Explanation G is a coordinate transformation from x y -plane to u v -plane then G ( u , v ) = ( x , y ) ---------------- (i) Given: G ( u , v ) = ( v ln u , u 2 v − 1 ) ---------- (ii) Comparing (i) and (ii), x = v ln u ; y = u 2 v − 1 Calculation: Jacobian of the transformation is J ( u , v ) = ∂ ( x , y ) ∂ ( u , v ) = | ∂ x ∂ u ∂ x ∂ v ∂ y ∂ u ∂ y ∂ v | --------- (iii) Where ∂ x ∂ u , ∂ x ∂ v , ∂ y ∂ u , ∂ y ∂ v are first order partial derivatives. ∂ x ∂ u = ∂ ∂ u ( v ln u ) = v ∂ ∂ u ( ln u ) = v ( 1 u ) = v u [While differentiating partially with respect to u treat v as constant] ∂ x ∂ u = v u ; [using ∂ ∂ u ( ln u ) = 1 u ] ∂ x ∂ v = ∂ ∂ v ( v ln u ) = ( ln u ) ∂ ∂ v ( v ) = ln u [While differentiating partially with respect to v treat u as constant] ∂ x ∂ v = ln u ∂ y ∂ u = ∂ ∂ u ( u 2 v − 1 ) = v − 1 ∂ ∂ u ( u 2 ) = v −...

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 16 Solutions

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The Jacobian

Solution Summary: The author explains that the Jacobian solution is J ( u, v ) = 1 2 and compares x to y.

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Chapter 16 Solutions