Transcript Calculus 7.1

Greg Kelly, Hanford High School, Richland, Washington
Fiona is very proudly showing her King of the Hill car to the
class. Diabolically, however, Andy has secretly attached a
small motor to the car to make it zig-zag by remote control.
While Fiona futilely chases the car back and forth, the car
goes first north at 10 ft/sec, stops, then south at 10 ft/sec and
continuing this torturous pattern over a period of 30 seconds
Manisha is barely able to stop laughing long enough to
make a graph of the car’s motion…
The velocity graph is shown below.
What isy the total distance traveled by the car?
200 feet
50  50  100  200

ft

sec
10 ft/sec north
50 ft
100ft
v=0



10 ft/sec north
v=0


x

50 ft
10 ft/sec south


seconds

The velocity graph is shown below.
What is the
displacement of the car?
y
50  50  100  100

ft

sec
100 feet north
10 ft/sec north
10 ft/sec north
50 ft
100ft
v=0

v=0


x

–50 ft


10 ft/sec south


seconds

Change in position is measured in both
displacement
and
The difference between
the starting and ending
positions of an object
distance
The total length of the
object’s path
If you were to drive 10 miles east and then 4 miles west
10 miles east
4 miles west
Displacement = 6 miles
Distance = 14 miles


Displacement:
v(t )
1
5
1
2

x

1 
2


2
1 1
1    2  1
2 2

y
0 v(t ) dt
velocity graph


v(t )
Distance Traveled:

5
x





|velocity| graph

0 v(t ) dt
1 1
1   2  4
2 2

Displacement:

1
1
2
x


1 
2


1 1
1    2  1
2 2
2

velocity graph

A displacement of –1 is not the
same as a displacement of 1.
The difference here is…?
2
direction
1
0
1
2
3
4
-1
-2
position graph
5
Distance Traveled:
1 1
1   2  4
2 2
In terms of driving a car on a round trip,
Displacement   V  t  dt
b
a
would be like asking how far it was from where your car
started to where it ended the trip.
Distance Traveled   V  t  dt
b
a
would be like asking how much mileage you put on
your car during the round trip. Your odometer wouldn’t
be accounting for which direction you were going.
This same technique of integrating rates of change to find
quantities is used in many different real-life problems.
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.
National Potato Consumption
C t   2.2 1.1t
How would we find the total
consumption?
total consumption   C  t  dt
How about specifically from the
beginning of 1972 to the end of
1973:
end of ‘73 (beginning of ‘74)
1
t
2.2

1.1
dt

2.2
t

1.1
2
ln1.1
4
4
 7.066
t
beginning of ‘72
2
million
bushels

National Potato Consumption
C t   2.2 1.1t
By the way, what is a great way to make sure that we
are on the right track here?
Units. Why?
We go from millions of
bushels/year to millions of
bushels.
How about specifically 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
National Potato Consumption
C t   2.2 1.1t
By the way, what is a great way to make sure that we
are on the right track here?
Units. Why?
We go from millions of
bushels/year to millions of
bushels.
How about specifically from the
beginning of 1972 to the end of
1973:
There will be other problems like this that involve calculating
quantity of product from some rate of production or
consumption.
And finally, remember…
In terms of driving a car on a round trip,
Displacement   V  t  dt
b
a
would be like asking how far it was from where your car
started to where it ended the trip.
Distance Traveled   V  t  dt
b
a
would be like asking how much mileage you put on
your car during the round trip. Your odometer wouldn’t
be accounting for which direction you were going.