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The problem asks us to find $\frac{\partial w}{\partial \theta}$ for $w = x^2y + z^2$, where $x = \rho \cos \theta \sin \phi$, $y = \rho \sin \theta \sin \phi$, and $z = \rho \cos \phi$. We need to evaluate the result at $\rho = 2$, $\theta = \pi$, and $\phi = \frac{\pi}{2}$.

AnalysisPartial DerivativesChain RuleMultivariable Calculus
2025/5/9

1. Problem Description

The problem asks us to find ∂w∂θ\frac{\partial w}{\partial \theta} for w=x2y+z2w = x^2y + z^2, where x=ρcos⁡θsin⁡ϕx = \rho \cos \theta \sin \phi, y=ρsin⁡θsin⁡ϕy = \rho \sin \theta \sin \phi, and z=ρcos⁡ϕz = \rho \cos \phi. We need to evaluate the result at ρ=2\rho = 2, θ=π\theta = \pi, and ϕ=π2\phi = \frac{\pi}{2}.

2. Solution Steps

First, we find the partial derivatives of ww with respect to x,y,zx, y, z:
∂w∂x=2x\frac{\partial w}{\partial x} = 2x
∂w∂y=x2\frac{\partial w}{\partial y} = x^2
∂w∂z=2z\frac{\partial w}{\partial z} = 2z
Next, we find the partial derivatives of x,y,zx, y, z with respect to θ\theta:
∂x∂θ=ρ(−sin⁡θ)sin⁡ϕ=−ρsin⁡θsin⁡ϕ\frac{\partial x}{\partial \theta} = \rho (-\sin \theta) \sin \phi = -\rho \sin \theta \sin \phi
∂y∂θ=ρ(cos⁡θ)sin⁡ϕ=ρcos⁡θsin⁡ϕ\frac{\partial y}{\partial \theta} = \rho (\cos \theta) \sin \phi = \rho \cos \theta \sin \phi
∂z∂θ=0\frac{\partial z}{\partial \theta} = 0
Using the chain rule, we have:
∂w∂θ=∂w∂x∂x∂θ+∂w∂y∂y∂θ+∂w∂z∂z∂θ\frac{\partial w}{\partial \theta} = \frac{\partial w}{\partial x} \frac{\partial x}{\partial \theta} + \frac{\partial w}{\partial y} \frac{\partial y}{\partial \theta} + \frac{\partial w}{\partial z} \frac{\partial z}{\partial \theta}
∂w∂θ=(2x)(−ρsin⁡θsin⁡ϕ)+(x2)(ρcos⁡θsin⁡ϕ)+(2z)(0)\frac{\partial w}{\partial \theta} = (2x)(-\rho \sin \theta \sin \phi) + (x^2)(\rho \cos \theta \sin \phi) + (2z)(0)
∂w∂θ=−2xρsin⁡θsin⁡ϕ+x2ρcos⁡θsin⁡ϕ\frac{\partial w}{\partial \theta} = -2x \rho \sin \theta \sin \phi + x^2 \rho \cos \theta \sin \phi
Now, we evaluate x,y,zx, y, z at ρ=2\rho = 2, θ=π\theta = \pi, and ϕ=π2\phi = \frac{\pi}{2}:
x=2cos⁡πsin⁡π2=2(−1)(1)=−2x = 2 \cos \pi \sin \frac{\pi}{2} = 2(-1)(1) = -2
y=2sin⁡πsin⁡π2=2(0)(1)=0y = 2 \sin \pi \sin \frac{\pi}{2} = 2(0)(1) = 0
z=2cos⁡π2=2(0)=0z = 2 \cos \frac{\pi}{2} = 2(0) = 0
Substitute these values into the expression for ∂w∂θ\frac{\partial w}{\partial \theta}:
∂w∂θ=−2(−2)(2)sin⁡πsin⁡π2+(−2)2(2)cos⁡πsin⁡π2\frac{\partial w}{\partial \theta} = -2(-2)(2) \sin \pi \sin \frac{\pi}{2} + (-2)^2 (2) \cos \pi \sin \frac{\pi}{2}
∂w∂θ=8(0)(1)+4(2)(−1)(1)\frac{\partial w}{\partial \theta} = 8 (0)(1) + 4 (2) (-1)(1)
∂w∂θ=0−8=−8\frac{\partial w}{\partial \theta} = 0 - 8 = -8

3. Final Answer

-8

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