Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

4th Edition

ISBN: 9780133178579

Author: Ross L. Finney

Publisher: PEARSON

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Differential Equations

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Question

Chapter 7.4, Problem 41E

To determine

To calculate: The values of p0 and k.

Expert Solution

Answer to Problem 41E

The values of p0 and k

The values of p0=1013 and k=−0.121

Explanation of Solution

Given information: Suppose that the pressure at sea level is 1013 millibars and that the pressure at an altitude of 20 km is 90 millibars

Differential Equation: dpdh=kp

Initial condition: p=p0 when h=0

Calculation:

Suppose that the pressure at sea level is 1013 millibars and that the pressure at an altitude of 20 km is 90 millibars

Differential Equation: dpdh=kp

Initial condition: p=p0 when h=0

Initial condition: p=1013 when h=0

dpdh=kpdpp=kdhlnp=kh+Cp=ekh+Cp=Cekhat h=0, p=10131013=Ce0C=1013p=1013ekhat h=20, p=9090=1013e20k20k=ln(901031)k=−0.121p=1013e−0.121h

To determine

To calculate: The atmospheric pressure at h=50 km .

Expert Solution

Answer to Problem 41E

The atmospheric pressure at h=50 km is p=2.38 millibars .

Explanation of Solution

Given information: Suppose that the pressure at sea level is 1013 millibars and that the pressure at an altitude of 20 km is 90 millibars

Differential Equation: dpdh=kp

Initial condition: p=p0 when h=0

Pressure

p=1013e−0.121h

Calculation:

Pressure

p=1013e−0.121h

pressure at h=50 km

p=1013e−0.121hp=1013e−0.121×50p=2.38 millibars

To determine

To calculate: at what altitude does the pressure equal 900 millibars .

Expert Solution

Answer to Problem 41E

at 0.977 km altitude the pressure equal 900 millibars .

Explanation of Solution

Given information: Suppose that the pressure at sea level is 1013 millibars and that the pressure at an altitude of 20 km is 90 millibars

Differential Equation: dpdh=kp

Initial condition: p=p0 when h=0

Pressure

p=1013e−0.121h

Calculation:

at what altitude does the pressure equal 900 millibars

Pressure

p=1013e−0.121h

p=900 millibarsp=1013e−0.121hp=900 millibars900=1013e−0.121h−0.121h=ln(9001031)h=0.977 km

Chapter 7.4, Problem 40E

Chapter 7.4, Problem 42E

Chapter 7 Solutions

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

Ch. 7.1 - Prob. 1QR Ch. 7.1 - Prob. 2QR Ch. 7.1 - Prob. 3QR Ch. 7.1 - Prob. 4QR Ch. 7.1 - Prob. 5QR Ch. 7.1 - Prob. 6QR Ch. 7.1 - Prob. 7QR Ch. 7.1 - Prob. 8QR Ch. 7.1 - Prob. 9QR Ch. 7.1 - Prob. 10QR

Ch. 7.1 - Prob. 11QR Ch. 7.1 - Prob. 12QR Ch. 7.1 - Prob. 1E Ch. 7.1 - Prob. 2E Ch. 7.1 - Prob. 3E Ch. 7.1 - Prob. 4E Ch. 7.1 - Prob. 5E Ch. 7.1 - Prob. 6E Ch. 7.1 - Prob. 7E Ch. 7.1 - Prob. 8E Ch. 7.1 - Prob. 9E Ch. 7.1 - Prob. 10E Ch. 7.1 - Prob. 11E Ch. 7.1 - Prob. 12E Ch. 7.1 - Prob. 13E Ch. 7.1 - Prob. 14E Ch. 7.1 - Prob. 15E Ch. 7.1 - Prob. 16E Ch. 7.1 - Prob. 17E Ch. 7.1 - Prob. 18E Ch. 7.1 - Prob. 19E Ch. 7.1 - Prob. 20E Ch. 7.1 - Prob. 21E Ch. 7.1 - Prob. 22E Ch. 7.1 - Prob. 23E Ch. 7.1 - Prob. 24E Ch. 7.1 - Prob. 25E Ch. 7.1 - Prob. 26E Ch. 7.1 - Prob. 27E Ch. 7.1 - Prob. 28E Ch. 7.1 - Prob. 29E Ch. 7.1 - Prob. 30E Ch. 7.1 - Prob. 31E Ch. 7.1 - Prob. 32E Ch. 7.1 - Prob. 33E Ch. 7.1 - Prob. 34E Ch. 7.1 - Prob. 35E Ch. 7.1 - Prob. 36E Ch. 7.1 - Prob. 37E Ch. 7.1 - Prob. 38E Ch. 7.1 - Prob. 39E Ch. 7.1 - Prob. 40E Ch. 7.1 - Prob. 41E Ch. 7.1 - Prob. 42E Ch. 7.1 - Prob. 43E Ch. 7.1 - Prob. 44E Ch. 7.1 - Prob. 45E Ch. 7.1 - Prob. 46E Ch. 7.1 - Prob. 47E Ch. 7.1 - Prob. 48E Ch. 7.1 - 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The values of p 0 and k.

Concept explainers

Videos

To calculate: The values of p0 and k.

Answer to Problem 41E

Explanation of Solution

To calculate: The atmospheric pressure at h=50 km .

Answer to Problem 41E

Explanation of Solution

To calculate: at what altitude does the pressure equal 900 millibars .

Answer to Problem 41E

Explanation of Solution

Chapter 7 Solutions

Additional Math Textbook Solutions