Explanation Given : f ( x , y ) = { ( x y ) p x 4 + y 4 ( x , y ) ≠ ( 0 , 0 ) 0 ( x , y ) = ( 0 , 0 ) Formula Used : Squeeze Theorem Proof : We show using the polar coordinates x = r cos θ , y = r sin θ that the limit of f(x, y) as ( x , y ) → ( 0 , 0 ) is zero for p>2. This will prove that f is continuous at the origin. Since f is a rational function with non-zero denominator for ( x , y ) ≠ ( 0 , 0 ) , f is continuous there We have lim ( x , y ) → ( 0 , 0 ) f ( x , y ) = lim r → 0 + ( r cos θ ) p ( r sin θ ) p ( r cos θ ) 4 + ( r sin θ ) 4 = lim r → 0 + ( r ) 2 p ( cos θ sin θ ) p r 4 ( cos 4 θ + sin 4 θ ) = lim r → 0 + ( r ) 2 ( p − 2 ) ( cos θ sin θ ) p ( cos 4 θ + sin 4 θ ) We use the following inequalities: | cos 4 θ sin 4 θ | ≤ 1 cos 4 θ + sin 4 θ = ( cos 2 θ + sin 2 θ ) 2 − 2 cos 2 θ sin 2 θ = 1 − 1 2 ( 2 cos θ sin θ ) 2 = 1 − 1 2 sin 2 2 θ ≥ 1 − 1 2 = 1 2 Therefore, 0 ≤ | ( r ) 2 ( p − 2 ) ( cos θ sin θ ) p ( cos 4 θ + sin 4 θ ) | ≤ ( r ) 2 ( p − 2 ) .1 1 2 = 2 r 2 ( p − 2 ) Since p-2>0, lim r → 0 + 2 r 2 ( p − 2 ) = 0 , hence by the Squeeze Theorem the limit in lim r → 0 + ( r ) 2 ( p − 2 ) ( cos θ sin θ ) p ( cos 4 θ + sin 4 θ ) is also zero...

4th Edition

ISBN: 9781319055844

Author: FRANZOSA

Publisher: VST

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Chapter 15, Problem 15CRE

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To prove: $f (x, y) = {\begin{cases} \frac{{(x y)}^{p}}{x^{4} + y^{4}} (x, y) \neq (0, 0) \\ 0 (x, y) = (0, 0) \end{cases}$ is continuous at all (x, y) if p>2 but is discontinuous at (0, 0) if $p \leq 2$ using polar coordinates.

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Chapter 15, Problem 14CRE

Chapter 15, Problem 16CRE

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Chapter 15 Solutions

EBK CALCULUS

Ch. 15.1 - Prob. 1PQ Ch. 15.1 - Prob. 2PQ Ch. 15.1 - Prob. 3PQ Ch. 15.1 - Prob. 4PQ Ch. 15.1 - Prob. 5PQ Ch. 15.1 - Prob. 1E Ch. 15.1 - Prob. 2E Ch. 15.1 - Prob. 3E Ch. 15.1 - Prob. 4E Ch. 15.1 - Prob. 5E

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To prove : f ( x , y ) = { ( x y ) p x 4 + y 4 ( x , y ) ≠ ( 0 , 0 ) 0 ( x , y ) = ( 0 , 0 ) is continuous at all (x, y) if p>2 but is discontinuous at (0, 0) if p ≤ 2 using polar coordinates.

Solution Summary: The author proves that f is continuous at all (x, y) if p>2 but discontinuous at (0,0) using polar coordinates.

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To prove: $f (x, y) = {\begin{cases} \frac{{(x y)}^{p}}{x^{4} + y^{4}} (x, y) \neq (0, 0) \\ 0 (x, y) = (0, 0) \end{cases}$ is continuous at all (x, y) if p>2 but is discontinuous at (0, 0) if $p \leq 2$ using polar coordinates.

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Chapter 15 Solutions

To prove : f ( x , y ) = { ( x y ) p x 4 + y 4 ( x , y ) ≠ ( 0 , 0 ) 0 ( x , y ) = ( 0 , 0 ) is continuous at all (x, y) if p&gt;2 but is discontinuous at (0, 0) if p ≤ 2 using polar coordinates.

Solution Summary: The author proves that f is continuous at all (x, y) if p>2 but discontinuous at (0,0) using polar coordinates.

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To prove: f(x,y)={(xy)px4+y4 (x,y)≠(0,0)0 (x,y)=(0,0) is continuous at all (x, y) if p>2 but is discontinuous at (0, 0) if p≤2 using polar coordinates.

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Chapter 15 Solutions

To prove : f ( x , y ) = { ( x y ) p x 4 + y 4 ( x , y ) ≠ ( 0 , 0 ) 0 ( x , y ) = ( 0 , 0 ) is continuous at all (x, y) if p>2 but is discontinuous at (0, 0) if p ≤ 2 using polar coordinates.

To prove: $f (x, y) = {\begin{cases} \frac{{(x y)}^{p}}{x^{4} + y^{4}} (x, y) \neq (0, 0) \\ 0 (x, y) = (0, 0) \end{cases}$ is continuous at all (x, y) if p>2 but is discontinuous at (0, 0) if $p \leq 2$ using polar coordinates.