Problem

If an initial amount $ A_0 $ of money is invested…

01:36

Problem 78 Hard Difficulty

If an object with mass $ m $ is dropped from rest, one model for its speed $ v $ after $ t $ seconds, taking air resistance into account, is
$$ v = \frac{mg}{c}(1 - e^{-ct/m}) $$
where $ g $ is the acceleration due to gravity and $ c $ is a positive constant. (In Chapter 9 we will be able to deduce this equation from the assumption that the air resistance is proportional to the speed of the object; $ c $ is the proportionality constant.)
(a) Calculate $ lim_{t\to \infty} v $. What is the meaning of this limit?
(b) For fixed $ t $, use l'Hospital's Rule to calculate $ lim_{c\to 0^+} v $. What can you conclude about the velocity of a falling object in a vacuum?

Answer

a)$\lim _{t \rightarrow \infty} \frac{m g}{c}\left(1-e^{-c t / m}\right)=\frac{m g}{c}(1-0)=\frac{m g}{c}$
b)$g t$

Watch More Solved Questions in Chapter 4

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Problem 60

Problem 61

Problem 62

Problem 63

Problem 64

Problem 65

Problem 66

Problem 67

Problem 68

Problem 69

Problem 70

Problem 71

Problem 72

Problem 73

Problem 74

Problem 75

Problem 76

Problem 77

Problem 78

Problem 79

Problem 80

Problem 81

Problem 82

Problem 83

Problem 84

Problem 85

Problem 86

Problem 87

Problem 88

Problem 89

Problem 90

Problem 91

Problem 92

Video Transcript

So for this problem we're having tea, go to infinity. Um and we see that when T goes to infinity, we end up getting things cancel out, and we get MG oversee that's the first portion. But then for part B, what we're gonna do is we're letting when we have the indeterminate form As C approaches zero from the right, what we end up having is to take the derivative of the top and bottom. So once we use the hotel's rule, we end up getting by treating see as a mhm variable, we end up getting the limit. Uh see approaches zero of energy times T over em of E. To the negative C. T. Over end. Then as we when we plug in uh C two B zero, we end up getting just G. T. As the final answer for that.

Carson M.

California Baptist University

Related Courses

Calculus 1 / AB

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

Section 4

Indeterminate Forms and l'Hospital's Rule

Problem

If an initial amount $ A_0 $ of money is invested…

Problem 78 Hard Difficulty

Answer

a)$\lim _{t \rightarrow \infty} \frac{m g}{c}\left(1-e^{-c t / m}\right)=\frac{m g}{c}(1-0)=\frac{m g}{c}$
b)$g t$

More Answers

Related Courses

Calculus: Early Transcendentals

Related Topics

Discussion

Top Calculus 2 / BC Educators

Grace H.

Catherine R.

Kristen K.

Joseph L.

Calculus 2 / BC Courses

Integration Review - Intro

Definite vs Indefinite

Recommended Videos

Watch More Solved Questions in Chapter 4

Video Transcript

Calculus: Early Transcendentals

Catherine R.

Kayleah T.

Kristen K.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

a)$\lim _{t \rightarrow \infty} \frac{m g}{c}\left(1-e^{-c t / m}\right)=\frac{m g}{c}(1-0)=\frac{m g}{c}$b)$g t$

Calculus: Early Transcendentals

Grace H.

Catherine R.

Kristen K.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

Catherine R.

Kayleah T.

Kristen K.

Joseph L.

a)$\lim _{t \rightarrow \infty} \frac{m g}{c}\left(1-e^{-c t / m}\right)=\frac{m g}{c}(1-0)=\frac{m g}{c}$
b)$g t$