Explanation Given: It is given that, r = 〈 x , y , x 〉 and e r = r ‖ r ‖ The function f ( x , y , z ) = F ( r ) depends only on the distance from the origin r = ‖ r ‖ = x 2 + y 2 + z 2 . Formula used: Chain rule: let f ( x ( s , t ) , y ( s , t ) , z ( s , t ) ) be a differentiable function of x, y and z. And x, y, z are differentiable functions of s and t. Then, ∂ f ∂ s = ∂ f ∂ x ∂ x ∂ s + ∂ f ∂ y ∂ y ∂ s + ∂ f ∂ z ∂ z ∂ s And ∂ f ∂ t = ∂ f ∂ x ∂ x ∂ t + ∂ f ∂ y ∂ y ∂ t + ∂ f ∂ z ∂ z ∂ t Power rule : d d x ( x n ) = n x n − 1 Consider f ( x , y , z ) = F ( r ) Apply the chain rule, ∂ f ∂ x = ∂ f ∂ r ∂ r ∂ x = F ′ ( r ) ⋅ ∂ r ∂ x …… (1) ∂ f ∂ y = ∂ f ∂ r ∂ r ∂ y = F ′ ( r ) ⋅ ∂ r ∂ y …… (2) ∂ f ∂ z = ∂ f ∂ r ∂ r ∂ z = F ′ ( r ) ⋅ ∂ r ∂ z …… (1) Now, it is given that r = ‖ r ‖ = x 2 + y 2 + z 2 Differentiate r partially with respect to x, y and z respectively

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ISBN: 9781319279813

Author: Rogawski

Publisher: MAC HIGHER

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Domain

In simple mathematics terms, a domain is defined as the space on which all the possible inputs for the function will lie. It can also be expressed as a set of departures for the required function.

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Chapter 14.6, Problem 35E

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∇f=F′(r)er

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Chapter 14.6, Problem 34E

Chapter 14.6, Problem 36E

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Chapter 14 Solutions

CALCULUS:EARLY TRANS.-SAPLINGPLUS

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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∇ f = F ′ ( r ) e r

Solution Summary: The author explains that the function f(x,y,z)=F (r ) depends only on the distance from the origin.

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∇f=F′(r)er

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Chapter 14 Solutions

∇ f = F ′ ( r ) e r

Solution Summary: The author explains that the function f(x,y,z)=F (r ) depends only on the distance from the origin.

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To show: ∇f=F′(r)er

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Chapter 14 Solutions

To show:

∇f=F′(r)er