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Mademuasel [1]
4 years ago
15

Determine whether or not the vector field is conservative. if it is conservative, find a function f such that f = ∇f. (if the ve

ctor field is not conservative, enter dne.) f(x, y, z) = 5y2z3 i + 10xyz3 j + 15xy2z2 k
Mathematics
2 answers:
Ludmilka [50]4 years ago
7 0

The vector field is conservative

<h3>Explanation: </h3>

The vector field is a vector assignment to each point in a subset of space.

Determine whether or not the vector field is conservative. if it is conservative, find a function f such that f = \Delta f. (if the vector field is not conservative, enter dne.) f(x, y, z) = 5y^2z^3 i + 10xyz^3 j + 15xy^2z^2 k

\vec F is conservative if we can find a scalar function  such that . This would require

\dfrac{\partial f}{\partial x}=5y^2z^3\\\dfrac{\partial f}{\partial y}=10xyz^3\\\dfrac{\partial f}{\partial z}=15xy^2z^2

Integrate both sides of the first PDE (partial differential equation) with respect to x :

f(x,y,z)=5xy^2z^3+g(y,z) (*)

Differentiate both sides of (*) with respect to y:

\dfrac{\partial f}{\partial y}=10xyz^3=10xyz^3+\dfrac{\partial g}{\partial y}

\implies\dfrac{\partial g}{\partial y}=0

\implies g(y,z)=h(z)

Differentiate both sides of (*) with respect to z:

\dfrac{\partial f}{\partial z}=15xy^2z^2=15xy^2z^2+\dfrac{\mathrm dh}{\mathrm dz}

\implies\dfrac{\mathrm dh}{\mathrm dz}=0

\implies h(z)=C

So we have

f(x,y,z)=5xy^2z^3+C

and so \vec F is indeed conservative.

Learn more about the vector field brainly.com/question/9792983

#LearnWithBrainly

ludmilkaskok [199]4 years ago
4 0

If there is some scalar function f(x,y,z) such that \nabla f=\mathbf f as given, then this f satisfies the following partial differential equations:

\dfrac{\partial f}{\partial x}=5y^2z^3

\dfrac{\partial f}{\partial y}=10xyz^3

\dfrac{\partial f}{\partial z}=15xy^2z^2

Integrate the first PDE with respect to x:

f(x,y,z)=5xy^2z^3+g(y,z)

Differentiate with respect to y:

\dfrac{\partial f}{\partial y}=10xyz^3+\dfrac{\partial g}{\partial y}=10xyz^3\implies\dfrac{\partial g}{\partial y}=0\implies g(y,z)=h(z)

f(x,y,z)=5xy^2z^3+h(z)

Differentiate with respect to z:

\dfrac{\partial f}{\partial z}=15xy^2z^2+\dfrac{\mathrm dh}{\mathrm dz}=15xy^2z^2\implies\dfrac{\mathrm dh}{\mathrm dz}=0\implies h(z)=C

f(x,y,z)=5xy^2z^3+C

So \mathbf f is indeed conservative.

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