Explanation Formula Used: Quotient rule d d t ( y x ) = x d y d t − y d x d t x 2 Area enclosed by C = ∮ C x d y = ∮ C y d x = 1 2 ∫ C x d y − y d x Given: ( x 2 + y 2 ) 2 = x y t = y x Calculation: We have been given, ( x 2 + y 2 ) 2 = x y . We have to find parameterization of the lemniscates. ( x 2 + y 2 ) 2 = x y t = y x y = x t ( x 2 + ( x t ) 2 ) 2 = x ( x t ) ( x 2 + x 2 t 2 ) 2 = x 2 t ( x 2 ( 1 + t 2 ) ) 2 = x 2 t x 4 ( 1 + t 2 ) 2 = x 2 t x 2 = t ( 1 + t 2 ) 2 x = t 1 + t 2 x = t 1 2 1 + t 2 Now let’s find y parameter, t = y x x = y t Therefore, ( x 2 + y 2 ) 2 = x y ( y 2 t 2 + y 2 ) 2 = y t ( y ) y 4 ( 1 t 2 + 1 ) 2 = y 2 t y 2 ( 1 + t 2 t 2 ) 2 = 1 t y 2 ( 1 + t 2 ) 2 t 4 = 1 t y 2 = 1 t ⋅ t 4 ( 1 + t 2 ) 2 y = t 3 2 1 + t 2 Therefore, Parameters of the lemniscates are x = t 1 2 1 + t 2 and y ...

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 18.1, Problem 24E

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To find: the parameterization of the lemniscates ${(x^{2} + y^{2})}^{2} = x y$ and area of one loop of it.

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Chapter 18.1, Problem 23E

Chapter 18.1, Problem 25E

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Chapter 18 Solutions

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the parameterization of the lemniscates ( x 2 + y 2 ) 2 = x y and area of one loop of it.

Solution Summary: The author explains how to find the parameterization of the lemniscates and the area of one loop of it.

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To find: the parameterization of the lemniscates ${(x^{2} + y^{2})}^{2} = x y$ and area of one loop of it.

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Chapter 18 Solutions

the parameterization of the lemniscates ( x 2 + y 2 ) 2 = x y and area of one loop of it.

Solution Summary: The author explains how to find the parameterization of the lemniscates and the area of one loop of it.

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To find: the parameterization of the lemniscates (x2+y2)2=xy and area of one loop of it.

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Chapter 18 Solutions

To find: the parameterization of the lemniscates ${(x^{2} + y^{2})}^{2} = x y$ and area of one loop of it.