Question

Math

Posted 2 months ago

```
$f(x, y, z)=x y-y^{3}+\tan (z)$
What is $\operatorname{div}(\operatorname{grad}(f))$ ?
$\operatorname{div}(\operatorname{grad}(f))=$
$\square$
```

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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 $f(x, y, z) = xy - y^3 + \tan(z)$, we first need to compute the gradient of $f$

step 2

The gradient of $f$, denoted as $\nabla f$, is the vector of partial derivatives of $f$ with respect to $x$, $y$, and $z$

step 3

Calculating the partial derivatives, we get $\nabla f = \left(\frac{\partial}{\partial x}(xy - y^3 + \tan(z)), \frac{\partial}{\partial y}(xy - y^3 + \tan(z)), \frac{\partial}{\partial z}(xy - y^3 + \tan(z))\right)$

step 4

This results in $\nabla f = (y, x - 3y^2, \sec^2(z))$

step 5

The divergence of a vector field $A = (A_x, A_y, A_z)$ is given by $\nabla \cdot A = \frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z}$

step 6

Applying this to the gradient vector, we get $\nabla \cdot \nabla f = \frac{\partial}{\partial x}(y) + \frac{\partial}{\partial y}(x - 3y^2) + \frac{\partial}{\partial z}(\sec^2(z))$

step 7

Simplifying, we find $\nabla \cdot \nabla f = 0 + (1 - 6y) + 2\sec^2(z)\tan(z)$

step 8

Therefore, the divergence of the gradient of $f$ is $\nabla \cdot \nabla f = -6y + 2\sec^2(z)\tan(z)$

Answer

$\nabla \cdot \nabla f = -6y + 2\sec^2(z)\tan(z)$

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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