Upload
others
View
1
Download
0
Embed Size (px)
Citation preview
The Definite Integral
Day 6
Motion Problems
Strategies for Finding Total Area
ARRIVAL---HW Questions Working in PODS
Additional Practice—Packet p. 13 and 14
•Make good use of your time!
•Practice makes perfect!
•Ask ME questions, ask your
CLASSMATES questions!
Area Problems
Review Problem ……
1. Set up an integral that represents the shaded region.
2. Evaluate the integral using the Fundamental Theorem of
Calculus THEN confirm your answer with “ fnInt”
f x( ) = x2
x2
0
3
ò dx = 9
Find the area of the shaded region f x( ) = x2
27 - 9 = 18
Discuss with your partner what we need to do
to find the area of the shaded region.
Important to Remember:
Integrals find area between the curve
and the x-axis.
Remember ….. • When evaluating integrals, “areas” beneath the x-axis are negative.
• When evaluating total area, all areas are positive.
a) Evaluate the integral
a) Find the total area of the graph
12 4
f(x)
12
0
( )f x dx
12 4 12
0 0 4
( ) ( ) ( ) 3 7 4f x dx f x dx f x dx
12 4 12
0 0 4
( ) ( ) ( ) 3 7 10f x dx f x dx f x dx
Practice—Find the area of the shaded region.
NO CALCULATOR
YOU TRY!!
Check your answers…
16
316
19
3
8
3
Today’s Learning Outcomes…
Apply antiderivatives to motion problems
Recognize the relationship between displacement and the total distance traveled by an object
Motion Problems
Remember …….
Given a position function:
Then Velocity: '
And Acceleration: ' ''( )
s t
v t s t
a t v t s t
Given the velocity function,
what could we determine …….
Given velocity v t
Position ( ) + C= s t v t dt
BECAUSE s(t) is the ANTIDERIVATIVE of v(t)
Remember our earlier example in this unit ……
( ) 55 from 2 : 00 to 5:00
Distance traveled = (55)(3) miles
v t mph
55
2 2
OR
55 55 55(5 2) 55(3) milesdt t
Graphically: What does this look like?
Position of object at time t.
Terminology …..
Displacement—how far away from the starting point the
object is at the end of a given time interval
Distance Traveled—amount of movement by the object in
the positive and negative direction
The graph below shows the velocity of a particle over time.
The area between the curve and the x-axis represents
distance traveled.
Area above the x-axis
Distance traveled in the positive direction
Movement away from the starting location
Area below the x-axis
Distance traveled in the negative direction
Movement back towards the starting location
The graph shows the velocity of a particle over time.
a) What is the displacement of the particle from 0 to 20
seconds?
1 1a) 13 6 + 7 -4 = 39 -14 = 25 feet displacement
2 2
The graph shows the velocity of a particle over time.
b) What is the total distance traveled from 0 to 20
seconds?
1 1b) 13 6 + 7 -4 = 39+14
2 2
= 53 feet total distance traveled
Another Example …… The velocity of a particle, in ft/sec, is given by 2 .
Find the displacement and total distance traveled from 2 to 4.
v t t
t
4 4
2 2 2v t dt t
42
2t
2 24 2 16 4 12
Displacement is 12 feet.
Velocity is positive in interval,
therefore total distance also 12 feet.
Before you start: Draw a sketch of the velocity
curve to understand visually what is involved.
The velocity of a particle, in ft/sec, is given by 2 6.
Find the displacement and total distance traveled from 2 to 7.
v t t
t
7 7
2 2 2 6v t dt t
72
26t t
2 27 6 7 2 6 2 7 8 15
Displacement is 15 feet.
But what about total distance?
Before you start: Draw a sketch of the velocity
curve to understand visually what is involved.
The velocity of a particle, in ft/sec, is given by 2 6.
Find the displacement and total distance traveled from 2 to 7.
v t t
t
7
2 v t dt
3 72 2
2 36 6t t t t
2 2 2 23 6 3 2 6 2 7 6 7 3 6 3
1 16 17
Total distance traveled is 17 feet.
3 7
2 3 v t dt v t dt
How confident are you in your
ability to… Apply antiderivatives to motion problems
Recognize the relationship between displacement and the total distance traveled by an object