Explanation Given: The equations: ( a ) c u r l ( ∇ f ) = 0 ( b ) d i v ( ∇ f ) = 0 ( c ) d i v ( c u r l ( F ) ) = 0 ( d ) ∇ ( d i v ( F ) ) = 0 Let us analyze each of the given equations one by one: ( a ) c u r l ( ∇ f ) = 0 We can write, ∇ f = 〈 f x , f y , f z 〉 Thus, c u r l ( ∇ f ) = c u r l 〈 f x , f y , f z 〉 c u r l ( ∇ f ) = | i j k δ δ x δ δ y δ δ z f x f y f z | = 〈 f y z − f z y , f x z − f z x , f y x − f x y 〉 c u r l ( ∇ f ) = 〈 0 , 0 , 0 〉 = 0 Therefore, this equation is an identity. ( b ) d i v ( ∇ f ) = 0 We can write, ∇ f = 〈 f x , f y , f z 〉 Thus, d i v ( ∇ f ) = 〈 δ δ x , δ δ y , δ δ z 〉 〈 f x , f y , f z 〉 d i v ( ∇ f ) = f x x + f y y + f z z This is not necessarily zero. Hence, this equation is not an identity. A counter example is f = x 2 y 2 z 2 ...

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Author: Rogawski

Publisher: MAC HIGHER

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Chapter 18, Problem 12CRE

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Chapter 18, Problem 11CRE

Chapter 18, Problem 13CRE

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Chapter 18 Solutions

CALCULUS W/SAPLING ACCESS >IC<

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Whether each of the given equations are identities or not. If any equation is not an identity, provide a counter example.

Solution Summary: The author analyzes each of the given equations one by one. If any equation is not an identity, provide a counter example.

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Whether each of the given equations are identities or not. If any equation is not an identity, provide a counter example.

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Chapter 18 Solutions