Average Velocity and Instantaneous Velocity v.03
The graph of distance dHtL as a function of time for uniformly accelerated motion, that is the motion you experience for a moment while being in the car
with the gas pedal pressed down 7/8 of the way, is the familiar parabola. Below is such a graph on the domain [0, 4]. On the horizontal axis we have time t,
in seconds, while the vertical axis represents the distance dHtL, in feet.
You will use the graph below to understand the concept of average velocity around t = 2, and instantaneous velocity at t = 2
Average velocity and instantaneous velocity from the right
H4, 1600L
1500
dHtL = 100 t2
1000
dHfeetL
~ 1200
~~2
~4 ~2
500
H2, 400L
0
0
1
2
3
4
tHsecondsL
Average velocity on the interval @2, 4D =
=
=
=
Total Distance traveled
Time interval
Distance after 4 seconds -Distancde after 2 seconds
4-2
dH4L-dH2L
4-2
1600-400
2
= 600 ft ê sec.
Geometrically, this number represents the slope of the line segment in the graph passing through the points H2, 400L and H4, 1600L. From the graph, this
slope is approximately
1200
2
º 600 ft ê sec
Use the same reasoning to find the average velocity on each of the intervals below. Include the units. Sketch the lines on the graph and estimate their slopes
from the graph. Those values should be approximations of the
Average velocity on [2, 3]
Average velocity on [2, 2.5]
Average velocity on [2, 2.3]
Average velocity on [2, 2.01]
What can you conjecture as to what happens when the right end values of the intervals in consideration are getting "closer" to 2 from the right?
Let's analyze what happens.
Notice that the intervals in consideration are of the form @2, 2 + hD where his getting closer to zero but positive.
Average velocity on the interval @2, 2 + hD =
=
=
dH2+hL-dH2L
H2+hL-2
, h>0
100 H2+hL2 -100 H2L2
h
400 h+100 h2
h
= 400 + 100 h, h > 0
The intervals above are of the form [2, 2 + h] for h = 1, h = 0.5, h = 0.3, h = 0.01. Now the question is, what happens when those values of h get
infinitely close to 0, or when h Ø 0+ ?. From the expression obtained it can be claimed that these values approach 400. This is the instantaneous velocity
at t = 2 using values of t to the right of t=2. This value is , denoted as v+ H2L or the velocity at t=2 from the right.
Average velocity and instantaneous velocity from the left
In a similar way let's calculate the instantaneous velocity using values to the left of 2.
2
Instantaneous Velocity v03.nb
Average velocity and instantaneous velocity from the left
In a similar way let's calculate the instantaneous velocity using values to the left of 2.
H4, 1600L
1500
dHtL = 100 t2
1000
dHfeetL
~ 1200
~2
500
~4 ~2
H2, 400L
~ 400
H0, 0L
0
0
1
2
3
4
tHsecondsL
Average velocity on the interval@0, 2D =
=
=
=
Total Distance traveled
Time interval
Distance after 2 seconds -Distancde after 0 seconds
2-0
dH2L-dH0L
2-0
400-0
2
= 200 ft ê sec.
Geometrically, this number represents the slope of the line segment in the graph passing through the points H0, 0L and H2, 400L. From the graph, this slope
is approximately
400
2
º 200 ft ê sec
Use the same reasoning to find the average velocity on each of the intervals below. Include the units. Sketch the lines on the graph and estimate their slopes
from the graph. Remember that the calculations using the formula for the function, and the estimates using the graph are not necessarily equal.
Average velocity on [1, 2]
Average velocity on [1.5, 2]
Average velocity on [1.9, 2]
Average velocity on [1.95, 2]
What can you conjecture as to what happens when the left end values of the intervals in consideration are getting "closer" to 2 from the left?
What happens now? .
Notice that the intervals in consideration are of the form @2 + h, 2D where his getting closer to zero but positive.
Notice that the intervals we are taking are of the form @2 + h, 2D, where h is negative and getting closer to zero. .
Average velocity on the interval @2 + h, 2D =
=
=
=
dH2L-dH2+hL
2-H2+hL
,h<0
100 H2L2 -100 H2+hL2
h
400-I400+400 h+h2 M
-h
-400 h-h2
-h
= 400 + h, h < 0
The intervals above are of the form [2 + h,2] for h = -1, h = -0.5, h = -0.1, h = -0.05. Now the question is, what happens when those values of h get
infinitely close to 0, or when h Ø 0- ?. From the expression obtained it can be claimed that these values approach 400, but are less than 400. This is the
instantaneous velocity at t = 2 using values of t to the left of t=2. This value is , denoted as v- H2L or the velocity at t=2 from the left.
Since the instantaneous velocity from the left and from the right are the same, it is said that the instantaneous velocity at 2 is 400 ft/sec.
Notation
Since the instantaneous velocity from the left and from the right are the same ( v+ H2L = v- H2L = 400) , it is written as vH2L = 400 ft ê sec which means that
the instantaneous velocity at t=2 is 400 ft/sec regardless of whether t Ø 2+ , or t Ø 2Instantaneous velocity at t=2 is just the instantaneous rate of change of the distance at t=2.
Instantaneous Velocity v03.nb
3
Notation
Since the instantaneous velocity from the left and from the right are the same ( v+ H2L = v- H2L = 400) , it is written as vH2L = 400 ft ê sec which means that
the instantaneous velocity at t=2 is 400 ft/sec regardless of whether t Ø 2+ , or t Ø 2Instantaneous velocity at t=2 is just the instantaneous rate of change of the distance at t=2.
Geometrically , the instantaneous velocity at t=2 is the slope of the tangent line to the graph at t=2.
1500
H4, 1600L
dHtL = 100 t2
dHfeetL
1000
500
H2, 400L
0
0
1
2
3
4
tHsecondsL
This animation summarizes this behavior
EXERCISE
Estimate the slope of the tangent line in the graph an compare it to the value of the instantaneous velocity , or simply velocity, at t = 2 found above.
EQUATION OF THE TANGENT LINE
Since the slope of the tangent line is known as well as a point on it, the point of tangency,
the equation of the line can be determine using the slope point form of a line.
In this case the slope at the point where t = 2 is the same as the velocity at t = 2. The point on the line is the point of tangency with the curve,
which is H2, 400L.
Thus, the equation of the tangent line is
y - 400 = 400 Ht - 2L point H2, 400L, slope 400
y = 400 Ht - 2L - 400
yTangent = 400 t - 400
COMPARING THE SHAPE OF THE GRAPH TO THE TANGENT LINE
What does the graph of the distance function look like nearby the point (2,400) compared to the tangent line at t=2? The graphs below show the graphs of
dHtL = 100 t2 and yTangent = 400 t + 400 in windows where the domain are intervals "shrinking" around t=2.
1000
1500
800
dHfeetL
dHfeetL
1000
500
600
400
200
0
0
1
2
3
0
4
tHsecondsL
400.4
400.2
dHfeetL
dHfeetL
450
350
1.5
2.0
tHsecondsL
500
400
1.0
400.0
399.8
399.6
2.5
3.0
4
Instantaneous Velocity v03.nb
1000
1500
800
dHfeetL
dHfeetL
1000
500
600
400
200
0
0
1
2
3
0
4
1.0
1.5
tHsecondsL
500
2.5
3.0
2.0005
2.0010
400.4
450
400.2
dHfeetL
dHfeetL
2.0
tHsecondsL
400
400.0
399.8
350
399.6
300
1.8
1.9
2.0
2.1
2.2
1.9990
1.9995
tHsecondsL
2.0000
tHsecondsL
EXERCISE 1
a. In your graphing calculator reproduce the graphs above. Make sure you in each case the window is the same as the one above.
b. In a paragraph, and in your own language, explain your observations about what happens to the graphs of the distance function and its tangent line at
t = 2, when the interval in the domain containing 2 "shrinks".
c. Now consider the secant line passing through the points H2, dH2LL and H2.5, dH2.5LL. You are going to compare the graph of d = dHtL at t = 2, and the
graph of the secant line you found. Can you draw the same observations as in part (b)? Explain.
EXERCISE 2
Use the graph of the function d = dHtL, same as above, to answer the question below.
1400
1200
dHfeetL
1000
800
600
400
200
0
0
1
2
3
4
tHsecondsL
a) Sketch the tangent lines to the graph of the distance function at the times t = 1, t = 3.
b) Find the slope of each of these tangent lines.
c) What are the units of the slopes of the tangent lines?
d) Can you determine from the graph, not from any calculation, at which of the two given times the instantaneous velocity is greater? Explain.
Instantaneous Velocity v03.nb
5
b) Find the slope of each of these tangent lines.
c) What are the units of the slopes of the tangent lines?
d) Can you determine from the graph, not from any calculation, at which of the two given times the instantaneous velocity is greater? Explain.
c) Find the equation of each of the tangent lines from (a)
EXERCISE 3
An object is moving along a straight line. An observer at point P calculates the distance to the object, after t seconds, using the
formula
dHtL =
3
t+1
+ 2, t ¥ 0
The graph of d = dHtL is given below
dHtL distance HftL
2.0
1.5
1.0
0.5
0
2
4
6
8
10
tHsecondsL
a. Is the object approaching or moving away from point P? How do you determine it from the graph?
c. What is the closest that the object gets to the observer? How about the farthest the object is from the observer?
d. Use the graph to estimate the average velocity on the intervals @2, 5D. Sketch the line that will provide this information.
e. Find the average velocity of the object on the interval @2, 5D using the formula for d = dHtL.
f. Sketch the graph of the tangent at t=2. Use it to estimate the velocity of the object at t=2.
g. Use the calculator to draw the graph of the function and the graph of the tangent line from (c). Zoom in at the point of tangency. Do the graph and the tangent line look like you would expect?
e. From (f) you know how fast the object is moving after 2 seconds. Use that information to predict at which time the object will
hit the observer.
d. Use the graph to estimate the average velocity on the intervals @2, 5D. Sketch the line that will provide this information.
6
Instantaneous Velocity v03.nb
e. Find the average velocity of the object on the interval @2, 5D using the formula for d = dHtL.
f. Sketch the graph of the tangent at t=2. Use it to estimate the velocity of the object at t=2.
g. Use the calculator to draw the graph of the function and the graph of the tangent line from (c). Zoom in at the point of tangency. Do the graph and the tangent line look like you would expect?
e. From (f) you know how fast the object is moving after 2 seconds. Use that information to predict at which time the object will
hit the observer.