Download 20 40 60 80 t 50 100 150 200

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!!