Math 20D
January 27
Homework, Section 6.2, pg. 414
Lecture Section 6.3
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 |
|
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) |
f(-2)=6 |
f(-1)=3 |
f(0)=2 |
3 |
6 |
|
area = distance (feet) |
6 |
3 |
2 |
3 |
6 |
Total area is approximation to distance = 6+3+2+3+6=20
|
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) |
f(-1)=3 |
f(0)=2 |
3 |
6 |
11 |
|
area = distance (feet) |
3 |
2 |
3 |
6 |
11 |
Total area is approximation to distance = 3+2+3+6+11 = 25
Maybe, a better approximation would be the average of 20 and 25 = 22.5
Or, maybe another improvement would be to have more rectangles
With 10 rectangles, we get an approximation of 23.125
With 30, 22.106, we get 21.79 feet per second