Explanation Given: x = 1 2 ( u + v ) y = 1 2 ( u − v ) and z = 2 u v w And ∂ ( x , y , z ) ∂ ( u , v , w ) = | ∂ x ∂ u ∂ x ∂ v ∂ x ∂ w ∂ y ∂ u ∂ y ∂ v ∂ y ∂ w ∂ z ∂ u ∂ z ∂ v ∂ z ∂ w | Formula used: Partial differentiation : If f = x y + 1 Then ∂ f ∂ x = y Determinant is determined as: Δ = | a b d c | = a c − b d Calculation: Tofind the value of jacobian ∂ ( x , y , z ) ∂ ( u , v , w ) = | ∂ x ∂ u ∂ x ∂ v ∂ x ∂ w ∂ y ∂ u ∂ y ∂ v ∂ y ∂ w ∂ z ∂ u ∂ z ∂ v ∂ z ∂ w | Put the value of x , y , z Thus we get, ∂ ( x , y , z ) ∂ ( u , v , w ) = | 1 2 . ∂ ( u + v ) ∂ u 1 2 . ∂ ( u + v ) ∂ v 1 2 . ∂ ( u + v ) ∂ w 1 2 . ∂ ( u − v ) ∂ u 1 2 . ∂ ( u − v ) ∂ v 1 2

Calculus: Early Transcendental Functions

7th Edition

ISBN: 9781337552516

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Chapter 14.8, Problem 37E

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To calculate: The value of jacobian ∂(x,y,z)∂(u,v,w) for the given change of variables x=12(u+v)y=12(u−v) and z=2uvw

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Chapter 14.8, Problem 36E

Chapter 14.8, Problem 38E

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Chapter 14 Solutions

Calculus: Early Transcendental Functions

Ch. 14.1 - Evaluate the iterated integral: 0433cosrdrd Ch. 14.1 - CONCEPT CHECK Region of Integration Sketch the...Ch. 14.1 - Evaluate the integral: 0x(2xy)dy Ch. 14.1 - Evaluate the integral: xx2yxdy Ch. 14.1 - Evaluate the integral: 04x2x2ydy Ch. 14.1 - Evaluate the integral: x3x(x2+3y2)dy Ch. 14.1 - Evaluate the integral: eyyylnxxdx;y0 Ch. 14.1 - Evaluate the integral: 1y21y2(x2+y2)dx Ch. 14.1 - Evaluate the integral: 0x2yeyxdy Ch. 14.1 - Evaluate the integral: y2sin3xcosydx

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The value of jacobian ∂ ( x , y , z ) ∂ ( u , v , w ) for the given change of variables x = 1 2 ( u + v ) y = 1 2 ( u − v ) and z = 2 u v w

Solution Summary: The author explains how to calculate the value of jacobian 12(x,y,z) for the given change of variables.

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To calculate: The value of jacobian ∂(x,y,z)∂(u,v,w) for the given change of variables x=12(u+v)y=12(u−v) and z=2uvw

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Chapter 14 Solutions