Section 6.3—Estimating Distance Traveled

Problem:  If we know the velocity at which an object has traveled over a time interval, can we find the distance that the object has traveled?

The program to draw the left and right hand rectangles may be obtained from the TI-83 Help Website.

Suppose that an object travels with velocity v(t) = t² + 2 feet per second on the interval [-2, 3] where t is in seconds.  How far does the object travel?

Reminder: distance = rate x time

area rectangle = height x base

If base is time and height is velocity, area is distance

time (seconds)	-2	-1	0	1	2	3
velocity (feet per second)	6	3	2	3	6	11

I have divided the x-axis on the interval [-2, 3] into 5 subintervals of equal length. On each subinterval, I went to the left hand endpoint, up to the function, and used that as the height of a rectangle.

rectangle	1	2	3	4	5
left endpoint	-2	-1	0	1	2
base width (seconds)	1	1	1	1	1
height (feet per second)	6	3	2	3	6
area = distance (feet)	6	3	2	3	6

Total area is approximation to distance  = 6 + 3 + 2 + 3 + 6 = 20 feet

For this graph, on each subinterval, I went to the right hand endpoint and then up to the curve of the function. I used that as the height of the rectangle.

rectangle	1	2	3	4	5
right endpoint	-1	0	1	2	3
base width (seconds)	1	1	1	1	1
height (feet per second)	3	2	3	4	11
area = distance (feet)	3	2	3	4	11

The distance traveled here is approximated by 3 + 2 + 3 + 4 + 11 = 23 feet

So my 2 approximations are 20 feet and 23 feet. So the actual distance is probably between 20 and 23 feet. So how can we improve this? More rectangles!

Let’s try 10 rectangles and use the right-hand endpoint:

rectangle	1	2	3	4	5	6	7	8	9	10
right endpoint	-1.5	-1	-0.5	0	0.5	1	1.5	2	2.5	3
base width (seconds)	0.5	0.5	0.5	0.5	0.5	0.5	0.5	0.5	0.5	0.5
height (feet per second)	5.25	4	3.25	3	3.25	4	5.25	7	9.25	12
area = distance (feet)	2.625	2	1.625	1.5	1.625	2	2.625	3.5	4.625	6

The approximation of distance is 2.625 + 2 + 1.625 + … + 6 = 28.125

The actual distance is 26.67 feet