Why Do We Need Derivatives? Bernd Schr oder logo1 Bernd Schr - - PowerPoint PPT Presentation

▶
why do we need derivatives
SMART_READER_LITE
LIVE PREVIEW

Why Do We Need Derivatives? Bernd Schr oder logo1 Bernd Schr - - PowerPoint PPT Presentation

Feb 23, 2023 106 likes •1.44k views

Introduction Velocities Tangent Lines Rates of Change Why Do We Need Derivatives? Bernd Schr oder logo1 Bernd Schr oder Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives? Introduction

slide-1
SLIDE 1

logo1 Introduction Velocities Tangent Lines Rates of Change

Why Do We Need Derivatives?

Bernd Schr¨

  • der

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-2
SLIDE 2

logo1 Introduction Velocities Tangent Lines Rates of Change

Driving From Kansas To Virginia

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-3
SLIDE 3

logo1 Introduction Velocities Tangent Lines Rates of Change

Driving From Kansas To Virginia

  • 1. Distance: 1,300mi

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-4
SLIDE 4

logo1 Introduction Velocities Tangent Lines Rates of Change

Driving From Kansas To Virginia

  • 1. Distance: 1,300mi, driving time: 3 days

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-5
SLIDE 5

logo1 Introduction Velocities Tangent Lines Rates of Change

Driving From Kansas To Virginia

  • 1. Distance: 1,300mi, driving time: 3 days
  • 2. vavg = distance traveled

time needed

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-6
SLIDE 6

logo1 Introduction Velocities Tangent Lines Rates of Change

Driving From Kansas To Virginia

  • 1. Distance: 1,300mi, driving time: 3 days
  • 2. vavg = distance traveled

time needed = 1,300 mi 3 days

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-7
SLIDE 7

logo1 Introduction Velocities Tangent Lines Rates of Change

Driving From Kansas To Virginia

  • 1. Distance: 1,300mi, driving time: 3 days
  • 2. vavg = distance traveled

time needed = 1,300 mi 3 days ≈ 433.33 mi day

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-8
SLIDE 8

logo1 Introduction Velocities Tangent Lines Rates of Change

Driving From Kansas To Virginia

  • 1. Distance: 1,300mi, driving time: 3 days
  • 2. vavg = distance traveled

time needed = 1,300 mi 3 days ≈ 433.33 mi day, useful to plan where to book hotels along the way, etc.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-9
SLIDE 9

logo1 Introduction Velocities Tangent Lines Rates of Change

Driving From Kansas To Virginia

  • 1. Distance: 1,300mi, driving time: 3 days
  • 2. vavg = distance traveled

time needed = 1,300 mi 3 days ≈ 433.33 mi day, useful to plan where to book hotels along the way, etc.

  • 3. Average velocity is always vavg = s

t

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-10
SLIDE 10

logo1 Introduction Velocities Tangent Lines Rates of Change

Driving From Kansas To Virginia

  • 1. Distance: 1,300mi, driving time: 3 days
  • 2. vavg = distance traveled

time needed = 1,300 mi 3 days ≈ 433.33 mi day, useful to plan where to book hotels along the way, etc.

  • 3. Average velocity is always vavg = s

t

  • 4. Suppose after driving 200 miles in the first four hours

(you’re in Missouri), a highway patrolman stops you

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-11
SLIDE 11

logo1 Introduction Velocities Tangent Lines Rates of Change

Driving From Kansas To Virginia

  • 1. Distance: 1,300mi, driving time: 3 days
  • 2. vavg = distance traveled

time needed = 1,300 mi 3 days ≈ 433.33 mi day, useful to plan where to book hotels along the way, etc.

  • 3. Average velocity is always vavg = s

t

  • 4. Suppose after driving 200 miles in the first four hours

(you’re in Missouri), a highway patrolman stops you and reveals that you’ve been going 65mph in an area in which the speed limit is 55mph.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-12
SLIDE 12

logo1 Introduction Velocities Tangent Lines Rates of Change

Driving From Kansas To Virginia

  • 1. Distance: 1,300mi, driving time: 3 days
  • 2. vavg = distance traveled

time needed = 1,300 mi 3 days ≈ 433.33 mi day, useful to plan where to book hotels along the way, etc.

  • 3. Average velocity is always vavg = s

t

  • 4. Suppose after driving 200 miles in the first four hours

(you’re in Missouri), a highway patrolman stops you and reveals that you’ve been going 65mph in an area in which the speed limit is 55mph. v

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-13
SLIDE 13

logo1 Introduction Velocities Tangent Lines Rates of Change

Driving From Kansas To Virginia

  • 1. Distance: 1,300mi, driving time: 3 days
  • 2. vavg = distance traveled

time needed = 1,300 mi 3 days ≈ 433.33 mi day, useful to plan where to book hotels along the way, etc.

  • 3. Average velocity is always vavg = s

t

  • 4. Suppose after driving 200 miles in the first four hours

(you’re in Missouri), a highway patrolman stops you and reveals that you’ve been going 65mph in an area in which the speed limit is 55mph. v = 200 mi 4 hrs

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-14
SLIDE 14

logo1 Introduction Velocities Tangent Lines Rates of Change

Driving From Kansas To Virginia

  • 1. Distance: 1,300mi, driving time: 3 days
  • 2. vavg = distance traveled

time needed = 1,300 mi 3 days ≈ 433.33 mi day, useful to plan where to book hotels along the way, etc.

  • 3. Average velocity is always vavg = s

t

  • 4. Suppose after driving 200 miles in the first four hours

(you’re in Missouri), a highway patrolman stops you and reveals that you’ve been going 65mph in an area in which the speed limit is 55mph. v = 200 mi 4 hrs = 50 mph

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-15
SLIDE 15

logo1 Introduction Velocities Tangent Lines Rates of Change

Driving From Kansas To Virginia

  • 1. Distance: 1,300mi, driving time: 3 days
  • 2. vavg = distance traveled

time needed = 1,300 mi 3 days ≈ 433.33 mi day, useful to plan where to book hotels along the way, etc.

  • 3. Average velocity is always vavg = s

t

  • 4. Suppose after driving 200 miles in the first four hours

(you’re in Missouri), a highway patrolman stops you and reveals that you’ve been going 65mph in an area in which the speed limit is 55mph. v = 200 mi 4 hrs = 50 mph ???

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-16
SLIDE 16

logo1 Introduction Velocities Tangent Lines Rates of Change

Driving From Kansas To Virginia

  • 1. Distance: 1,300mi, driving time: 3 days
  • 2. vavg = distance traveled

time needed = 1,300 mi 3 days ≈ 433.33 mi day, useful to plan where to book hotels along the way, etc.

  • 3. Average velocity is always vavg = s

t

  • 4. Suppose after driving 200 miles in the first four hours

(you’re in Missouri), a highway patrolman stops you and reveals that you’ve been going 65mph in an area in which the speed limit is 55mph. v = 200 mi 4 hrs = 50 mph ???

  • 5. Even sensible average velocity can look strange

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-17
SLIDE 17

logo1 Introduction Velocities Tangent Lines Rates of Change

Driving From Kansas To Virginia

  • 1. Distance: 1,300mi, driving time: 3 days
  • 2. vavg = distance traveled

time needed = 1,300 mi 3 days ≈ 433.33 mi day, useful to plan where to book hotels along the way, etc.

  • 3. Average velocity is always vavg = s

t

  • 4. Suppose after driving 200 miles in the first four hours

(you’re in Missouri), a highway patrolman stops you and reveals that you’ve been going 65mph in an area in which the speed limit is 55mph. v = 200 mi 4 hrs = 50 mph ???

  • 5. Even sensible average velocity can look strange:

433.33 mi day

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-18
SLIDE 18

logo1 Introduction Velocities Tangent Lines Rates of Change

Driving From Kansas To Virginia

  • 1. Distance: 1,300mi, driving time: 3 days
  • 2. vavg = distance traveled

time needed = 1,300 mi 3 days ≈ 433.33 mi day, useful to plan where to book hotels along the way, etc.

  • 3. Average velocity is always vavg = s

t

  • 4. Suppose after driving 200 miles in the first four hours

(you’re in Missouri), a highway patrolman stops you and reveals that you’ve been going 65mph in an area in which the speed limit is 55mph. v = 200 mi 4 hrs = 50 mph ???

  • 5. Even sensible average velocity can look strange:

433.33 mi day = 433.33 mi 24 hr

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-19
SLIDE 19

logo1 Introduction Velocities Tangent Lines Rates of Change

Driving From Kansas To Virginia

  • 1. Distance: 1,300mi, driving time: 3 days
  • 2. vavg = distance traveled

time needed = 1,300 mi 3 days ≈ 433.33 mi day, useful to plan where to book hotels along the way, etc.

  • 3. Average velocity is always vavg = s

t

  • 4. Suppose after driving 200 miles in the first four hours

(you’re in Missouri), a highway patrolman stops you and reveals that you’ve been going 65mph in an area in which the speed limit is 55mph. v = 200 mi 4 hrs = 50 mph ???

  • 5. Even sensible average velocity can look strange:

433.33 mi day = 433.33 mi 24 hr ≈ 18.06mi hr

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-20
SLIDE 20

logo1 Introduction Velocities Tangent Lines Rates of Change

Driving From Kansas To Virginia

  • 1. Distance: 1,300mi, driving time: 3 days
  • 2. vavg = distance traveled

time needed = 1,300 mi 3 days ≈ 433.33 mi day, useful to plan where to book hotels along the way, etc.

  • 3. Average velocity is always vavg = s

t

  • 4. Suppose after driving 200 miles in the first four hours

(you’re in Missouri), a highway patrolman stops you and reveals that you’ve been going 65mph in an area in which the speed limit is 55mph. v = 200 mi 4 hrs = 50 mph ???

  • 5. Even sensible average velocity can look strange:

433.33 mi day = 433.33 mi 24 hr ≈ 18.06mi hr

  • 6. We need a way to define instantaneous velocity.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-21
SLIDE 21

logo1 Introduction Velocities Tangent Lines Rates of Change

Tangent Lines

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-22
SLIDE 22

logo1 Introduction Velocities Tangent Lines Rates of Change

Tangent Lines

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-23
SLIDE 23

logo1 Introduction Velocities Tangent Lines Rates of Change

Tangent Lines

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-24
SLIDE 24

logo1 Introduction Velocities Tangent Lines Rates of Change

Tangent Lines

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-25
SLIDE 25

logo1 Introduction Velocities Tangent Lines Rates of Change

Tangent Lines

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-26
SLIDE 26

logo1 Introduction Velocities Tangent Lines Rates of Change

Tangent Lines

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-27
SLIDE 27

logo1 Introduction Velocities Tangent Lines Rates of Change

Tangent Lines

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-28
SLIDE 28

logo1 Introduction Velocities Tangent Lines Rates of Change

Tangent Lines

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-29
SLIDE 29

logo1 Introduction Velocities Tangent Lines Rates of Change

Tangent Lines

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-30
SLIDE 30

logo1 Introduction Velocities Tangent Lines Rates of Change

Definition.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-31
SLIDE 31

logo1 Introduction Velocities Tangent Lines Rates of Change

  • Definition. Let s be a position function.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-32
SLIDE 32

logo1 Introduction Velocities Tangent Lines Rates of Change

  • Definition. Let s be a position function. We define the average

velocity in the interval [a,b] by vavg := s(b)−s(a) b−a .

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-33
SLIDE 33

logo1 Introduction Velocities Tangent Lines Rates of Change

Find the Instantaneous Velocity of a Coin with Vertical Position s(t) = −4.9t2 +20[m] at Time t = 2[s]

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-34
SLIDE 34

logo1 Introduction Velocities Tangent Lines Rates of Change

Find the Instantaneous Velocity of a Coin with Vertical Position s(t) = −4.9t2 +20[m] at Time t = 2[s]

b s(b)−s(2) b−2

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-35
SLIDE 35

logo1 Introduction Velocities Tangent Lines Rates of Change

Find the Instantaneous Velocity of a Coin with Vertical Position s(t) = −4.9t2 +20[m] at Time t = 2[s]

b s(b)−s(2) b−2 2.1

  • 20.09

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-36
SLIDE 36

logo1 Introduction Velocities Tangent Lines Rates of Change

Find the Instantaneous Velocity of a Coin with Vertical Position s(t) = −4.9t2 +20[m] at Time t = 2[s]

b s(b)−s(2) b−2 2.1

  • 20.09

2.01

  • 19.649

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-37
SLIDE 37

logo1 Introduction Velocities Tangent Lines Rates of Change

Find the Instantaneous Velocity of a Coin with Vertical Position s(t) = −4.9t2 +20[m] at Time t = 2[s]

b s(b)−s(2) b−2 2.1

  • 20.09

2.01

  • 19.649

2.001

  • 19.6049

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-38
SLIDE 38

logo1 Introduction Velocities Tangent Lines Rates of Change

Find the Instantaneous Velocity of a Coin with Vertical Position s(t) = −4.9t2 +20[m] at Time t = 2[s]

b s(b)−s(2) b−2 2.1

  • 20.09

2.01

  • 19.649

2.001

  • 19.6049

2.0001

  • 19.6005

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-39
SLIDE 39

logo1 Introduction Velocities Tangent Lines Rates of Change

Find the Instantaneous Velocity of a Coin with Vertical Position s(t) = −4.9t2 +20[m] at Time t = 2[s]

b s(b)−s(2) b−2 2.1

  • 20.09

2.01

  • 19.649

2.001

  • 19.6049

2.0001

  • 19.6005

2 undefined

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-40
SLIDE 40

logo1 Introduction Velocities Tangent Lines Rates of Change

Find the Instantaneous Velocity of a Coin with Vertical Position s(t) = −4.9t2 +20[m] at Time t = 2[s]

b s(b)−s(2) b−2 2.1

  • 20.09

2.01

  • 19.649

2.001

  • 19.6049

2.0001

  • 19.6005

2 undefined

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-41
SLIDE 41

logo1 Introduction Velocities Tangent Lines Rates of Change

Find the Instantaneous Velocity of a Coin with Vertical Position s(t) = −4.9t2 +20[m] at Time t = 2[s]

b s(b)−s(2) b−2 2.1

  • 20.09

2.01

  • 19.649

2.001

  • 19.6049

2.0001

  • 19.6005

2 undefined 1.9

  • 19.11

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-42
SLIDE 42

logo1 Introduction Velocities Tangent Lines Rates of Change

Find the Instantaneous Velocity of a Coin with Vertical Position s(t) = −4.9t2 +20[m] at Time t = 2[s]

b s(b)−s(2) b−2 2.1

  • 20.09

2.01

  • 19.649

2.001

  • 19.6049

2.0001

  • 19.6005

2 undefined 1.99

  • 19.551

1.9

  • 19.11

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-43
SLIDE 43

logo1 Introduction Velocities Tangent Lines Rates of Change

Find the Instantaneous Velocity of a Coin with Vertical Position s(t) = −4.9t2 +20[m] at Time t = 2[s]

b s(b)−s(2) b−2 2.1

  • 20.09

2.01

  • 19.649

2.001

  • 19.6049

2.0001

  • 19.6005

2 undefined 1.999

  • 19.5951

1.99

  • 19.551

1.9

  • 19.11

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-44
SLIDE 44

logo1 Introduction Velocities Tangent Lines Rates of Change

Find the Instantaneous Velocity of a Coin with Vertical Position s(t) = −4.9t2 +20[m] at Time t = 2[s]

b s(b)−s(2) b−2 2.1

  • 20.09

2.01

  • 19.649

2.001

  • 19.6049

2.0001

  • 19.6005

2 undefined 1.9999

  • 19.5995

1.999

  • 19.5951

1.99

  • 19.551

1.9

  • 19.11

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-45
SLIDE 45

logo1 Introduction Velocities Tangent Lines Rates of Change

Find the Instantaneous Velocity of a Coin with Vertical Position s(t) = −4.9t2 +20[m] at Time t = 2[s]

b s(b)−s(2) b−2 2.1

  • 20.09

2.01

  • 19.649

2.001

  • 19.6049

2.0001

  • 19.6005

2 undefined (−19.6 “feels right”) 1.9999

  • 19.5995

1.999

  • 19.5951

1.99

  • 19.551

1.9

  • 19.11

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-46
SLIDE 46

logo1 Introduction Velocities Tangent Lines Rates of Change

Definition.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-47
SLIDE 47

logo1 Introduction Velocities Tangent Lines Rates of Change

  • Definition. Let s be a position function.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-48
SLIDE 48

logo1 Introduction Velocities Tangent Lines Rates of Change

  • Definition. Let s be a position function. We define the

instantaneous velocity at a by vinst := lim

b→a

s(b)−s(a) b−a .

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-49
SLIDE 49

logo1 Introduction Velocities Tangent Lines Rates of Change

Find the Instantaneous Velocity of a Coin with Vertical Position s(t) = −4.9t2 +20[m] at Time t = 2[s].

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-50
SLIDE 50

logo1 Introduction Velocities Tangent Lines Rates of Change

Find the Instantaneous Velocity of a Coin with Vertical Position s(t) = −4.9t2 +20[m] at Time t = 2[s].

lim

b→2

s(b)−s(2) b−2

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-51
SLIDE 51

logo1 Introduction Velocities Tangent Lines Rates of Change

Find the Instantaneous Velocity of a Coin with Vertical Position s(t) = −4.9t2 +20[m] at Time t = 2[s].

lim

b→2

s(b)−s(2) b−2 = lim

b→2

−4.9b2 +20−

  • −4.9·22 +20
  • b−2

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-52
SLIDE 52

logo1 Introduction Velocities Tangent Lines Rates of Change

Find the Instantaneous Velocity of a Coin with Vertical Position s(t) = −4.9t2 +20[m] at Time t = 2[s].

lim

b→2

s(b)−s(2) b−2 = lim

b→2

−4.9b2 +20−

  • −4.9·22 +20
  • b−2

= lim

b→2

−4.9b2 +4.9·22 b−2

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-53
SLIDE 53

logo1 Introduction Velocities Tangent Lines Rates of Change

Find the Instantaneous Velocity of a Coin with Vertical Position s(t) = −4.9t2 +20[m] at Time t = 2[s].

lim

b→2

s(b)−s(2) b−2 = lim

b→2

−4.9b2 +20−

  • −4.9·22 +20
  • b−2

= lim

b→2

−4.9b2 +4.9·22 b−2 = lim

b→2

−4.9

  • b2 −22

b−2

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-54
SLIDE 54

logo1 Introduction Velocities Tangent Lines Rates of Change

Find the Instantaneous Velocity of a Coin with Vertical Position s(t) = −4.9t2 +20[m] at Time t = 2[s].

lim

b→2

s(b)−s(2) b−2 = lim

b→2

−4.9b2 +20−

  • −4.9·22 +20
  • b−2

= lim

b→2

−4.9b2 +4.9·22 b−2 = lim

b→2

−4.9

  • b2 −22

b−2 = lim

b→2

−4.9(b+2)(b−2) b−2

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-55
SLIDE 55

logo1 Introduction Velocities Tangent Lines Rates of Change

Find the Instantaneous Velocity of a Coin with Vertical Position s(t) = −4.9t2 +20[m] at Time t = 2[s].

lim

b→2

s(b)−s(2) b−2 = lim

b→2

−4.9b2 +20−

  • −4.9·22 +20
  • b−2

= lim

b→2

−4.9b2 +4.9·22 b−2 = lim

b→2

−4.9

  • b2 −22

b−2 = lim

b→2

−4.9(b+2)(b−2) b−2 = lim

b→2−4.9(b+2)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-56
SLIDE 56

logo1 Introduction Velocities Tangent Lines Rates of Change

Find the Instantaneous Velocity of a Coin with Vertical Position s(t) = −4.9t2 +20[m] at Time t = 2[s].

lim

b→2

s(b)−s(2) b−2 = lim

b→2

−4.9b2 +20−

  • −4.9·22 +20
  • b−2

= lim

b→2

−4.9b2 +4.9·22 b−2 = lim

b→2

−4.9

  • b2 −22

b−2 = lim

b→2

−4.9(b+2)(b−2) b−2 = lim

b→2−4.9(b+2) = −19.6

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-57
SLIDE 57

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-58
SLIDE 58

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-59
SLIDE 59

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-60
SLIDE 60

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-61
SLIDE 61

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-62
SLIDE 62

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-63
SLIDE 63

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-64
SLIDE 64

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-65
SLIDE 65

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-66
SLIDE 66

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-67
SLIDE 67

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-68
SLIDE 68

logo1 Introduction Velocities Tangent Lines Rates of Change

Definition.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-69
SLIDE 69

logo1 Introduction Velocities Tangent Lines Rates of Change

  • Definition. Let f be a function.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-70
SLIDE 70

logo1 Introduction Velocities Tangent Lines Rates of Change

  • Definition. Let f be a function. The secant line of f through
  • a,f(a)
  • and
  • b,f(b)
  • Bernd Schr¨
  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-71
SLIDE 71

logo1 Introduction Velocities Tangent Lines Rates of Change

  • Definition. Let f be a function. The secant line of f through
  • a,f(a)
  • and
  • b,f(b)
  • is the unique straight line that goes

through the points

  • a,f(a)
  • and
  • b,f(b)
  • .

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-72
SLIDE 72

logo1 Introduction Velocities Tangent Lines Rates of Change

  • Definition. Let f be a function. The secant line of f through
  • a,f(a)
  • and
  • b,f(b)
  • is the unique straight line that goes

through the points

  • a,f(a)
  • and
  • b,f(b)
  • .

Definition.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-73
SLIDE 73

logo1 Introduction Velocities Tangent Lines Rates of Change

  • Definition. Let f be a function. The secant line of f through
  • a,f(a)
  • and
  • b,f(b)
  • is the unique straight line that goes

through the points

  • a,f(a)
  • and
  • b,f(b)
  • .
  • Definition. Let f be a function. The tangent line of f at
  • a,f(a)
  • Bernd Schr¨
  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-74
SLIDE 74

logo1 Introduction Velocities Tangent Lines Rates of Change

  • Definition. Let f be a function. The secant line of f through
  • a,f(a)
  • and
  • b,f(b)
  • is the unique straight line that goes

through the points

  • a,f(a)
  • and
  • b,f(b)
  • .
  • Definition. Let f be a function. The tangent line of f at
  • a,f(a)
  • (if it exists) is the unique line that goes through
  • a,f(a)
  • whose slope is

lim

b→a

f(b)−f(a) b−a .

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-75
SLIDE 75

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-76
SLIDE 76

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

lim

b→−1

f(b)−f(−1) b−(−1)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-77
SLIDE 77

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

lim

b→−1

f(b)−f(−1) b−(−1) = lim

b→−1

b3 −4b+1−4 b+1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-78
SLIDE 78

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

lim

b→−1

f(b)−f(−1) b−(−1) = lim

b→−1

b3 −4b+1−4 b+1 = lim

b→−1

b3 −4b−3 b+1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-79
SLIDE 79

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

lim

b→−1

f(b)−f(−1) b−(−1) = lim

b→−1

b3 −4b+1−4 b+1 = lim

b→−1

b3 −4b−3 b+1 = lim

h→0

(−1+h)3 −4(−1+h)−3 (−1+h)+1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-80
SLIDE 80

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

lim

b→−1

f(b)−f(−1) b−(−1) = lim

b→−1

b3 −4b+1−4 b+1 = lim

b→−1

b3 −4b−3 b+1 = lim

h→0

(−1+h)3 −4(−1+h)−3 (−1+h)+1 = lim

h→0

−1+3h−3h2 +h3 +4−4h−3 h

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-81
SLIDE 81

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

lim

b→−1

f(b)−f(−1) b−(−1) = lim

b→−1

b3 −4b+1−4 b+1 = lim

b→−1

b3 −4b−3 b+1 = lim

h→0

(−1+h)3 −4(−1+h)−3 (−1+h)+1 = lim

h→0

−1+3h−3h2 +h3 +4−4h−3 h = lim

h→0

h3 −3h2 −h h

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-82
SLIDE 82

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

lim

b→−1

f(b)−f(−1) b−(−1) = lim

b→−1

b3 −4b+1−4 b+1 = lim

b→−1

b3 −4b−3 b+1 = lim

h→0

(−1+h)3 −4(−1+h)−3 (−1+h)+1 = lim

h→0

−1+3h−3h2 +h3 +4−4h−3 h = lim

h→0

h3 −3h2 −h h = lim

h→0

h

  • h2 −3h−1
  • h

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-83
SLIDE 83

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

lim

b→−1

f(b)−f(−1) b−(−1) = lim

b→−1

b3 −4b+1−4 b+1 = lim

b→−1

b3 −4b−3 b+1 = lim

h→0

(−1+h)3 −4(−1+h)−3 (−1+h)+1 = lim

h→0

−1+3h−3h2 +h3 +4−4h−3 h = lim

h→0

h3 −3h2 −h h = lim

h→0

h

  • h2 −3h−1
  • h

= lim

h→0h2 −3h−1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-84
SLIDE 84

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

lim

b→−1

f(b)−f(−1) b−(−1) = lim

b→−1

b3 −4b+1−4 b+1 = lim

b→−1

b3 −4b−3 b+1 = lim

h→0

(−1+h)3 −4(−1+h)−3 (−1+h)+1 = lim

h→0

−1+3h−3h2 +h3 +4−4h−3 h = lim

h→0

h3 −3h2 −h h = lim

h→0

h

  • h2 −3h−1
  • h

= lim

h→0h2 −3h−1 = −1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-85
SLIDE 85

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-86
SLIDE 86

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

Slope: −1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-87
SLIDE 87

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

Slope: −1, point: (−1,4).

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-88
SLIDE 88

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

Slope: −1, point: (−1,4). y = mx+b

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-89
SLIDE 89

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

Slope: −1, point: (−1,4). y = mx+b 4

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-90
SLIDE 90

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

Slope: −1, point: (−1,4). y = mx+b 4 = (−1)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-91
SLIDE 91

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

Slope: −1, point: (−1,4). y = mx+b 4 = (−1)(−1)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-92
SLIDE 92

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

Slope: −1, point: (−1,4). y = mx+b 4 = (−1)(−1)+b

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-93
SLIDE 93

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

Slope: −1, point: (−1,4). y = mx+b 4 = (−1)(−1)+b b = 3

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-94
SLIDE 94

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

Slope: −1, point: (−1,4). y = mx+b 4 = (−1)(−1)+b b = 3 y = −x+3

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-95
SLIDE 95

logo1 Introduction Velocities Tangent Lines Rates of Change

Compute the Equation of the Tangent Line of f(x) = x3 −4x+1 at a = −1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-96
SLIDE 96

logo1 Introduction Velocities Tangent Lines Rates of Change

Estimate the Slope of the Tangent Line at x = 1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-97
SLIDE 97

logo1 Introduction Velocities Tangent Lines Rates of Change

Estimate the Slope of the Tangent Line at x = 1

✲ ✻

−3 −2 −1 1 2 3 −1 1 2 3 4 5 Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-98
SLIDE 98

logo1 Introduction Velocities Tangent Lines Rates of Change

Estimate the Slope of the Tangent Line at x = 1

✲ ✻

−3 −2 −1 1 2 3 −1 1 2 3 4 5 Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-99
SLIDE 99

logo1 Introduction Velocities Tangent Lines Rates of Change

Estimate the Slope of the Tangent Line at x = 1

✲ ✻

−3 −2 −1 1 2 3 −1 1 2 3 4 5

slope ≈ 2

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-100
SLIDE 100

logo1 Introduction Velocities Tangent Lines Rates of Change

Definition.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-101
SLIDE 101

logo1 Introduction Velocities Tangent Lines Rates of Change

  • Definition. Let f be a function.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-102
SLIDE 102

logo1 Introduction Velocities Tangent Lines Rates of Change

  • Definition. Let f be a function. We define the average rate of

change of f in the interval [a,b]

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-103
SLIDE 103

logo1 Introduction Velocities Tangent Lines Rates of Change

  • Definition. Let f be a function. We define the average rate of

change of f in the interval [a,b] by ravg := f(b)−f(a) b−a .

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-104
SLIDE 104

logo1 Introduction Velocities Tangent Lines Rates of Change

  • Definition. Let f be a function. We define the average rate of

change of f in the interval [a,b] by ravg := f(b)−f(a) b−a . Definition.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-105
SLIDE 105

logo1 Introduction Velocities Tangent Lines Rates of Change

  • Definition. Let f be a function. We define the average rate of

change of f in the interval [a,b] by ravg := f(b)−f(a) b−a .

  • Definition. Let f be a function.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-106
SLIDE 106

logo1 Introduction Velocities Tangent Lines Rates of Change

  • Definition. Let f be a function. We define the average rate of

change of f in the interval [a,b] by ravg := f(b)−f(a) b−a .

  • Definition. Let f be a function. We define the instantaneous

rate of change of f at a

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-107
SLIDE 107

logo1 Introduction Velocities Tangent Lines Rates of Change

  • Definition. Let f be a function. We define the average rate of

change of f in the interval [a,b] by ravg := f(b)−f(a) b−a .

  • Definition. Let f be a function. We define the instantaneous

rate of change of f at a by rinst(a) := lim

b→a

f(b)−f(a) b−a .

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-108
SLIDE 108

logo1 Introduction Velocities Tangent Lines Rates of Change

The Charge of a Laptop Battery is Measured Every 6min, Giving the Data

  • Below. Estimate the Rate of Change of the Charge (the Current) at t = 30min.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-109
SLIDE 109

logo1 Introduction Velocities Tangent Lines Rates of Change

The Charge of a Laptop Battery is Measured Every 6min, Giving the Data

  • Below. Estimate the Rate of Change of the Charge (the Current) at t = 30min.

time t[min] charge q[mAh] 6 400 12 760 18 1080 24 1360 30 1600 36 1800 42 1960 48 2080 54 2160 60 2200

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-110
SLIDE 110

logo1 Introduction Velocities Tangent Lines Rates of Change

The Charge of a Laptop Battery is Measured Every 6min, Giving the Data

  • Below. Estimate the Rate of Change of the Charge (the Current) at t = 30min.

time t[min] charge q[mAh] 6 400 12 760 18 1080 24 1360 30 1600 36 1800 42 1960 48 2080 54 2160 60 2200

q(36min)−q(30min) 36min−30min

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-111
SLIDE 111

logo1 Introduction Velocities Tangent Lines Rates of Change

The Charge of a Laptop Battery is Measured Every 6min, Giving the Data

  • Below. Estimate the Rate of Change of the Charge (the Current) at t = 30min.

time t[min] charge q[mAh] 6 400 12 760 18 1080 24 1360 30 1600 36 1800 42 1960 48 2080 54 2160 60 2200

q(36min)−q(30min) 36min−30min = 1800mAh−1600mAh 0.6h−0.5h

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-112
SLIDE 112

logo1 Introduction Velocities Tangent Lines Rates of Change

The Charge of a Laptop Battery is Measured Every 6min, Giving the Data

  • Below. Estimate the Rate of Change of the Charge (the Current) at t = 30min.

time t[min] charge q[mAh] 6 400 12 760 18 1080 24 1360 30 1600 36 1800 42 1960 48 2080 54 2160 60 2200

q(36min)−q(30min) 36min−30min = 1800mAh−1600mAh 0.6h−0.5h = 2000mA

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-113
SLIDE 113

logo1 Introduction Velocities Tangent Lines Rates of Change

The Charge of a Laptop Battery is Measured Every 6min, Giving the Data

  • Below. Estimate the Rate of Change of the Charge (the Current) at t = 30min.

time t[min] charge q[mAh] 6 400 12 760 18 1080 24 1360 30 1600 36 1800 42 1960 48 2080 54 2160 60 2200

q(36min)−q(30min) 36min−30min = 1800mAh−1600mAh 0.6h−0.5h = 2000mA = 2A

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-114
SLIDE 114

logo1 Introduction Velocities Tangent Lines Rates of Change

The Charge of a Laptop Battery is Measured Every 6min, Giving the Data

  • Below. Estimate the Rate of Change of the Charge (the Current) at t = 30min.

time t[min] charge q[mAh] 6 400 12 760 18 1080 24 1360 30 1600 36 1800 42 1960 48 2080 54 2160 60 2200

q(36min)−q(30min) 36min−30min = 1800mAh−1600mAh 0.6h−0.5h = 2000mA = 2A q(30min)−q(24min) 30min−24min

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-115
SLIDE 115

logo1 Introduction Velocities Tangent Lines Rates of Change

The Charge of a Laptop Battery is Measured Every 6min, Giving the Data

  • Below. Estimate the Rate of Change of the Charge (the Current) at t = 30min.

time t[min] charge q[mAh] 6 400 12 760 18 1080 24 1360 30 1600 36 1800 42 1960 48 2080 54 2160 60 2200

q(36min)−q(30min) 36min−30min = 1800mAh−1600mAh 0.6h−0.5h = 2000mA = 2A q(30min)−q(24min) 30min−24min = 1600mAh−1360mAh 0.5h−0.4h

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-116
SLIDE 116

logo1 Introduction Velocities Tangent Lines Rates of Change

The Charge of a Laptop Battery is Measured Every 6min, Giving the Data

  • Below. Estimate the Rate of Change of the Charge (the Current) at t = 30min.

time t[min] charge q[mAh] 6 400 12 760 18 1080 24 1360 30 1600 36 1800 42 1960 48 2080 54 2160 60 2200

q(36min)−q(30min) 36min−30min = 1800mAh−1600mAh 0.6h−0.5h = 2000mA = 2A q(30min)−q(24min) 30min−24min = 1600mAh−1360mAh 0.5h−0.4h = 2400mA

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-117
SLIDE 117

logo1 Introduction Velocities Tangent Lines Rates of Change

The Charge of a Laptop Battery is Measured Every 6min, Giving the Data

  • Below. Estimate the Rate of Change of the Charge (the Current) at t = 30min.

time t[min] charge q[mAh] 6 400 12 760 18 1080 24 1360 30 1600 36 1800 42 1960 48 2080 54 2160 60 2200

q(36min)−q(30min) 36min−30min = 1800mAh−1600mAh 0.6h−0.5h = 2000mA = 2A q(30min)−q(24min) 30min−24min = 1600mAh−1360mAh 0.5h−0.4h = 2400mA = 2.4A

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-118
SLIDE 118

logo1 Introduction Velocities Tangent Lines Rates of Change

The Charge of a Laptop Battery is Measured Every 6min, Giving the Data

  • Below. Estimate the Rate of Change of the Charge (the Current) at t = 30min.

time t[min] charge q[mAh] 6 400 12 760 18 1080 24 1360 30 1600 36 1800 42 1960 48 2080 54 2160 60 2200

q(36min)−q(30min) 36min−30min = 1800mAh−1600mAh 0.6h−0.5h = 2000mA = 2A q(30min)−q(24min) 30min−24min = 1600mAh−1360mAh 0.5h−0.4h = 2400mA = 2.4A 2.4A+2A 2

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-119
SLIDE 119

logo1 Introduction Velocities Tangent Lines Rates of Change

The Charge of a Laptop Battery is Measured Every 6min, Giving the Data

  • Below. Estimate the Rate of Change of the Charge (the Current) at t = 30min.

time t[min] charge q[mAh] 6 400 12 760 18 1080 24 1360 30 1600 36 1800 42 1960 48 2080 54 2160 60 2200

q(36min)−q(30min) 36min−30min = 1800mAh−1600mAh 0.6h−0.5h = 2000mA = 2A q(30min)−q(24min) 30min−24min = 1600mAh−1360mAh 0.5h−0.4h = 2400mA = 2.4A 2.4A+2A 2 = 2.2A

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-120
SLIDE 120

logo1 Introduction Velocities Tangent Lines Rates of Change

The Charge of a Laptop Battery is Measured Every 6min, Giving the Data

  • Below. Estimate the Rate of Change of the Charge (the Current) at t = 30min.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-121
SLIDE 121

logo1 Introduction Velocities Tangent Lines Rates of Change

The Charge of a Laptop Battery is Measured Every 6min, Giving the Data

  • Below. Estimate the Rate of Change of the Charge (the Current) at t = 30min.

✲ 12 6 24 18 36 30 48 42 60 54 t[min] 500 1000 1500 2000 Q[mAh] s s s s s s s s s s ✻

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-122
SLIDE 122

logo1 Introduction Velocities Tangent Lines Rates of Change

The Charge of a Laptop Battery is Measured Every 6min, Giving the Data

  • Below. Estimate the Rate of Change of the Charge (the Current) at t = 30min.

✲ 12 6 24 18 36 30 48 42 60 54 t[min] 500 1000 1500 2000 Q[mAh] s s s s s s s s s s ✻

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-123
SLIDE 123

logo1 Introduction Velocities Tangent Lines Rates of Change

The Charge of a Laptop Battery is Measured Every 6min, Giving the Data

  • Below. Estimate the Rate of Change of the Charge (the Current) at t = 30min.

✲ 12 6 24 18 36 30 48 42 60 54 t[min] 500 1000 1500 2000 Q[mAh] s s s s s s s s s s ✻

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-124
SLIDE 124

logo1 Introduction Velocities Tangent Lines Rates of Change

Estimate the Slope of the Tangent Line at x = 1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-125
SLIDE 125

logo1 Introduction Velocities Tangent Lines Rates of Change

Estimate the Slope of the Tangent Line at x = 1

✲ ✻

−3 −2 −1 1 2 3 −1 1 2 3 4 5 Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-126
SLIDE 126

logo1 Introduction Velocities Tangent Lines Rates of Change

Estimate the Slope of the Tangent Line at x = 1

✲ ✻

−3 −2 −1 1 2 3 −1 1 2 3 4 5

✉

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-127
SLIDE 127

logo1 Introduction Velocities Tangent Lines Rates of Change

Estimate the Slope of the Tangent Line at x = 1

✲ ✻

−3 −2 −1 1 2 3 −1 1 2 3 4 5

✉ ✉

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-128
SLIDE 128

logo1 Introduction Velocities Tangent Lines Rates of Change

Estimate the Slope of the Tangent Line at x = 1

✲ ✻

−3 −2 −1 1 2 3 −1 1 2 3 4 5

✉ ✉

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-129
SLIDE 129

logo1 Introduction Velocities Tangent Lines Rates of Change

Estimate the Slope of the Tangent Line at x = 1

✲ ✻

−3 −2 −1 1 2 3 −1 1 2 3 4 5

✉ ✉

4.25−3 2−1 = 1.25

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-130
SLIDE 130

logo1 Introduction Velocities Tangent Lines Rates of Change

Estimate the Slope of the Tangent Line at x = 1

✲ ✻

−3 −2 −1 1 2 3 −1 1 2 3 4 5

✉ ✉

4.25−3 2−1 = 1.25

✉

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-131
SLIDE 131

logo1 Introduction Velocities Tangent Lines Rates of Change

Estimate the Slope of the Tangent Line at x = 1

✲ ✻

−3 −2 −1 1 2 3 −1 1 2 3 4 5

✉ ✉

4.25−3 2−1 = 1.25

✉

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-132
SLIDE 132

logo1 Introduction Velocities Tangent Lines Rates of Change

Estimate the Slope of the Tangent Line at x = 1

✲ ✻

−3 −2 −1 1 2 3 −1 1 2 3 4 5

✉ ✉

4.25−3 2−1 = 1.25

✉

0−3 0−1 = 3

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?

slide-133
SLIDE 133

logo1 Introduction Velocities Tangent Lines Rates of Change

Estimate the Slope of the Tangent Line at x = 1

✲ ✻

−3 −2 −1 1 2 3 −1 1 2 3 4 5

✉ ✉

4.25−3 2−1 = 1.25

✉

0−3 0−1 = 3 1 2(1.25+3) = 2.125

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Why Do We Need Derivatives?