Transcript 3.4 Rates of change

3.4 RATES OF CHANGE
RATES OF CHANGE

Example 1:
 Find
rate ofinchange
the
Area
of a circle
with
If r is the
measured
inches of
and
A is
measured
in square
inches,
would be appropriate for dA/dr?
respectwhat
to itsunits
radius.
A  r 2
dA
 2r
dr
 Evaluate
in2/in
These are units,
so we do not
cancel them!!!
the rate of change of A at r = 5 and r = 10.
dA
 2 (5)  10
dr
dA
 2 (10 )  20
dr
MOTION

Displacement of an object is how far an object
has moved over time.
s  f (t  t )  f (t )

Average velocity is the slope of a displacement
vs. time graph.
vavg
d s1  s2


t t1  t2
MOTION

Instantaneous Velocity is the velocity at a
certain point.
 Instantaneous
Velocity is the first derivative of the
position function.
ds
v
dt
 Speed
is the absolute value of velocity.
MOTION

Acceleration is the rate at which a particle’s
velocity changes.
 Measures
how quickly the body picks up or loses
speed.
 2nd derivative of the position function!!!
2
d s
a 2
dt
EXAMPLE

A particle moves along a line so that its position at any time
t ≥ 0 is given by the function s(t) = 2t2 – 5t + 3 where s is
measured in meters and t is measured in seconds.
a.) Find the displacement of the particle during the first 2 seconds.
s  s(2)  s(0)  1  3  2 meters
b.) Find the average velocity of the particle during the first 6
seconds.
vavg
s (6)  s (0) 45  3 42



 7 meters
second
60
60
6
EXAMPLE

A particle moves along a line so that its position at any time
t ≥ 0 is given by the function s(t) = 2t2 – 5t + 3 where s is
measured in meters and t is measured in seconds.
c.) Find the instantaneous velocity of the particle at 6 seconds.
velocity s' (t )  4t  5 t 6
m
 19
s
d.) Find the acceleration of the particle when t = 6.
acceleration  s' ' (t )  v' (t )
 4 t 6
m
4 2
s
EXAMPLE

A particle moves along a line so that its position at any time
t ≥ 0 is given by the function s(t) = 2t2 – 5t + 3 where s is
measured in meters and t is measured in seconds.
e.) When does the particle change directions?
v(t )  0
4t  5  0
5
t  seconds
4
It is important to understand the relationship between a position graph, velocity
and acceleration:
acc neg
vel pos &
decreasing
acc neg
vel neg &
decreasing
acc zero
vel pos &
constant
distance
velocity
zero
acc pos
vel pos &
increasing
acc zero
vel neg &
constant
acc pos
vel neg &
increasing
acc zero,
velocity zero
time
FREE-FALL
Gravitational
Constants:
Free-fall equation:
ft
g  32
sec 2
1 2
s  gt
2
m
g  9.8
sec 2
cm
g  980
sec 2
s is the position at any time t during the fall
g is the acceleration due to Earth’s gravity
(gravitational constant)
VERTICAL MOTION

Example: A dynamite blast propels a heavy rock straight
up with a launch velocity of 160 ft/sec. It reaches a
height of s = 160t – 16t2 ft after t seconds.
a.) How high does the rock go?
Find when position = 0 and divide by 2 (symmetric path)
s(t )  160t 16t  0
16t (10  t )  0
16t  0
t  0 (when blast occurs)
2
10  t  0
t  10 (when rock hitsground)
Since it takes 10 seconds for the rock to hit the ground, it
takes it 5 seconds to reach it max height.
s(5)  160(t ) 16(5)2  400feet
VERTICAL MOTION

Example: A dynamite blast propels a heavy rock straight
up with a launch velocity of 160 ft/sec. It reaches a
height of s = 160t – 16t2 ft after t seconds.
a.) How high does the rock go?
Find when velocity = 0 (this is when the rock changes direction)
v(t )  160 32t  0
t 5
s(5)  160(5) 16(5)2  400feet
VERTICAL MOTION
Example: A dynamite blast propels a heavy rock straight
up with a launch velocity of 160 ft/sec. It reaches a
height of s = 160t – 16t2 ft after t seconds.
b.) What is the velocity and speed of the rock when it is 256 ft
above the ground on the way up?

At what time is the rock 256 ft above the ground on the way up?
t 20
s(t )  256  160t 16t 2
16t 2  160t  256  0
t2
16(t 2 10t 16)  0
v(t )  160 32t
16(t  2)(t  8)  0
m
v ( 2)  96
s
t 8  0
t 8
m
speed  velo  96
s
VERTICAL MOTION
Example: A dynamite blast propels a heavy rock straight
up with a launch velocity of 160 ft/sec. It reaches a
height of s = 160t – 16t2 ft after t seconds.
b.) What is the velocity and speed of the rock when it is 256 ft
above the ground on the way down?

At what time is the rock 256 ft above the ground on the way down?
t 20
s(t )  256  160t 16t 2
t2
16t 2  160t  256  0
16(t 2 10t 16)  0
16(t  2)(t  8)  0
t 8  0
t 8
v(t )  160 32t
m
v(8)  96
s
m
speed  velo  96
s
VERTICAL MOTION
Example: A dynamite blast propels a heavy rock straight
up with a launch velocity of 160 ft/sec. It reaches a
height of s = 160t – 16t2 ft after t seconds.
c.) What is the acceleration of the rock at any time t at any time
t during its flight?

feet
a (t )  32
sec 2
from Economics:
Marginal cost is the first derivative of the cost function, and
represents an approximation of the cost of producing one more
unit.
Marginal revenue is the first derivative of the revenue function,
and represents an approximation of the revenue of selling one
more unit.