 / (2 + 0.2) = 1/2.2
Similarly, we can calculate the velocities for the other subintervals:
[0.2, 0.4]: v(0.2) = 1/2.2, v(0.4) = 1/2.4
[0.4, 0.6]: v(0.4) = 1/2.4, v(0.6) = 1/2.6
[0.6, 0.8]: v(0.6) = 1/2.6, v(0.8) = 1/2.8
[0.8, 1]: v(0.8) = 1/2.8, v(1) = 1/3
Next, we can calculate the distance traveled during each subinterval by multiplying the velocity by the interval size: Δs = v * Δt.
For each subinterval, we have:
[0, 0.2]: Δs(0.2) = (1/2) * 0.2 = 0.1
[0.2, 0.4]: Δs(0.4) = (1/2.2) * 0.2 = 0.090909...
[0.4, 0.6]: Δs(0.6) = (1/2.4) * 0.2 = 0.083333...
[0.6, 0.8]: Δs(0.8) = (1/2.6) * 0.2 = 0.076923...
[0.8, 1]: Δs(1) = (1/3) * 0.2 = 0.066666...
Now, we can sum up all the distances to get an estimate of the total distance traveled:
Total distance = Δs(0.2) + Δs(0.4) + Δs(0.6) + Δs(0.8) + Δs(1)
= 0.1 + 0.090909... + 0.083333... + 0.076923... + 0.066666...
≈ 0.101
Therefore, an estimate of the distance the bug crawls during the hour is approximately 0.101 meters.
To find an overestimate and underestimate, we can choose two extreme cases:
1. Overestimate: Consider the largest value of the velocity within the interval [0, 1] times the total time. In this case, the largest value occurs at t = 0, which is v(0) = 1/2. Using this value, the overestimate distance is:
Overestimate distance = v(0) * (1 - 0) = (1/2) * 1 = 0.5 meters.
2. Underestimate: Consider the smallest value of the velocity within the interval [0, 1] times the total time. In this case, the smallest value occurs at t = 1, which is v(1) = 1/3. Using this value, the underestimate distance is:
Underestimate distance = v(1) * (1 - 0) = (1/3) * 1 = 0.333 meters.
Therefore, the overestimate distance is 0.5 meters and the underestimate distance is 0.333 meters.