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If z=r2-x2, find(∂z∂r)θ,(∂z∂θ)r,∂2z∂r∂θ,(∂z∂x)y

Short Answer

Expert verified

The value of∂z∂rθ=2rsin2θ,∂z∂θr=r2sin2θ,∂2z∂r.∂θ=2rsin2θand∂z∂xy=0

∂z∂rθ=2rsin2θ,∂z∂θr=r2sin2θ,∂2z∂r.∂θ=2rsin2θand∂z∂xy=0∂z∂rθ=2rsin2θ,∂z∂θr=r2sin2θ,∂2z∂r.∂θ=2rsin2θand∂z∂xy=0.

Step by step solution

01

Given Information

Given that,z=r2-x2.

02

Formula Used

The value of r2=x2+y2 ...(1)

x=rcosθy=rsinθ
03

Solving the given equation

The given equation

z=r2-x2 ...(3)

From equation 1 and 2,

z=x2+y2-x2z=y2

Substitute the value of y in the above equation.

z=rsinθ2

Differentiate with respect to r,

∂z∂r=2rsin2θ∂z∂rθ=2rsin2θ

We know thatz=r2sin2θ

Differentiate with respect toθ

∂z∂θ=2r2sinθ.cosθ∂z∂θ=r2sin2θ∂z∂θr=r2sin2θ

Solving further

∂2z∂r.∂θ=∂∂r∂z∂θ∂2z∂r.∂θ=∂∂rr2sin2θ∂2z∂r.∂θ=2rsin2θ

We know thatz=y2

∂z∂x=0∂z∂xy=0

Hence∂z∂rθ=2rsin2θ,∂z∂θr=r2sin2θ,∂2z∂r.∂θ=2rsin2θand∂z∂xy=0 .

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