Transcript Calculus 7.1 - Ms. Waldron

QuickTime™ and a
decompressor
are needed to see this picture.

4
2,3
always positive
2,1,1,2
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.
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.
4
16
S(4) =  2t - 2 dx = 3 + 8 = 11
1
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?
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)
3
= 244.8
1
= .068 mile
3600
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

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
9
4x dx = 2x 2
0
= 162 N m