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3.4 Velocity, Speed, and Rates of Change
downward-256
2, 8
−∞,144( ⎤⎦
5,144( )
X=3, 7
x=158−∞,5⎛
⎝⎜
⎞
⎠⎟
64
-32
Consider a graph of displacement (distance traveled) vs. time.
time (hours)
distance(miles)
Average velocity can be found by taking:
change in position
change in time
s
t
Δ=Δ
tΔ
sΔA
B
( ) ( )ave
f t t f tsV
t t
+ Δ −Δ= =
Δ Δ
The speedometer in your car does not measure average velocity, but instantaneous velocity.
( ) ( ) ( )0
limt
f t t f tdsV t
dt tΔ →
+ Δ −= =
Δ
(The velocity at one moment in time.)
→
Velocity is the first derivative of position.
→
Example: Free Fall Equation
21
2s g t=
GravitationalConstants:
2
ft32
secg =
2
m9.8
secg =
2
cm980
secg =
2132
2s t= ⋅
216 s t=
32 ds
V tdt
= =
Speed is the absolute value of velocity.
→
Acceleration is the derivative of velocity.
dva
dt=
2
2
d s
dt= example: 32v t=
32a =
If distance is in: feet
Velocity would be in:feet
sec
Acceleration would be in:ft
sec sec
2
ft
sec=
→
time
distance
acc posvel pos &increasing
acc zerovel pos &constant
acc negvel pos &decreasing
velocityzero
acc negvel neg &decreasing acc zero
vel neg &constant
acc posvel neg &increasing
acc zero,velocity zero
It is important to understand the relationship between a position graph, velocity and acceleration:
→
Rates of Change:
Average rate of change =( ) ( )f x h f x
h
+ −
Instantaneous rate of change = ( ) ( ) ( )0
limh
f x h f xf x
h→
+ −′ =
These definitions are true for any function.
( x does not have to represent time. )
→
Example 1:
For a circle:
2A rπ=
2dA dr
dr drπ=
2dA
rdr
π=
Instantaneous rate of change of the area withrespect to the radius.
For tree ring growth, if the change in area is constant then dr must get smaller as r gets larger.
2 dA r drπ=
→
A particle P moves back and forth on the number line. The graph below shows the position of P as a function of time.
a) Describe the motion of the particle over time.
b) Graph the particle’s velocity and speed (where defined).
4
-4
2 4 6
Particle Motion
Particle is moving right when P‘(t) > 0 or (0,1)
Particle is moving left when P’(t) < 0 or (2,3), (5,6)
Particle is standing still when P’(t) = 0 or (1,2), (3,5)
4
-4
2 4 6
2
-2
Particle Motion
from Economics:
Marginal cost is the first derivative of the cost function, and represents an approximation of the cost of producing one more unit.
→
Example 13:Suppose it costs: ( ) 3 26 15c x x x x= − +
to produce x stoves. ( ) 23 12 15c x x x′ = − +
If you are currently producing 10 stoves, the 11th stove will cost approximately:
( ) 210 3 10 12 10 15c′ = ⋅ − ⋅ +
300 120 15= − +
$195=
marginal costThe actual cost is: ( ) ( )11 10C C−
( ) ( )3 2 3 211 6 11 15 11 10 6 10 15 10= − ⋅ + ⋅ − − ⋅ + ⋅
770 550= − $220= actual cost
Note that this is not a great approximation – Don’t let that bother you.
→
Note that this is not a great approximation – Don’t let that bother you.
Marginal cost is a linear approximation of a curved function. For large values it gives a good approximation of the cost of producing the next item.
π
3.5 Derivatives of Trig Functions
3π4
97.403°
32
Domain: all realsRange: −1, 1⎡
⎣⎢⎢
⎤
⎦⎥⎥
Domain: x≠kπ2 (k odd integer)
Range: all reals
±12
0
Multiply by 1+cos h1+cos h and use
identity 1-cos2h=sin2h
y = 12x - 35
12
π
2
π0
2
π−
π−
Consider the function ( )siny θ=
We could make a graph of the slope: θ slope
1−
0
1
0
1−Now we connect the dots!
The resulting curve is a cosine curve.
( )sin cosd
x xdx
=
→
π
2
π0
2
π−
π−
We can do the same thing for ( )cosy θ= θ slope
0
1
0
1−
0The resulting curve is a sine curve that has been reflected about the x-axis.
( )cos sind
x xdx
= −
→
We can find the derivative of tangent x by using the quotient rule.
tand
xdx
sin
cos
d x
dx x
( )2
cos cos sin sin
cos
x x x x
x
⋅ − ⋅ −
2 2
2
cos sin
cos
x x
x
+
2
1
cos x
2sec x
( ) 2tan secd
x xdx
=
→
Derivatives of the remaining trig functions can be determined the same way.
sin cosd
x xdx
=
cos sind
x xdx
=−
2tan secd
x xdx
=
2cot cscd
x xdx
=−
sec sec tand
x x xdx
= ⋅
csc csc cotd
x x xdx
=− ⋅
π
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