Transcript Calculus 7.1 - Ms. Waldron

Photo by Vickie Kelly, 2006
Greg Kelly, Hanford High School, Richland, Washington
A honey bee makes several trips from the hive to a flower
garden. The velocity graph is shown below.
What is the total distance traveled by the bee?
200  200  200  100  700
700 feet
100
ft
min 50
0
200ft
200ft
2
4
6
8
10
minutes
-50
200ft
100ft
-100

What is the displacement of the bee?
200  200  200  100  100
100 feet towards the hive
100
ft
min 50
0
200ft
200ft
2
4
6
8
10
minutes
-50
-200ft
-100ft
-100

To find the displacement (position shift) from the velocity
function, we just integrate the function. The negative
areas below the x-axis subtract from the total
displacement.
Displacement   V  t  dt
b
a
To find distance traveled we have to use absolute value.
Distance Traveled   V  t  dt
b
a
Find the roots of the velocity equation and integrate in
pieces, just like when we found the area between a curve
and the x-axis. (Take the absolute value of each integral.)
Or you can use your calculator to integrate the absolute
value of the velocity function.

2
Displacement:
1
1
2
1
0
1
2
1 3
2
4
5
1 1
1    2  1
2 2
2
-1
Distance Traveled:
velocity graph
-2
1 1
1   2  4
2 2
2
1
0
1
2
3
4
-1
-2
position graph
5
Every AP exam I have seen
has had at least one
problem requiring students
to interpret velocity and
position graphs.

Linear Motion
16
V(t) = 2t - 2 , 1  t  4
t
V(t) is the velocity in m/sec of a particle moving along the x-axis
and starting at the position, s(0) = 8.
a) Determine when the particle is moving to the right, to the left,
and stopped.
b) Find the particle’s displacement for the given time interval
and its final position.
c) Find the total distance traveled by the particle.
Linear Motion
16
V(t) = 2t - 2 , 1  t  4
t
V(t) is the velocity in m/sec of a particle moving along the x-axis
and starting at the position, s(0) = 8.
a) Determine when the particle is moving to the right, to the left,
and stopped.
Particle is moving left on 1 < t < 2, stopped at t =
2 and moving right on 2 < t < 4.
b) Find the particle’s displacement for the given time interval
and its final position.
S(4) =

4
1
16
2t - 2 dx = 3 + 8 = 11
t
c) Find the total distance traveled by the particle.
Total distance =

4
1
16
x +
dx = 13
x
2
Effects of Acceleration
A car moving with initial velocity of 5 mph accelerates at
the rate of a(t) = 2.4 t mph per second for 8 seconds.
a) How fast is the car going when the 8 seconds are up?
b) How far did the car travel during those 8 seconds?
Effects of Acceleration
A car moving with initial velocity of 5 mph accelerates at
the rate of a(t) = 2.4 t mph per second for 8 seconds.
a) How fast is the car going when the 8 seconds are up?
Velocity = 5 +

8
0
2.4 t dt = 5 + 1.2 t 2 ]80 = 81.8 mph
b) How far did the car travel during those 8 seconds?

8
0
v(t) dt =
  5 + 1.2t  dt
8
2
0
8
= 
 5t + .4t 
0
= 244.8 mph/(seconds per hour)
1
= 244.8
= .068 mile
3600
3
In the linear motion equation:
dS
 V t 
dt
V(t) is a function of time.
dS  V t  dt
For a very small change in time, V(t) can be
considered a constant.
S  V t  t
We add up all the small changes in S to get
the total distance.
S  V1  t  V2  t  V3  t  
S  V1  V2  V3  t

S  V t  t
We add up all the small changes in S to get
the total distance.
S  V1  t  V2  t  V3  t  
S  V1  V2  V3  t
k
S  Vn  t
n 1

S  Vn  t
As the number of subintervals becomes
infinitely large (and the width becomes
infinitely small), we have integration.
n 1
S   V  t  dt

This same technique is used in many different real-life
problems.

Example 5:
National Potato Consumption
The rate of potato consumption
for a particular country was:
C t   2.2 1.1t
where t is the number of years
since 1970 and C is in millions
of bushels per year.
For a small
t , the rate of consumption is constant.
The amount consumed during that short time is C  t   t .

Example 5:
National Potato Consumption
C t   2.2 1.1t
The amount consumed during that short time is C  t   t .
We add up all these small
amounts to get the total
consumption:
total consumption   C  t  dt
From the beginning of 1972 to
the end of 1973:
1
t
2.2

1.1
dt

2.2
t

1.1
2
ln1.1
4
4
 7.066
t
2
million
bushels

Work:
work  force  distance
Calculating the work is easy
when the force and distance are
constant.
When the amount of force
varies, we get to use calculus!

Hooke’s law for springs:
F  kx
k = spring
constant
x = distance that
the spring is
extended beyond
its natural length

Hooke’s law for springs:
F=10 N
x=2 M
F  kx
Example 7:
It takes 10 Newtons to stretch a
spring 2 meters beyond its natural
length.
10  k  2
5k
F  5 x
How much work is done stretching
the spring to 4 meters beyond its
natural length?

F(x)
How much work is done stretching
the spring to 4 meters beyond its
natural length?
x=4 M
For a very small change in x, the
force is constant.
dw  F  x  dx
dw  5 x dx
 dw   5 x dx
4
W   5x dx
0
F  x   5x
5 2
W  x
2
4
0
W  40
newton-meters
W  40
joules
p
A Bit of Work
It takes a force of 16 N to stretch a spring 4 m beyond
its natural length. How much work is done in
stretching the spring 9 m from its natural length?
A Bit of Work
It takes a force of 16 N to stretch a spring 4 m beyond
its natural length. How much work is done in
stretching the spring 9 m from its natural length?
F  4  = 16
= 4k
so k = 4 N/m and F(x) = 4x for this spring.
Work done =

9
0
4x dx = 2x
2
9
0
= 162 N m