Explanation Given: Use the inverse in exercise 9,that is G − 1 ( x , y ) = ( 3 x − y , − 5 x + 2 y ) Calculation: Since ( x , y ) = ( 2 , 1 ) . Therefore x = 2 and y = 1 Now, G − 1 ( 2 , 1 ) = ( 3 × 2 − 1 , − 5 × 2 + 2 × 1 ) G − 1 ( 2 , 1 ) = ( 6 − 1 , − 10 + 2 ) G − 1 ( 2 , 1 ) = ( 5 , − 8 ) Hence, The point is ( 5 , − 8 ) in the u v - plane which maps to ( 2 , 1 ) in the x y -plane. (b) Given: Use the inverse in exercise 9,that is G − 1 ( x , y ) = ( 3 x − y , − 5 x + 2 y ) Calculation: Since ( x , y ) = ( − 2 , 1 ) and ( 3 , 4 ) ...

4th Edition

ISBN: 9781319434540

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 16.6, Problem 10E

To determine

To find:

Use the inverse in exercise 9 to find

(a)A point in the $u v$ plane mapping to $(2, 1)$ .

(b)a segment in the $u v$ plane to the segment joining $(- 2, 1) and (3, 4)$ .

Expert Solution & Answer

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Chapter 16.6, Problem 9E

Chapter 16.6, Problem 11E

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Chapter 16 Solutions

ACHIEVE STANDALONE ACCESS F/ CALC 4E

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Knowledge Booster

Use the inverse in exercise 9 to find (a)A point in the u v plane mapping to ( 2 , 1 ) . (b)a segment in the u v plane to the segment joining ( − 2 , 1 ) and ( 3 , 4 ) .

Solution Summary: The author explains how to find a point in the u v plane mapping to ( 2, 1 ).

To find:

Use the inverse in exercise 9 to find

(a)A point in the $u v$ plane mapping to $(2, 1)$ .

(b)a segment in the $u v$ plane to the segment joining $(- 2, 1) and (3, 4)$ .

Want to see the full answer?

Chapter 16 Solutions

Use the inverse in exercise 9 to find (a)A point in the u v plane mapping to ( 2 , 1 ) . (b)a segment in the u v plane to the segment joining ( − 2 , 1 ) and ( 3 , 4 ) .

Solution Summary: The author explains how to find a point in the u v plane mapping to ( 2, 1 ).

To find: Use the inverse in exercise 9 to find (a)A point in the uv plane mapping to (2,1). (b)a segment in the uv plane to the segment joining (−2,1) and (3,4).

Want to see the full answer?

Chapter 16 Solutions

To find:

Use the inverse in exercise 9 to find

(a)A point in the $u v$ plane mapping to $(2, 1)$ .

(b)a segment in the $u v$ plane to the segment joining $(- 2, 1) and (3, 4)$ .