Explanation Given information: The equation ∂ z ∂ x = − F x F z , ∂ z ∂ y = − F y F z Formula used: when x, y and z are related by an equation F ( x , y , z ) = 0 , ( ∂ z ∂ x ) y means same as ( ∂ z ∂ x ) and we have an equation which gives ∂ z ∂ x = − F x F z , ∂ z ∂ y = − F y F z Calculation: To prove the cyclic relation ( ∂ z ∂ x ) y ( ∂ x ∂ y ) z ( ∂ y ∂ z ) x = − 1 First from the equation ∂ z ∂ x = − F x F z , ∂ z ∂ y = − F y F z , we get ∂ z ∂ x = − F x F z , & To determine (b) To prove the cyclic relation ( ∂ z ∂ x ) y ( ∂ x ∂ y ) z ( ∂ y ∂ z ) x = − 1 When, F ( x , y , z ) = x + y + z = 0 To determine (c) To prove the cyclic relation ( ∂ T ∂ P ) P ( ∂ P ∂ V ) T ( ∂ V ∂ T ) P = − 1 For variables P, V& T in the ideal gas law P V − n R T = 0

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ISBN: 9781319050740

Author: Jon Rogawski, Colin Adams, Robert Franzosa

Publisher: W. H. Freeman

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Chapter 14.6, Problem 42E

To determine

(a)

To prove the cyclic relation (∂z∂x)y(∂x∂y)z(∂y∂z)x=−1

With the use of equation ∂z∂x=−FxFz,∂z∂y=−FyFz

To determine

(b)

To prove the cyclic relation (∂z∂x)y(∂x∂y)z(∂y∂z)x=−1

When, F(x,y,z)=x+y+z=0

To determine

(c)

To prove the cyclic relation (∂T∂P)P(∂P∂V)T(∂V∂T)P=−1

For variables P, V& T in the ideal gas law PV−nRT=0

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Chapter 14.6, Problem 41E

Chapter 14.6, Problem 43E

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According to Newton's law of universal gravitation, the force F between two bodies of constant mass GmM m and M is given by the formula F = , where G is the gravitational constant and d is the d² distance between the bodies. a. Suppose that G, m, and M are constants. Find the rate of change of force F with respect to distance d. F' (d) 2GmM b. Find the rate of change of force F with gravitational constant G = 6.67 × 10-¹¹ Nm²/kg², on two bodies 5 meters apart, each with a mass of 250 kilograms. Answer in scientific notation, rounding to 2 decimal places. -6.67x10 N/m syntax incomplete.

Chapter 14 Solutions

Calculus: Early Transcendentals

Ch. 14.1 - Prob. 1PQ Ch. 14.1 - Prob. 2PQ Ch. 14.1 - Prob. 3PQ Ch. 14.1 - Prob. 4PQ Ch. 14.1 - Prob. 5PQ Ch. 14.1 - Prob. 1E Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E Ch. 14.1 - Prob. 5E

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(a) To prove the cyclic relation ( ∂ z ∂ x ) y ( ∂ x ∂ y ) z ( ∂ y ∂ z ) x = − 1 With the use of equation ∂ z ∂ x = − F x F z , ∂ z ∂ y = − F y F z

Solution Summary: The author explains how to prove the cyclic relation with the equation partial x=f_x, y, and z.

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(a)

To prove the cyclic relation (∂z∂x)y(∂x∂y)z(∂y∂z)x=−1

With the use of equation ∂z∂x=−FxFz,∂z∂y=−FyFz

(b)

To prove the cyclic relation (∂z∂x)y(∂x∂y)z(∂y∂z)x=−1

When, F(x,y,z)=x+y+z=0

(c)

To prove the cyclic relation (∂T∂P)P(∂P∂V)T(∂V∂T)P=−1

For variables P, V& T in the ideal gas law PV−nRT=0

Want to see the full answer?

Chapter 14 Solutions

(a) To prove the cyclic relation ( ∂ z ∂ x ) y ( ∂ x ∂ y ) z ( ∂ y ∂ z ) x = − 1 With the use of equation ∂ z ∂ x = − F x F z , ∂ z ∂ y = − F y F z

Solution Summary: The author explains how to prove the cyclic relation with the equation partial x=f_x, y, and z.

Videos

(a) To prove the cyclic relation (∂z∂x)y(∂x∂y)z(∂y∂z)x=−1 With the use of equation ∂z∂x=−FxFz,∂z∂y=−FyFz

(b) To prove the cyclic relation (∂z∂x)y(∂x∂y)z(∂y∂z)x=−1 When, F(x,y,z)=x+y+z=0

(c) To prove the cyclic relation (∂T∂P)P(∂P∂V)T(∂V∂T)P=−1 For variables P, V& T in the ideal gas law PV−nRT=0

Want to see the full answer?

Chapter 14 Solutions

(a)

To prove the cyclic relation (∂z∂x)y(∂x∂y)z(∂y∂z)x=−1

With the use of equation ∂z∂x=−FxFz,∂z∂y=−FyFz

(b)

To prove the cyclic relation (∂z∂x)y(∂x∂y)z(∂y∂z)x=−1

When, F(x,y,z)=x+y+z=0

(c)

To prove the cyclic relation (∂T∂P)P(∂P∂V)T(∂V∂T)P=−1

For variables P, V& T in the ideal gas law PV−nRT=0