Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Tutorial: Cumulative change, rates of change, and Riemann sums, through the study of deaths due to Ebola Math 1310: CSM SOLUTIONS IN RED Goal: To study (again) a model for deaths due to an Ebola outbreak, and to use this model to understand aspects of derivatives, antiderivatives, and Riemann sums A certain 90-day-long outbreak of Ebola in the Democratic Republic of Congo (DRC), in 1995, may be modeled reasonably well by the equation 1654 D(t) = 21 ✓ ✓ 2(t arctan 45) 21 ◆ + arctan ✓ 30 7 ◆◆ , where t is elapsed time, in days, since the start of the outbreak, and D(t) is total number of deaths since the start of the outbreak, at time t. A sketch of D(t) appears on the axes below. 200 150 100 50 20 40 60 80 t 1. The tth entry in the list below gives the actual number N (t) of cumulative deaths from this outbreak, at the end of day t. Carefully plot t versus N (t) on the above axes. See red dots above. 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 6, 9, 11, 13, 13, 13, 13, 13, 13, 15, 16, 17, 19, 23, 24, 26, 28, 33, 33, 33, 33, 34, 36, 38, 43, 45, 52, 61, 74, 88, 99, 104, 112, 119, 120, 121, 124, 130, 135, 136, 146, 153, 159, 165, 170, 176, 179, 182, 183, 185, 187, 191, 192, 193, 193, 194, 195, 196, 200, 204, 207, 208, 208, 210, 210, 210, 211, 211, 211, 211, 211, 214, 214, 214, 214, 214, 214, 214 2. How well does the mathematical model D(t) for the number of deaths represent the actual cumulative death count N (t)? To the extent that D(t) and N (t) are di↵erent, what are the major di↵erences? Quite well, though the actual count seems to sag a bit below the model in a couple of places, for t in the 30s and 50s. 3. Find a formula for the instantaneous death rate R(t), in deaths per day, assuming that the actual death count is given by the above function D(t). It may help to recall that d 1 arctan t = . dt 1 + t2 ✓ ✓ ◆ ✓ ◆◆ d 1654 2(t 45) 30 0 R(t) = D (t) = arctan + arctan dt 21 21 7 ✓ ◆ 1654 d 2(t 45) = arctan 21 dt 21 ✓ ◆ 1654 1 d 2(t 45) = · ✓ ◆2 · 21 dt 21 2(t 45) 1+ 21 1654 2 1 = · · ✓ ◆2 21 21 2(t 45) 1+ 21 3308 = 2 . 21 + 4(t 45)2 The function R(t) that you found in problem 3 is sketched on the axes below. (Note the bell shape!!) R(t) 10 8 6 4 2 0 20 40 60 80 Note that we’ve dashed in a rectangle over each of the first three intervals of length 5, on the above t axis. The height of each rectangle is just the value of the function R(t) at the right endpoint of the interval in question. 4. Continue the process of drawing rectangles over each of the above subintervals of length 5, on the t axis, with the height of each rectangle being the value of R(t) at the rightmost edge of the rectangle. See above. 5. Let T (t0 ) denote the total area of the rectangles you’ve sketched in, between t = 0 and t = t0 . Compute T (20), T (30), T (40), T (60), T (80), and T (90). (Recall that each rectangle has base length 5, and height given by the value of R(t) at the right endpoint of the rectangle.) (The total areas you are considering here are called Riemann sums; more on these later in class.) T (20) = 5R(5) + 5R(10) + 5R(15) + 5R(20) = 15.2316 T (30) = 5R(5) + 5R(10) + 5R(15) + 5R(20) + · · · + 5R(30) = 35.6695 T (40) = 5R(5) + 5R(10) + 5R(15) + 5R(20) + · · · + 5R(40) = 85.9096 T (60) = 5R(5) + 5R(10) + 5R(15) + 5R(20) + · · · + 5R(60) = 185.989 T (80) = 5R(5) + 5R(10) + 5R(15) + 5R(20) + · · · + 5R(80) = 206.907 T (90) = 5R(5) + 5R(10) + 5R(15) + 5R(20) + · · · + 5R(90) = 211.261 t 6. Compare the above numbers T (20), T (30), T (40), T (60), T (80), and T (90) to the numbers D(20), D(30), D(40), D(60), D(80), and D(90) you get by plugging in the appropriate t-values into the formula for D(t) above. Do you see a correspondence between these sequences of numbers? Do you have any idea why this correspondence should be true? (This correspondence amounts to a HUGE theorem, called the Fundamental Theorem of Calculus, which we’ll discuss in class shortly.) We compute that D(20) = 13.2632, D(30) = 30.0472, D(40) = 70.6609, D(60) = 181.281, D(80) = 206.427, D(90) = 211.328 Note that these values of D are quite close (especially for t values of T given on the previous page. 60) to the corresponding The BIG IDEA behind what we’re seeing here (and behind the Fundamental Theorem of Calculus) is this. Suppose a quantity D(t) has a constant rate of change R over some interval. Then total change in D(t) over the interval = rate of change R over the interval ⇥ length of interval (for example: if your speed is a constant 60 mph for an interval of length 1 sec = 1/3600 of an hour, then your total distance traveled over that second is 60 ⇥ 1/3600 = 1/60 of a mile (=88 feet, since 5280/60=88). So in this situation, the total change in D is the area under the graph of R, on the interval in question. Now if the rate of change of D(t) is not constant over an interval, we can break the interval up into small subintervals, over which R(t) should be relatively constant, since nice functions don’t change all that much over small intervals. The total change in D(t) over the BIG interval will be the sum of the changes over the smaller subintervals, and since the change over each subinterval is (as argued above) roughly equal to the area under the graph of the derivative on that subinterval, we conclude: the total change in D(t) over an interval is roughly equal to the sum of the areas of the rectangles under the graph of R(t) over the smaller subintervals. The Fundamental Theorem of Calculus tells us that the error in approximating this total change by the areas of these rectangles gets smaller and smaller as we use more and more, thinner and thinner, rectangles!!