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Use Theorem 12.34 to find the indicated derivatives in Exercises 31–36. Be sure to simplify your answers.

∂w∂θwhenw=x2+zey,x=Òϲõ¾±²ÔÏ•³¦´Ç²õθ,y=Òϲõ¾±²ÔÏ•²õ¾±²Ôθ,z=Òϳ¦´Ç²õÏ•

Short Answer

Expert verified

The value of∂w∂θ=ÒÏeÒÏsinÏ•sinθsinÏ•-2sinÏ•cosθsinθ+ÒÏ2sin2Ï•cos3θ+ÒÏcosÏ•cosθ

Step by step solution

01

Step 1. Given Information.

w=x2+zeyx=Òϲõ¾±²ÔÏ•³¦´Ç²õθy=Òϲõ¾±²ÔÏ•²õ¾±²Ôθz=Òϳ¦´Ç²õÏ•

02

Step 2. Calculation.

By Theorem 12.34, we have

∂w∂θ=∂w∂x·∂x∂θ+∂w∂y·∂y∂θ+∂w∂z·∂z∂θ-------(1)

So first we find ∂w∂x,∂x∂θ,∂w∂y,∂y∂θ,∂w∂z,∂z∂θ

So we have

∂w∂x=ey2x∂w∂y=x2+zey∂w∂z=ey·1=ey∂x∂θ=-ÒÏsinÏ•sinθ∂y∂θ=ÒÏsinÏ•sinθ∂z∂θ=0

03

Step 3. Calculation. 

Use these above values in (1) we get,

∂w∂θ=∂w∂x·∂x∂θ+∂w∂y·∂y∂θ+∂w∂z·∂z∂θ∂w∂θ=2xey-ÒÏsinÏ•sinθ+eyx2+zÒÏsinÏ•cosθ+ey0∂w∂θ=ÒÏeysinÏ•-2xsinθ+x2+zcosθ

So from here, putting the value of xandyin terms of ÒÏ,Ï•,θwe get

∂w∂θ=ÒÏeysinÏ•-2xsinθ+x2+zcosθ∂w∂θ=ÒÏeÒÏsinÏ•sinθsinÏ•-2sinÏ•cosθsinθ+ÒÏ2sin2Ï•cos3θ+ÒÏcosÏ•cosθ

04

Step 4. Conclusion. 

The value of∂w∂θ=ÒÏeÒÏsinÏ•sinθsinÏ•-2sinÏ•cosθsinθ+ÒÏ2sin2Ï•cos3θ+ÒÏcosÏ•cosθ

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