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Westkost [7]
3 years ago
11

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

Mathematics
1 answer:
Jet001 [13]3 years ago
4 0

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.

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