The gradient in spherical coordinate system for a scalar function f is

The Correct Answer and Explanation is:

Let’s break down the problem step by step.

The given function in Cartesian coordinates is:

f(x,y,z)=x2z2+xy2f(x, y, z) = x^2 z^2 + x y^2f(x,y,z)=x2z2+xy2

We are asked to express this function in spherical coordinates and then calculate the gradient and evaluate it at a specific point.

Step A: Expressing f(x,y,z)f(x, y, z)f(x,y,z) in spherical coordinates

To convert f(x,y,z)f(x, y, z)f(x,y,z) to spherical coordinates, recall the relations between Cartesian and spherical coordinates:x=rsin⁡θcos⁡ϕx = r \sin\theta \cos\phix=rsinθcosϕy=rsin⁡θsin⁡ϕy = r \sin\theta \sin\phiy=rsinθsinϕz=rcos⁡θz = r \cos\thetaz=rcosθ

Substitute these into the function:f(x,y,z)=(rsin⁡θcos⁡ϕ)2(rcos⁡θ)2+(rsin⁡θcos⁡ϕ)(rsin⁡θsin⁡ϕ)2f(x, y, z) = (r \sin\theta \cos\phi)^2 (r \cos\theta)^2 + (r \sin\theta \cos\phi) (r \sin\theta \sin\phi)^2f(x,y,z)=(rsinθcosϕ)2(rcosθ)2+(rsinθcosϕ)(rsinθsinϕ)2

Simplifying each term:

1. (rsin⁡θcos⁡ϕ)2(rcos⁡θ)2=r4sin⁡2θcos⁡2ϕcos⁡2θ(r \sin\theta \cos\phi)^2 (r \cos\theta)^2 = r^4 \sin^2\theta \cos^2\phi \cos^2\theta(rsinθcosϕ)2(rcosθ)2=r4sin2θcos2ϕcos2θ
2. (rsin⁡θcos⁡ϕ)(rsin⁡θsin⁡ϕ)2=r3sin⁡3θcos⁡ϕsin⁡2ϕ(r \sin\theta \cos\phi) (r \sin\theta \sin\phi)^2 = r^3 \sin^3\theta \cos\phi \sin^2\phi(rsinθcosϕ)(rsinθsinϕ)2=r3sin3θcosϕsin2ϕ

Thus, the function in spherical coordinates becomes:f(r,θ,ϕ)=r4sin⁡2θcos⁡2ϕcos⁡2θ+r3sin⁡3θcos⁡ϕsin⁡2ϕf(r, \theta, \phi) = r^4 \sin^2\theta \cos^2\phi \cos^2\theta + r^3 \sin^3\theta \cos\phi \sin^2\phif(r,θ,ϕ)=r4sin2θcos2ϕcos2θ+r3sin3θcosϕsin2ϕ

Step B: Finding the gradient ∇f(r,θ,ϕ)\nabla f(r, \theta, \phi)∇f(r,θ,ϕ)

The gradient in spherical coordinates is given by:∇f(r,θ,ϕ)=r^∂f∂r+θ^1r∂f∂θ+ϕ^1rsin⁡θ∂f∂ϕ\nabla f(r, \theta, \phi) = \hat{r} \frac{\partial f}{\partial r} + \hat{\theta} \frac{1}{r} \frac{\partial f}{\partial \theta} + \hat{\phi} \frac{1}{r \sin\theta} \frac{\partial f}{\partial \phi}∇f(r,θ,ϕ)=r^∂r∂f​+θ^r1​∂θ∂f​+ϕ^​rsinθ1​∂ϕ∂f​

Now, let’s compute the partial derivatives.

1. ∂f∂r\frac{\partial f}{\partial r}∂r∂f​

This will be:∂∂r(r4sin⁡2θcos⁡2ϕcos⁡2θ+r3sin⁡3θcos⁡ϕsin⁡2ϕ)\frac{\partial}{\partial r} \left( r^4 \sin^2\theta \cos^2\phi \cos^2\theta + r^3 \sin^3\theta \cos\phi \sin^2\phi \right)∂r∂​(r4sin2θcos2ϕcos2θ+r3sin3θcosϕsin2ϕ)

2. ∂f∂θ\frac{\partial f}{\partial \theta}∂θ∂f​

For this derivative, we will differentiate each term with respect to θ\thetaθ.

3. ∂f∂ϕ\frac{\partial f}{\partial \phi}∂ϕ∂f​

Here, we will differentiate each term with respect to ϕ\phiϕ.

Step C: Calculating ∇f(2,π/4,π/3)\nabla f(2, \pi/4, \pi/3)∇f(2,π/4,π/3)

Once we have the gradient expression, we substitute r=2r = 2r=2, θ=π/4\theta = \pi/4θ=π/4, and ϕ=π/3\phi = \pi/3ϕ=π/3 into the formula for ∇f\nabla f∇f.

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Given the complexity of the derivatives, it’s ideal to first work through the detailed computations of the partial derivatives for rrr, θ\thetaθ, and ϕ\phiϕ.