INTEGRAL CALCULUS: AN INTRODUCTION THE DISTANCE PROBLEM In physics, recall that in a velocity-time graph, the area under the graph represents the distance traveled by an object. Why is this so? Recall that the velocity of a body is expressed as a ratio of distance to velocity:
VELOCITY
DISTANCE TIME
=
(1)
Thus, if the velocity is constant, then the distance traveled is given by:
DISTANCE = VELOCITY ×
(2)
TIME
However, if an object has varying velocities, then equation (2) will no longer be applicable. Instead, we find the distance traveled by such object by computing the distance traveled within each time interval (assuming that the velocity attained is approximately equal in each interval). Then we sum up all the individual distances. Consider this illustration:
t (s)
0
0.5
1.0
1.5
2.0
2.5
3.0
v (ft/s)
0
6.2
10.8
14.9
18.1
19.4
20.2
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 above. Let's assume we want to find the lower and upper estimates for the distance that she traveled during these three seconds. First, we have to ensure that the time and speed have consistent units. In this case, the time is in seconds (s), and the velocity is in feet per second (ft/s). So, we’re okay. Here, the time intervals are equal (0.5s). Thus we say that
∆t = 0.5 and the speeds will be represented by f(tn). The task is to find the distance traveled by the runner in three seconds. On the following page is a graph of the runner’s motion in those three seconds. Therefore, a = t0 = 0 and b = tn = 3. So, we have
t0
=
0.0
t4
=
2.0
t1
=
0.5
t5
=
2.5
t2
=
1.0
t6
=
3.0
t3
=
1.5
y
20
15
10
5
x 1
2
3
4
In the same vein,
f(t0)
=
f(t2)
0
f(t1)
=
6.2
= 10.8
f(t3)
=
14.9
f(t4)
= 18.1
f(t5)
=
19.4
f(t6)
= 20.2
Now, if we assume the initial velocity is 6.2 ft/s, then the distance traveled in the first 0.5 seconds is
f(t1)∆t
=
6.2 × 0.5
= 3.1 feet
Similarly, during the second and third time interval, we also assume that the velocity is approximately constant. Therefore, the distance traveled in the second time interval is
f(t2)∆t =
10.8 × 0.5 = 5.4 feet
and so on. If we then add up the estimates for the other time intervals, the result is an estimate for the total distance traveled:
D1
=
f(t1)∆t + f(t2)∆t + f(t3)∆t + f(t4)∆τ + f(t5)∆t + f(t6)∆t
D1
=
(6.2 × 0.5) + (10.8 × 0.5) + (14.9 × 0.5) + (18.1 × 0.5) + (19.4 × 0.5) + (20.2 × 0.5)
D1
=
3.1 + 5.4 + 7.45 + 9.05 + 9.7 + 10.1
=
44.8 ft.
Therefore, the runner traveled a distance of approximately 44.8 ft. The sum can be expressed in Sigma Notation as
n
D1
=
∑ i=1
f(ti)∆ t
(3)
Thus, this sum represents an upper estimate for the total distance traveled by the runner in three seconds. On the other hand, a lower estimate for the distance traveled is given (in Sigma Notation) by the sum:
n
D2
=
∑ i=1
f(ti-1)∆ t
(4)
Which is equal to
D2
=
f(t0)∆t + f(t1)∆t + f(t2)∆t + f(t3)∆t + f(t4)∆t + f(t5)∆t
D2
=
(0 × 0.5) + (6.2 × 0.5) + (10.8 × 0.5) + (14.9 × 0.5)
D2
=
0 + 3.1 + 5.4 + 7.45 + 9.05 + 9.7
=
34.7 ft.
+ (18.1 × 0.5) + (19.4 × 0.5)
Therefore, the runner traveled a distance of approximately 34.7 ft. At this point, we can say that the actual distance covered by the runner lies between 34.7 feet and 44.8 feet. We would, therefore, be able to obtain a much better estimate if we reduce the time interval to as low as 0.01 seconds (or even less, if possible)!! This procedure is quite similar to that used in estimating areas. In fact, the similarity becomes obvious when we sketch a “velocity-time graph” of the runner’s motion in those three seconds. We then project rectangles whose heights represent the velocities for each time interval:
(The rectangles are used to estimate the distances covered by the runner in the respective time intervals). Note that the curve touches the right hand endpoints of the rectangles. For instance, the area of the first rectangle (R1) in the graph is: 6.2 × 0.5 = 3.1 ft Similarly, the area the second rectangle (R2) in figure 2 is: 10.8 × 0.5 = 7.45 ft So, generally, the sum of areas of the approximating rectangles represent the total distance covered. We can also use left hand endpoints to estimate the total distance traveled:
(The rectangles are used to estimate the distances covered by the runner in the respective time intervals). Note that the curve touches the left hand endpoints of the rectangles. Alternatively, we could use midpoints to find the estimates (see the next page): Note that the curve touches the midpoints of the rectangles. In any case, the more we reduce the time intervals, the better our estimates become. Thus, we can obtain a precise value for the distance traveled by the runner by computing the limits of the sums of the individual distances:
n
D
=
∑ i=1
f(ti)∆ t
and/or
n
D
=
∑ i=1
f(ti-1)∆ t
I hope the illustrations given so far have helped you understand the concept of areas and distances. Notice that in this section (and the previous sections), we've been dealing with many sums. In the following section, we take a close look at these sums.
calculus4engineeringstudents.com