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Derivative as a Rate of Change Chapter 3 Section 4

Derivative as a Rate of Change

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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. - PowerPoint PPT Presentation

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Page 1: Derivative as a Rate of Change

Derivative as a Rate of Change

Chapter 3 Section 4

Page 2: Derivative as a Rate of Change

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.

Page 3: Derivative as a Rate of Change

Example• The area A of a circle is related to its diameter by

the equation• 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

• Thus, when D = 10 meters the area is changing with respect to the diameter at the rate of

2

4DA

2

2444

22 DDD

dD

dAD

dD

dA

mmD

/71.152

)10(

22

Page 4: Derivative as a Rate of Change

Motion Along a Line

• Displacement of object over time Δs = f(t + Δt) – f(t)

• Average velocity of object over time interval

t

tfttf

t

svaverage

)()(

timetravel

nt displaceme

Page 5: Derivative as a Rate of Change

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)

Page 6: Derivative as a Rate of Change
Page 7: Derivative as a Rate of Change

Speed

• Rate of progress regardless of direction

Page 8: Derivative as a Rate of Change

Graph of velocity f ’(t)

Page 9: Derivative as a Rate of Change

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

Page 10: Derivative as a Rate of Change

Example 1: Galileo Free Fall

• Galileo’s Free Fall Equations distance falleng is acceleration due to Earth’s gravity

(appx: 32 ft/sec2 or 9.8 m/sec2)– Same constant acceleration– No jerk

2

2

1gts

012

1 2

dt

dgt

dt

d

dt

dgt

dt

ds

dt

d

dt

dj

Page 11: Derivative as a Rate of Change

Example 2: Free Fall Example

• How many meters does the ball fall in the first 2 seconds?

• Free Fall equation s = 4.9t2 in meters

s(2) – s(0) = 4.9(2)2 - 4.9(0)2 = 19.6 m

22 9.4)8.9(2

1tts

Page 12: Derivative as a Rate of Change

Example 2: Free Fall Example

• What is its velocity, speed and acceleration when t = 2?– Velocity = derivative of position at any time t

– So at time t = 2, the velocity is

ttdt

dtstv 8.99.4)(')( 2

sec/6.19)2(8.98.99.4)(')( 2 mttdt

dtstv

Page 13: Derivative as a Rate of Change

Example 2: Free Fall Example

• What is its velocity, speed and acceleration when t = 2?– Velocity = derivative of position at any time t

– So at time t = 2, the speed is

ttdt

dtstv 8.99.4)(')( 2

sec/6.19sec/6.19 mmvelocityspeed

Page 14: Derivative as a Rate of Change

Example 2: Free Fall Example

• What is its velocity, speed and acceleration when t = 2?– Velocity = derivative of position at any time t

– The acceleration at any time t

– So at t = 2, acceleration is (no air resistance)

ttdt

dtstv 8.99.4)(')( 2

2sec/8.98.9)('')(')( mtdt

dtstvta

2sec/8.9)( mta

Page 15: Derivative as a Rate of Change

Derivatives of Trigonometric Functions

Chapter 3 Section 5

Page 16: Derivative as a Rate of Change

Derivatives

Page 17: Derivative as a Rate of Change

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?

Page 18: Derivative as a Rate of Change

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: s = 5cos(t)– Velocity: 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

Page 19: Derivative as a Rate of Change

Chain Rule

Chapter 6 Section 6

Page 20: Derivative as a Rate of Change
Page 21: Derivative as a Rate of Change

Implicit Differentiation

Chapter 3 Section 7

Page 22: Derivative as a Rate of Change

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 functionsx2 + 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

Page 23: Derivative as a Rate of Change

Circle Example

Page 24: Derivative as a Rate of Change

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.

Page 25: Derivative as a Rate of Change

Folium of Descartes• Find the slope of the 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

0933 xyyx

Page 26: Derivative as a Rate of Change

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

0933 xyyx

0)4)(2(942 33 072648

07272 00

0)2)(4(924 33

072864 07272

00

Page 27: Derivative as a Rate of Change

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

Page 28: Derivative as a Rate of Change

0933 xyyx

0933

dx

dxy

dx

dy

dx

dx

dx

d

0)(933 22

x

dx

dy

dx

dyx

dx

dyyx

PRODUCT RULE

09933 22 ydx

dyx

dx

dyyx

Page 29: Derivative as a Rate of Change

09933 22 ydx

dyx

dx

dyyx

Factor out dy/dx

09393 22 yxxydx

dy

22 3993 xyxydx

dy

xy

xy

dx

dy

93

392

2

xy

xy

dx

dy

3

32

2

Divide out 3

Page 30: Derivative as a Rate of Change

xy

xy

dx

dy

3

32

2

Evaluate at (2,4) and (4,2)

5

4

10

8

616

412

)2(34

2)4(3

3

32

2

)4,2(2

2

)4,2(

xy

xy

dx

dy

Slope at the point (2,4)

4

5

8

10

124

166

)4(32

4)2(3

3

32

2

)2,4(2

2

)2,4(

xy

xy

dx

dy

Slope at the point (4,2)

Page 31: Derivative as a Rate of Change

0933 xyyx

Page 32: Derivative as a Rate of Change

Find Tangents5

4)4,2(

dx

dy

11 xxmyy

25

44 xy

5

8

5

44 xy

45

8

5

4 xy

5

20

5

8

5

4 xy

5

12

5

4 xy

4

5)2,4(

dx

dy

44

52 xy

54

52 xy

34

5 xy

Page 33: Derivative as a Rate of Change

5

12

5

4 xy

34

5 xy

Page 34: Derivative as a Rate of Change

Folium of Descartes• Can you find the slope of

the folium of Descartes

• At what point other than the origin does the folium have a horizontal tangent?– Can you find this?

0933 xyyx

Page 35: Derivative as a Rate of Change

Derivatives of Inverse Functions and Logarithms

Chapter 3 Section 8

Page 36: Derivative as a Rate of Change
Page 37: Derivative as a Rate of Change
Page 38: Derivative as a Rate of Change
Page 39: Derivative as a Rate of Change

Examples 1 & 2

Page 40: Derivative as a Rate of Change
Page 41: Derivative as a Rate of Change