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Question
Math
Posted 7 months ago

f(x,y,z)=xy−y3+tan⁡(z)
f(x, y, z)=x y-y^{3}+\tan (z)


What is div⁡(grad⁡(f))\operatorname{div}(\operatorname{grad}(f)) ?
div⁡(grad⁡(f))=
\operatorname{div}(\operatorname{grad}(f))=

□\square
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Answer from Sia
Posted 7 months ago
Solution by Steps
step 1
To find the divergence of the gradient of the function f(x,y,z)=xy−y3+tan⁡(z)f(x, y, z) = xy - y^3 + \tan(z), we first need to compute the gradient of ff
step 2
The gradient of ff, denoted as ∇f\nabla f, is the vector of partial derivatives of ff with respect to xx, yy, and zz
step 3
Calculating the partial derivatives, we get ∇f=(∂∂x(xy−y3+tan⁡(z)),∂∂y(xy−y3+tan⁡(z)),∂∂z(xy−y3+tan⁡(z)))\nabla f = \left(\frac{\partial}{\partial x}(xy - y^3 + \tan(z)), \frac{\partial}{\partial y}(xy - y^3 + \tan(z)), \frac{\partial}{\partial z}(xy - y^3 + \tan(z))\right)
step 4
This results in ∇f=(y,x−3y2,sec⁡2(z))\nabla f = (y, x - 3y^2, \sec^2(z))
step 5
The divergence of a vector field A=(Ax,Ay,Az)A = (A_x, A_y, A_z) is given by ∇⋅A=∂Ax∂x+∂Ay∂y+∂Az∂z\nabla \cdot A = \frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z}
step 6
Applying this to the gradient vector, we get ∇⋅∇f=∂∂x(y)+∂∂y(x−3y2)+∂∂z(sec⁡2(z))\nabla \cdot \nabla f = \frac{\partial}{\partial x}(y) + \frac{\partial}{\partial y}(x - 3y^2) + \frac{\partial}{\partial z}(\sec^2(z))
step 7
Simplifying, we find ∇⋅∇f=0+(1−6y)+2sec⁡2(z)tan⁡(z)\nabla \cdot \nabla f = 0 + (1 - 6y) + 2\sec^2(z)\tan(z)
step 8
Therefore, the divergence of the gradient of ff is ∇⋅∇f=−6y+2sec⁡2(z)tan⁡(z)\nabla \cdot \nabla f = -6y + 2\sec^2(z)\tan(z)
Answer
∇⋅∇f=−6y+2sec⁡2(z)tan⁡(z)\nabla \cdot \nabla f = -6y + 2\sec^2(z)\tan(z)
Key Concept
Divergence of a Gradient
Explanation
The divergence of the gradient of a scalar field is a scalar quantity that represents the rate at which the quantity described by the field is expanding or contracting at a given point. It is computed by taking the dot product of the del operator with the gradient vector of the field.

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