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PtichkaEL [24]
2 years ago
8

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) = ye−xi + e−xj + 2zk
Mathematics
1 answer:
Akimi4 [234]2 years ago
8 0
A three-dimensional vector field is conservative if it is also irrotational, i.e. its curl is \mathbf 0. We have

\nabla\times\mathbf f(x,y,z)=-2e^{-x}\,\mathbf k

so this vector field is not conservative.

- - -

Another way of determining the same result: We want to find a scalar function f(x,y,z) such that its gradient is equal to the given vector field, \mathbf f(x,y,z):

\nabla f(x,y,z)=\mathbf f(x,y,z)

For this to happen, we need to satisfy

\begin{cases}f_x=ye^{-x}\\f_y=e^{-x}\\f_z=2z\end{cases}

From the first equation, integrating with respect to x yields

f_x=ye^{-x}\implies f(x,y,z)=-ye^{-x}+g(y,z)

Note that g *must* be a function of y,z only.

Now differentiate with respect to y and we have

f_y=-e^{-x}+g_y=e^{-x}\implies g_y=2e^{-x}\implies g(y,z)=2ye^{-x}+\cdots

but this contradicts the assumption that g(y,z) is independent of x. So, the scalar potential function does not exist, and therefore the vector field is not conservative.
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Answer:

(-21,-19)

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Step-by-step explanation:

We are given the equation of circle

x^2+y^2+42x+38y-47=0

General equation of circle:

x^2+y^2+2gx+2fy+c=0

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Radius: \sqrt{g^2+f^2-c}

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Standard form of circle:

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The centre of circle at the point (-21,-19) and its radius is \sqrt{849}.

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