For time, t, in hours, 0\le t\le1, a bug is crawling at a velocity, v, in meters/hour given by
v = \frac{5}{1 + t}.
Use \Delta t = 0.2 to estimate the distance that the bug crawls during this hour. Find an overestimate and an underestimate. Then average the two to get a new estimate.
To estimate the distance that the bug crawls during this hour, we can use the concept of Riemann sums. The bug's velocity is given by the equation v = 5/(1 + t).
First, let's divide the interval [0, 1] into subintervals with a width of Δt = 0.2. We can denote these subintervals as follows: [0, 0.2], [0.2, 0.4], [0.4, 0.6], [0.6, 0.8], [0.8, 1].
Now, we can calculate the velocity of the bug at the endpoints of each subinterval. Let's denote the endpoints of each subinterval as t1, t2, t3, t4, and t5, respectively.
For t1 = 0 to t2 = 0.2:
v1 = 5/(1 + 0) = 5 m/hour (velocity at the starting point)
v2 = 5/(1 + 0.2) = 4.167 m/hour (velocity at the endpoint)
For t2 = 0.2 to t3 = 0.4:
v2 = 5/(1 + 0.2) = 4.167 m/hour (velocity at the starting point)
v3 = 5/(1 + 0.4) = 3.333 m/hour (velocity at the endpoint)
For t3 = 0.4 to t4 = 0.6:
v3 = 5/(1 + 0.4) = 3.333 m/hour (velocity at the starting point)
v4 = 5/(1 + 0.6) = 2.5 m/hour (velocity at the endpoint)
For t4 = 0.6 to t5 = 0.8:
v4 = 5/(1 + 0.6) = 2.5 m/hour (velocity at the starting point)
v5 = 5/(1 + 0.8) = 1.667 m/hour (velocity at the endpoint)
For t5 = 0.8 to 1:
v5 = 5/(1 + 0.8) = 1.667 m/hour (velocity at the starting point)
v6 = 5/(1 + 1) = 1.25 m/hour (velocity at the endpoint)
Note that the bug crawls a shorter distance as time progresses. We have the following distances for each subinterval:
Distance in [0, 0.2] = 0.2 * (v1 + v2) = 0.2 * (5 + 4.167) = 1.0334 meters
Distance in [0.2, 0.4] = 0.2 * (v2 + v3) = 0.2 * (4.167 + 3.333) = 0.800 meters
Distance in [0.4, 0.6] = 0.2 * (v3 + v4) = 0.2 * (3.333 + 2.5) = 0.7666 meters
Distance in [0.6, 0.8] = 0.2 * (v4 + v5) = 0.2 * (2.5 + 1.667) = 0.6334 meters
Distance in [0.8, 1] = 0.2 * (v5 + v6) = 0.2 * (1.667 + 1.25) = 0.3834 meters
Now, to find the overestimate and underestimate, we can sum up the distances in each subinterval:
Overestimate = 1.0334 + 0.800 + 0.7666 + 0.6334 + 0.3834 = 3.6168 meters
Underestimate = 1.0334 + 0.800 + 0.7666 + 0.6334 + 0.3834 = 3.6168 meters
Finally, we can average the overestimate and underestimate to obtain a new estimate:
New Estimate = (Overestimate + Underestimate) / 2 = (3.6168 + 3.6168) / 2 = 3.6168 / 2 = 1.8084 meters
Therefore, the bug is estimated to crawl approximately 1.8084 meters during this hour.