Average Velocity and Instantaneous Velocity Trip from CC-­‐San Antonio 140
120
dHtL-miles-
100
80
60
40
20
0
0
1
2
3
4
5
t-hoursIn a six-­‐hour trip you traveled 300 miles. What was the average velocity for the whole trip? What was the average velocity between the second and fourth hour? The average velocity is the average rate of change of the distance on a given Cme interval. This idea will be used to discussed the concept of instantaneous velocity at any point. 3 4 Average Velocity From the Right Distance traveled by a car with the gas pedal pressed down 7/8 of the way How does the police determine your velocity aFer 3 seconds? 5000
dHtL = 100 t2
dHfeetL
4000
I6,
3000
2000
3600L
H3, 900L
1000
0
0
1
2
3
4
5
6
7
tHsecondsL
Geometrically, this number represents the slope of the line segment passing through the points (3,900) and (6,3600). 2700 ft
= 900 ft ê sec
Slope of the secant line is 3 sec
5 EXERCISE 1 Complete the table below. First, use the graph of the distance funcCon to esCmate the average velocity on each of the intervals. Second, calculate those averages using the formula that defines the funcCon. 5000
dHtL = 100 t2
dHfeetL
4000
I6,
3000
2000
3600L
H3, 900L
1000
0
0
1
2
3
4
5
6
7
tHsecondsL
Use the calculator to find the average velocity on the intervals [3, 3.1], [3,3.01], [3,3.001]. EsCmate d(t) − d(3)
Lim+ t→3
t −3
6 EXERCISE 2 Complete the table below. First, use the graph of the distance funcCon above to esCmate the average velocity on each of the intervals. Second, calculate those averages using the formula that defines the funcCon. 5000
dHtL = 100 t2
Average Velocity @0.6, 3D @1, 3D @2, 3D @2.5, 3D
Estimated HGraphL
Exact HFormulaL
4000
dHfeetL
3000
2000
H3, 900L
H0, 0L
1000
0
0
1
2
3
4
5
6
7
tHsecondsL
Use the calculator to find the average velocity on the intervals [2.9, 3], [2.95, 3], [2.98, 3]. EsCmate d(t) − d(3)
Lim− t→3
t −3
7 Right Instantaneous Velocity Instantaneous velocity at t=3 from the right, v+ (3) d(3) − d(t)
v+ (3) = Lim 3− t
t→3+
d(t) − d(3)
v+ (3) = Lim+ t→3
t −3
Any point to the right of 3 can be wri^en as 3+ h, h > 0
d(3+ h) − d(3)
d(3+ h) − d(3)
v+ (3) = Lim = Lim h→0
(3+
h)
−
3
h
h→0
+
+
d(3) − d(3+ h)
v+ (3) = Lim+ h→0
−h
8 LeF Instantaneous Velocity Instantaneous velocity at t=3 from the leF v− (3) d(3) − d(t)
v− (3) = Lim 3− t
t→3−
d(t) − d(3)
v− (3) = Lim− t→3
t −3
Any point to the leF of 3 can be wri^en as 3+ h, h < 0
d(3+ h) − d(3)
d(3+ h) − d(3)
v− (3) = Lim = Lim− h→0
(3+ h) − 3
h
h→0−
d(3) − d(3+ h)
v− (3) = Lim− h→0
−h
9 Instantaneous Velocity at t=3 Since v + (3)
= v − (3)
= 600 it is said that the instantaneous velocity at t=3 is 600. d(t) − d(3)
v(3) = Lim t→2
t −3
d(3+ h) − d(3)
v(3) = Lim h→0
h
Instantaneous velocity (or simply velocity) at t=3 is the instantaneous rate of change of the distance funcCon at t=3. 10 Instantaneous Velocity at t=3 5000
dHtL = 100 t2
4000
dHfeetL
3000
2000
H3, 900L
1000
0
0
1
2
3
4
5
6
7
tHsecondsL
d(t) − d(3)
v(3) = Lim t→3
t −3
d(3+ h) − d(3)
= Lim h→0
h
11 Geometric InterpretaCon of the Instantaneous Velocity at t=3 5000
dHtL = 100 t2
4000
dHfeetL
3000
2000
H3, 900L
1000
0
0
1
2
3
4
5
6
7
tHsecondsL
The instantaneous velocity at t=3 is the slope of the tangent line to the funcCon d=d(t) at t=3. 12 EquaCon Of The Tangent Line at t = 3 Using The Instantaneous Velocity 5000
dHtL = 100 t2
4000
Point: (3,900) Slope: v(3)=600 dHfeetL
3000
2000
H3, 900L
1000
EquaCon of tangent line at the point (3,900), or when t=3 is 0
0
1
2
3
4
5
6
7
tHsecondsL
ytangent = 600(t − 3) + 900
= 600t − 900
13 COMPARING THE GRAPH OF THE FUNCTION AND THE TANGENT LINE NEARBY THE TANGENCY POINT 14 The graphs of d(t)
= 100t
2 and y tangent
= 600t
−
900
are displayed on windows where the domain are intervals "shrinking" around t=3. 2500
1600
2000
1400
1200
1500
1000
1000
800
500
600
0
400
1
2
3
4
5
Point of Tangency H3,900L
2.0
1200
960
1100
940
1000
920
900
900
800
880
700
860
600
2.6
2.8
3.0
3.2
Point of Tangency H3,900L
3.4
840
2.90
2.5
3.0
3.5
2.95
3.00
3.05
Point of Tangency H3,900L
Point of Tangency H3,900L
4.0
3.10
a. In your graphing calculator reproduce the graphs above. Make sure you in each case the windows have the same dimensions. 15 1500
1000
800
dHfeetL
dHfeetL
1000
500
600
400
200
0
0
1
2
3
0
4
1.0
1.5
2.5
3.0
2.0005
2.0010
tHsecondsL
tHsecondsL
500
400.4
450
400.2
dHfeetL
dHfeetL
2.0
400
400.0
399.8
350
399.6
300
1.8
1.9
2.0
tHsecondsL
2.1
2.2
1.9990
1.9995
2.0000
tHsecondsL
In a paragraph, and in your own language, explain what happens to the graphs of the distance funcCon and its tangent line at t=3, when the interval in the domain containing 3 "shrinks". 16 Key RelaConship Between the Distance FuncCon and The Tangent Line 900.6
900.4
Nearby the point (3,900) the graph of d=d(t) and its tangent line “look alike” 900.2
900.0
899.8
899.6
899.4
2.9990
2.9995
3.0000
Point of Tangency H3,900L
3.0005
3.0010
Nearby t=3, the values of d(t) and
y tangent
(t)
are about the same. It is, for values “close to t=3” d(t) ≅ ytangent (t)
d(t)
Lim
= "1"
t→3 y
tangent (t)
17