Problem

In the theory of relativity, the energy of a part…

04:27

Problem 55 Easy Difficulty

In the theory of relativity, the mass of a particle is
$$ m = \dfrac{m_0}{\sqrt{1 - v^2/c^2}} $$
where $ m_0 $ is the rest mass of the particle, $ m $ is the mass when the particle moves with speed $ v $ relative to the observer, and $ c $ is the speed of light. Sketch the graph of $ m $ as a function of $ v $.

Answer

see solution

Topics

Derivatives

Differentiation

Volume

Book Cover for Calculus: Early Tra…

Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

Section 5

Summary of Curve Sketching

Discussion

You must be signed in to discuss.

Top Calculus 2 / BC Educators

Catherine R.

Missouri State University

Anna Marie V.

Campbell University

Caleb E.

Baylor University

Kristen K.

University of Michigan - Ann Arbor

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 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

Video Transcript

were sketching our function m as a function of V. So I just called the f a V. Um, so this is our function. So first, to find her domain, we have that zero to seize our domain because it must be positive. Under the radical intercepts we have from zero to em initial, because if we set our extra zero, we just have a minute Shal over one symmetry none. Assam toots. We have a vertical ass in tow vehicle, See? And the reason for that is if he does equal, see, we have one. So one minus 10 and we cannot have our function over zero. All right, so of prime of the equals. So we're doing the first derivative to see where it's increasing. This is her first derivative C squared, minus B squared, cubed. All right. And we see that this is increasing over our entire domain, which is from zero to see. So there are no local men's or Max is. And if we find a double prime Seacon cavity you see, see on initial times to V squared plus C squared over V squared minus C squared. That's all squared times radical C squared minus B squared. And we see that's also above zero everywhere over our our domain zero to see. So that means it's Kong cave up over our domain. Right? And now we can graft or function. So you know it's gonna be our X axis is gonna be V. And our Y axis is gonna be because here M is equal to F V. All right, so we haven't intercept at zero. I am initial. We haven't asked him to at sea, and our graph will approach but not touch, See?

Jamie M.

Topics

Derivatives

Differentiation

Volume

Book Cover for Calculus: Early Tra…

Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

Section 5

Summary of Curve Sketching

Top Calculus 2 / BC Educators

Catherine R.

Missouri State University

Anna Marie V.

Campbell University

Caleb E.

Baylor University

Kristen K.

University of Michigan - Ann Arbor