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Posted 2 months ago

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

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

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Answer from Sia
Posted 2 months ago
Solution by Steps
step 1
To find the divergence of the gradient of the function f(x,y,z)=xyy3+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(xyy3+tan(z)),y(xyy3+tan(z)),z(xyy3+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,x3y2,sec2(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=Axx+Ayy+Azz\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(x3y2)+z(sec2(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+(16y)+2sec2(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+2sec2(z)tan(z)\nabla \cdot \nabla f = -6y + 2\sec^2(z)\tan(z)
f=6y+2sec2(z)tan(z)\nabla \cdot \nabla f = -6y + 2\sec^2(z)\tan(z)
Key Concept
Divergence of a Gradient
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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