Question

Determine whether or not the vector field is conservative. if it is conservative, find a function f such that f = ∇f. (if the vector field is not conservative, enter dne.) f(x, y, z) = 5y2z3 i + 10xyz3 j + 15xy2z2 k

Answer 1

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.

Answer 2

The vector field is conservative

Explanation:

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.

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