Question

Consider f and c below. f(x, y, z) = yzexzi + exzj + xyexzk,

Answer 1

Answer:

$f(x,y,z)=ye^{xz}+C$  

Step-by-step explanation:

We can write the given expression as :

$\vec f(x,y,z)=yze^{xz}\,\vec\imath+e^{xz}\,\vec\jmath+xye^{xz}\,\vec k$

As given,   f = ∇f.

∇f = $\dfrac{\partial f}{\partial x}i$  + $\dfrac{\partial f}{\partial y}j$  +$\dfrac{\partial f}{\partial z}k$ 

We can write the partial derivative with respect to x, y and z.

$\dfrac{\partial f}{\partial x}=yze^{xz}$       ___(Equation 1)

$\dfrac{\partial f}{\partial y}=e^{xz}$               ______(Equation 2)

$\dfrac{\partial f}{\partial z}=xye^{xz}$             ______(Equation 3)

Take equation 2 and integrate with respect to y,

$\dfrac{\partial f}{\partial y}=e^{xz}$

$f(x,y,z)=ye^{xz}+a(x,z)$           ----------Equation 4

Derivate both sides w.r.t x , we get :

$\frac{d}{dx}(yze^{xz})=yze^{xz}+\dfrac{\partial a}{\partial x}$

or

$\dfrac{\partial a}{\partial x}=0$

integrate

a(x,z)=b(z)

put in equation 4 ,

we get :

$f(x,y,z)=ye^{xz}+b(z)$

take derivative wrt z

$\frac{d}{dz} (ye^{xz}+b(z))\impliesxye^{xz}=xye^{xz}+\frac{db}{dz}$

we can take here:

$\frac{db}{dz} = 0$

integrate:

$\int\ {\frac{db}{dz} } \, =\int0$

b(z) = C

The function can be written as :

from equation 4 :

$f(x,y,z)=ye^{xz}+C$  

Where C is a constant.