Problem

Car tires need to be inflated properly because ov…

02:19

Problem 53 Medium Difficulty

Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure $ P $ of the gas is inversely proportional to the volume $ V $ of the gas.
(a) Suppose that the pressure of a sample of air that occupies $ 0.106 m^3 at 25^{\circ} C $ is 50 kPa. Write $ V $ as a function of $ P. $
(b) Calculate $ dV\mid dP $ when $ P $ = 50 kPA. What is the meaning of the derivative? What are its units?

Answer

(a) $$P=\frac{5.3}{V} \text { and } V=\frac{5.3}{P}$$
(b) $$P=50, \frac{d V}{d P}=-\frac{5.3}{50^{2}}=-0.00212$$

Discussion

You must be signed in to discuss.

Top Calculus 1 / AB Educators

Watch More Solved Questions in Chapter 3

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

Video Transcript

it's clear. So when you married here. So we have PV equals K. He has a constant. We're gonna substitute 50 for P and 500.106 for ve me get Kate is equal to 5.3 and B is equal to 5.3 Overpay for part B V is equal to 5.3. Overpay p is equal to 50 Killer Paschal's. We're going to find the derivative of RV. This becomes equal to negative 5.3 he the negative to power. This could be rewritten as negative 5.3 over a piece square. Then we plug in 50 when we got negative 0.212 meters cubed over Killa Pasko's. And this represents the rate of change of the volume in respect to the pressure.

Clarissa N.

Topics

Derivatives

Differentiation

Calculus: Early Transcendentals

Chapter 3

Differentiation Rules

Section 1

Derivatives of Polynomials and Exponential Functions

Problem

Car tires need to be inflated properly because ov…

Problem 53 Medium Difficulty

Answer

(a) $$P=\frac{5.3}{V} \text { and } V=\frac{5.3}{P}$$
(b) $$P=50, \frac{d V}{d P}=-\frac{5.3}{50^{2}}=-0.00212$$

More Answers

Topics

Calculus: Early Transcendentals

Discussion

Top Calculus 1 / AB Educators

Catherine R.

Heather Z.

Caleb E.

Joseph L.

Calculus 1 / AB Bootcamp

Precalculus Review - Intro

Functions on the Real Line - Overview

Recommended Videos

Watch More Solved Questions in Chapter 3

Video Transcript

Topics

Calculus: Early Transcendentals

Top Calculus 1 / AB Educators

Catherine R.

Heather Z.

Caleb E.

Joseph L.

Calculus 1 / AB Bootcamp

Precalculus Review - Intro

Functions on the Real Line - Overview

Recommended Videos

Problem

Car tires need to be inflated properly because ov…

Problem 53 Medium Difficulty

Answer

(a) $$P=\frac{5.3}{V} \text { and } V=\frac{5.3}{P}$$(b) $$P=50, \frac{d V}{d P}=-\frac{5.3}{50^{2}}=-0.00212$$

More Answers

Topics

Calculus: Early Transcendentals

Discussion

Top Calculus 1 / AB Educators

Catherine R.

Heather Z.

Caleb E.

Joseph L.

Calculus 1 / AB Bootcamp

Precalculus Review - Intro

Functions on the Real Line - Overview

Recommended Videos

Watch More Solved Questions in Chapter 3

Video Transcript

Topics

Calculus: Early Transcendentals

Top Calculus 1 / AB Educators

Catherine R.

Heather Z.

Caleb E.

Joseph L.

Calculus 1 / AB Bootcamp

Precalculus Review - Intro

Functions on the Real Line - Overview

Recommended Videos

(a) $$P=\frac{5.3}{V} \text { and } V=\frac{5.3}{P}$$
(b) $$P=50, \frac{d V}{d P}=-\frac{5.3}{50^{2}}=-0.00212$$