Question

Consider the vector field f(x,y,z)=(4z+5y)i+(4z+5x)j+(4y+4x)k.
a.find a function f such that f=∇f and f(0,0,0)=0.

Answer 1

$\nabla f(x,y,z)=\mathbf f(x,y,z)=\dfrac{\partial f}{\partial x}\,\mathbf i+\dfrac{\partial f}{\partial y}\,\mathbf j+\dfrac{\partial f}{\partial z}\,\mathbf k$

$\dfrac{\partial f}{\partial x}=4z+5y$
$\implies f(x,y,z)=4xz+5xy+g(y,z)$

$\dfrac{\partial f}{\partial y}=4z+5x=5x+\dfrac{\partial g}{\partial y}$
$\implies\dfrac{\partial g}{\partial y}=4z\implies g(y,z)=4yz+h(z)$

$\dfrac{\partial f}{\partial z}=4y+4x=4x+\dfrac{\partial g}{\partial z}$
$\implies4y=4y+\dfrac{\mathrm dh}{\mathrm dz}\implies\dfrac{\mathrm dh}{\mathrm dz}=0\implies h(z)=C$

$f(x,y,z)=4xz+5xy+4yz+C$

$f(0,0,0)=0\implies C=0\implies f(x,y,z)=4xz+5xy+4yz$

Answer 2

The gradient of a function f, denoted as ∇f , is the collection of all its partial derivatives into a vector.

Function  $f(x,y,z)=5xy+4yz+4xz$

Gradient of vector field is given by,

                $\bigtriangledown f=i\frac{\partial f }{\partial x}+j\frac{\partial f }{\partial y}+k\frac{\partial f }{\partial z}$

And function is given that,

               $f(x,y,z)=(4z+5y)i+(4z+5x)j+(4y+4x)k.$

Since,      f = ∇f

After comparing, We get

      $\frac{\partial f}{\partial x}=4z+5y , f(x,y,z)=4xz+5xy$

     $\frac{\partial f}{\partial y}=4z+5x , f(x,y,z)=4yz+5xy$

     $\frac{\partial f}{\partial z}=4y+4x , f(x,y,z)=4yz+4xz$

After observing above equations,

We have,  $f(x,y,z)=5xy+4yz+4xz+c$

Since, $f(0,0,0)=0$ , So that, $c=0$

Thus,   $f(x,y,z)=5xy+4yz+4xz$

Learn more:

https://brainly.com/question/15306869