Transcript Acceleration - mathematicalminds.net

Derivative as a Rate of Change
Chapter 3 Section 4
Usually omit
instantaneous
Interpretation: The rate of change at which f is changing at the point x
Interpretation: Instantaneous rate are limits of average rates.
Example
• The area A of a circle is related to its diameter by

A

D
the equation
4
• How fast does the area change with respect to
the diameter when the diameter is 10 meters?
• The rate of change of the area with respect to
the diameter dA   D 2    dA D 2    2 D  D
2
dD  4

4 dD
4
2
• Thus, when D = 10 meters the area is changing
with respect to the diameter at the rate of
D
2

 (10)
2
 15.71 m 2 / m
Motion Along a Line
• Displacement of object over time
Δs = f(t + Δt) – f(t)
• Average velocity of object over time interval
vaverage
displacement s f (t  t )  f (t )



travel time
t
t
Velocity
• Find the body’s velocity at the exact instant t
– How fast an object is moving along a horizontal line
– Direction of motion (increasing >0 decreasing <0)
Speed
• Rate of progress regardless of direction
Graph of velocity f ’(t)
Acceleration
• The rate at which a body’s velocity changes
– How quickly the body picks up or loses speed
– A sudden change in acceleration is called jerk
• Abrupt changes in acceleration
Example 1: Galileo Free Fall
• Galileo’s Free Fall Equation
1 2
s  gt
2
s distance fallen
g is acceleration due to Earth’s gravity
(appx: 32 ft/sec2 or 9.8 m/sec2)
– Same constant acceleration
– No jerk
d  d  ds  1 2    d  d
 d


j     gt       gt   1  0
dt  dt  dt  2
   dt  dt
 dt
Example 2: Free Fall Example
• How many meters does the ball fall in
the first 2 seconds?
1
s  (9.8)t 2  4.9t 2
2
• Free Fall equation s = 4.9t2 in meters
s(2) – s(0) = 4.9(2)2 - 4.9(0)2 = 19.6 m
Example 2: Free Fall Example
• What is its velocity, speed and acceleration
when t = 2?
– Velocity = derivative of position at any time t
v(t )  s ' (t ) 


d
4.9t 2  9.8t
dt
– So at time t = 2, the velocity is
v(t )  s ' (t ) 


d
4.9t 2  9.8t  9.8(2)  19.6 m / sec
dt
Example 2: Free Fall Example
• What is its velocity, speed and acceleration
when t = 2?
– Velocity = derivative of position at any time t
v(t )  s ' (t ) 


d
4.9t 2  9.8t
dt
– So at time t = 2, the speed is
speed  velocity  19.6 m / sec  19.6 m / sec
Example 2: Free Fall Example
• What is its velocity, speed and acceleration
when t = 2?
– Velocity = derivative of position at any time t
v(t )  s ' (t ) 


d
4.9t 2  9.8t
dt
– The acceleration at any time t
a (t )  v' (t )  s ' ' (t ) 
d
9.8t   9.8 m / sec 2
dt
– So at t = 2, acceleration is
(no air resistance)
a (t )  9.8 m / sec 2
Derivatives of
Trigonometric Functions
Chapter 3 Section 5
Derivatives
Application: Simple Harmonic Motion
• Motion of an object/weight bobbing
freely up and down with no resistance
on an end of a spring
• Periodic, repeats motion
• A weight hanging from a spring is
stretched down 5 units beyond its rest
position and released at time t = 0 to
bob up and down. Its position at any
later time t is
s = 5 cos(t)
• What are its velocity and acceleration
at time t?
Application: Simple Harmonic Motion
• Its position at any later time t is s = 5 cos(t)
– Amp = 5
– Period = 2
• What are its velocity and acceleration at time t?
– Position:
– Velocity:
s = 5cos(t)
s’ = -5sin(t)
• Speed of weight is 0, when t = 0
– Acceleration: s’’ = -5 cos(t)
• Opposite of position value, gravity pulls down, spring pulls up
Chain Rule
Chapter 6 Section 6
Implicit Differentiation
Chapter 3 Section 7
Implicit Differentiation
• So far our functions have been y = f(x) in one variable
such as y = x2 + 3
– This is explicit differentiation
• Other types of functions
x2 + y2 = 25 or y2 – x = 0
• Implicit relation between the variables x and y
• Implicit Differentiation
– Differentiate both sides of the equation with respect to x,
treating y as a differentiable function of x (always put
dy/dx after derive y term)
– Collect the terms with dy/dx on one side of the equation
and solve for dy/dx
Circle Example
Folium of Descartes
• The curve was first proposed
by Descartes in 1638. Its claim to
fame lies in an incident in the
development of calculus.
• Descartes challenged Fermat to
find the tangent line to the curve
at an arbitrary point since Fermat
had recently discovered a
method for finding tangent lines.
• Fermat solved the problem
easily, something Descartes was
unable to do.
• Since the invention of calculus,
the slope of the tangent line can
be found easily using implicit
differentiation.
Folium of Descartes
• Find the slope of the folium
of Descartes
x 3  y 3  9 xy  0
• Show that the points (2,4)
and (4,2) lie on the curve
and find their slopes and
tangent line to curve
Folium of Descartes
• Show that the points (2,4) and (4,2) lie on the
curve and find their slopes and tangent line to
curve
3
3
x  y  9xy  0
23  43  9(2)(4)  0
8  64  72  0
72  72  0
00
43  23  9(4)(2)  0
64  8  72  0
72  72  0
00
Folium of Descartes
• Show that the points (2,4) and (4,2) lie on the
curve and find their slopes and tangent line
to curve
– Find slope of curve by implicit differentiation by
finding dy/dx
x  y  9xy  0
3
 
3
 
d 3 d 3 d
d
x 
y  9 xy   0
dx
dx
dx
dx
dy  dy
d

3x  3 y
 9 x  y ( x)   0
dx  dx
dx

2
2
PRODUCT RULE
dy
dy
3x  3 y
 9x  9 y  0
dx
dx
2
2
dy
dy
3x  3 y
 9x  9 y  0
dx
dx
2
2
Factor out dy/dx


dy
3 y 2  9 x  3x 2  9 y  0
dx


dy
3 y 2  9 x  9 y  3x 2
dx
dy 9 y  3 x
 2
dx 3 y  9 x
2
dy 3 y  x
 2
dx y  3 x
2
Divide out 3
dy 3 y  x 2
 2
dx y  3x
Evaluate at (2,4) and (4,2)
dy
dx
( 2, 4 )
3y  x2
 2
y  3x
( 2, 4 )
3(4)  2 2 12  4 8 4
 2



4  3(2) 16  6 10 5
Slope at the point (2,4)
dy
dx
( 4, 2 )
3y  x2
 2
y  3x
( 4, 2 )
3(2)  4 2 6  16  10 5
 2



2  3(4) 4  12  8 4
Slope at the point (4,2)
x3  y 3  9xy  0
dy
dx
( 2, 4 )
4

5
4
y  4  x  2
5
4
8
y4 x
5
5
4
8
y  x 4
5
5
4
8 20
y  x 
5
5 5
4
12
y  x
5
5
Find Tangents
y  y1  m x  x1 
dy
dx
y2
( 4, 2 )

5
4
5
x  4
4
5
y 2  x 5
4
5
y  x 3
4
4
12
y  x
5
5
5
y  x3
4
Folium of Descartes
• Can you find the slope of
the folium of Descartes
x 3  y 3  9 xy  0
• At what point other than
the origin does the folium
have a horizontal tangent?
– Can you find this?
Derivatives of Inverse Functions
and Logarithms
Chapter 3 Section 8
Examples 1 & 2