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The speed of a rur intervals is given in the table. Find lower and upper estimates for the distance that she traveled during these three seconds. incr steadily during thế first thré se0 race. Her speed at econd t (s) 0 0.5 1.0 2.5 3.0 1.5 2.0 v (ft/s) o 6.2 | 10.8 14.9 18.1 19.4 20.2 Step 1 We will first sketch an approximate graph of the velocity function of the runner using the given table. We then shade the region whose area represents the total distance the runner travels in the first three seconds. Doing so gives the following result. t (s) 0 0.5 1.0 v (ft/s) 0 1.5 2.0 2.5 3.0 6.2 10.8 14.9 18.1 19.4 20.2 25 20 15 10 0.5 1.0 1.5 2.0 2.5 3.0 We will approximate the area of the shaded region using the method of approximating rectangles. Recall that this can be done using either the left endpoints or the right endpoints of the subintervals. One will give an estimate that is smaller than the actual value, and the other will give an estimate that is larger than the actual value. In the given table we note that there are six clear subintervals: [0, 0.5), [o.5, 1.0]. [1.0, 1.5], [1.5, 2.0], [2.0, 2.5), and [2.5, 3.0]. The width of each of these subintervals is 0.5 0.5 . Therefore, the

The speed of a rur intervals is given in the table. Find lower and upper estimates for the distance that she traveled during these three seconds. incr steadily during thế first thré se0 race. Her speed at econd t (s) 0 0.5 1.0 2.5 3.0 1.5 2.0 v (ft/s) o 6.2 | 10.8 14.9 18.1 19.4 20.2 Step 1 We will first sketch an approximate graph of the velocity function of the runner using the given table. We then shade the region whose area represents the total distance the runner travels in the first three seconds. Doing so gives the following result. t (s) 0 0.5 1.0 v (ft/s) 0 1.5 2.0 2.5 3.0 6.2 10.8 14.9 18.1 19.4 20.2 25 20 15 10 0.5 1.0 1.5 2.0 2.5 3.0 We will approximate the area of the shaded region using the method of approximating rectangles. Recall that this can be done using either the left endpoints or the right endpoints of the subintervals. One will give an estimate that is smaller than the actual value, and the other will give an estimate that is larger than the actual value. In the given table we note that there are six clear subintervals: [0, 0.5), [o.5, 1.0]. [1.0, 1.5], [1.5, 2.0], [2.0, 2.5), and [2.5, 3.0]. The width of each of these subintervals is 0.5 0.5 . Therefore, the

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Step 3 We will now approximate the area using right endpoints. Recall that, using right endpoints, the area A of the region that lies under the graph of the continuous function v is the limit of the sum of the areas of the n approximating rectangles as follows. A = lim R, = lim v(t, JAt + v(t,)at +. n Here we wish to find R, where At = 0.5. Therefore, we have the following. R = v(t, )(0.5) + v(t,)X(0.5) + v(t,)(0.5) + v(e,)(0.5) + v(e,)(0.5) + vle,M0.5) We refer again to the given table. t (s) 0 0.5 v (ft/s) 0 6.2 10.8 1.0 1.5 2.0 2.5 3.0 V 14.9 18.1 19.4 20.2 Substituting the appropriate values of v from the table gives us the following result. = 44.8 44.8 Step 4 We have determined the estimations L = 34.7 and R, = 44.8, where both are measured in feet. We note that L, provides the -Select--- v estimate and R, provides the -Select--- v estimate. To conclude, state the lower and upper estimates for the distance (in ft) that the runner traveled. lower ft upper ft
Transcribed Image Text:Step 3 We will now approximate the area using right endpoints. Recall that, using right endpoints, the area A of the region that lies under the graph of the continuous function v is the limit of the sum of the areas of the n approximating rectangles as follows. A = lim R, = lim v(t, JAt + v(t,)at +. n Here we wish to find R, where At = 0.5. Therefore, we have the following. R = v(t, )(0.5) + v(t,)X(0.5) + v(t,)(0.5) + v(e,)(0.5) + v(e,)(0.5) + vle,M0.5) We refer again to the given table. t (s) 0 0.5 v (ft/s) 0 6.2 10.8 1.0 1.5 2.0 2.5 3.0 V 14.9 18.1 19.4 20.2 Substituting the appropriate values of v from the table gives us the following result. = 44.8 44.8 Step 4 We have determined the estimations L = 34.7 and R, = 44.8, where both are measured in feet. We note that L, provides the -Select--- v estimate and R, provides the -Select--- v estimate. To conclude, state the lower and upper estimates for the distance (in ft) that the runner traveled. lower ft upper ft
The speed of a runner increased steadily during the first three seconds of a race. Her speed at half-second intervals is given in the table. Find lower and upper estimates for the distance that she traveled during these three seconds. t (s) 0 0.5 v (ft/s) o 1.0 1.5 2.0 2.5 3.0 6.2 10.8 14.9 18.1 19.4 20.2 Step 1 We will first sketch an approximate graph of the velocity function of the runner using the given table. We then shade the region whose area represents the total distance the runner travels in the first three seconds. Doing so gives the following result. t (s) 0 0.5 v (ft/s) o 1.0 1.5 2.0 2.5 3.0 V 6.2 10.8 14.9 18.1 19.4 20.2 25 20 15 10 0.5 1.0 1.5 2.0 2.5 3.0 We will approximate the area of the shaded region using the method of approximating rectangles. Recall that this can be done using either the left endpoints or the right endpoints of the subintervals. One will give an estimate that is smaller than the actual value, and the other will give an estimate that is larger than the actual value. In the given table we note that there are six clear subintervals: [0, 0.5], [0.5, 1.0], [1.0, 1.5], [1.5, 2.0], [2.0, 2.5], and [2.5, 3.0]. The width of each of these subintervals is 0.5 0.5. Therefore, the change in t over each subinterval, At, can be stated as follows. At = 0.5 Step 2 Recall that, using left endpoints, the area A of the region that lies under the graph of the continuous function v is the limit of the sum of the areas of the n approximating rectangles as follows. A = lim L, = lim v(e,)At + v{e, )at + ... + vlt, - 1)a: Here we wish to find L, where At = 0.5. Therefore, we have the following. 4 = v(t,X0.5) + v(t,)(0.5) + v(t,)(0.5) + v(t,)(0.5) + v(t,(0.5) + v(t,X0.5) We refer again to the given table. t (s) 0 0.5 1.0 1.5 2.0 2.5 3.0 v (ft/s) o 6.2 10.8 14.9 18.1 19.4 20.2 Substituting the appropriate values of v from the table gives us the following result. = 34.7 34.7
Transcribed Image Text:The speed of a runner increased steadily during the first three seconds of a race. Her speed at half-second intervals is given in the table. Find lower and upper estimates for the distance that she traveled during these three seconds. t (s) 0 0.5 v (ft/s) o 1.0 1.5 2.0 2.5 3.0 6.2 10.8 14.9 18.1 19.4 20.2 Step 1 We will first sketch an approximate graph of the velocity function of the runner using the given table. We then shade the region whose area represents the total distance the runner travels in the first three seconds. Doing so gives the following result. t (s) 0 0.5 v (ft/s) o 1.0 1.5 2.0 2.5 3.0 V 6.2 10.8 14.9 18.1 19.4 20.2 25 20 15 10 0.5 1.0 1.5 2.0 2.5 3.0 We will approximate the area of the shaded region using the method of approximating rectangles. Recall that this can be done using either the left endpoints or the right endpoints of the subintervals. One will give an estimate that is smaller than the actual value, and the other will give an estimate that is larger than the actual value. In the given table we note that there are six clear subintervals: [0, 0.5], [0.5, 1.0], [1.0, 1.5], [1.5, 2.0], [2.0, 2.5], and [2.5, 3.0]. The width of each of these subintervals is 0.5 0.5. Therefore, the change in t over each subinterval, At, can be stated as follows. At = 0.5 Step 2 Recall that, using left endpoints, the area A of the region that lies under the graph of the continuous function v is the limit of the sum of the areas of the n approximating rectangles as follows. A = lim L, = lim v(e,)At + v{e, )at + ... + vlt, - 1)a: Here we wish to find L, where At = 0.5. Therefore, we have the following. 4 = v(t,X0.5) + v(t,)(0.5) + v(t,)(0.5) + v(t,)(0.5) + v(t,(0.5) + v(t,X0.5) We refer again to the given table. t (s) 0 0.5 1.0 1.5 2.0 2.5 3.0 v (ft/s) o 6.2 10.8 14.9 18.1 19.4 20.2 Substituting the appropriate values of v from the table gives us the following result. = 34.7 34.7
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