Determine if the vector field is conservative or not

[mathmari]

Hey!

Determine if the vector field $\overrightarrow{F}=y\hat{i}+(x+z)\hat{j}-y\hat{k}$ is conservative or not.The vector field $\overrightarrow{F}=M\hat{i}+N\hat{j}+P\hat{k}$ is conservative if $$\frac{\partial{M}}{\partial{y}}=\frac{\partial{N}}{\partial{x}}, \frac{\partial{M}}{\partial{z}}=\frac{\partial{P}}{\partial{x}}, \frac{\partial{N}}{\partial{z}}=\frac{\partial{P}}{\partial{y}}$$

In this case:
$$\frac{\partial{M}}{\partial{y}}=1=\frac{\partial{N}}{\partial{x}}, \frac{\partial{M}}{\partial{z}}=0=\frac{\partial{P}}{\partial{x}}, \frac{\partial{N}}{\partial{z}}=1 \neq \frac{\partial{P}}{\partial{y}}=-1$$

Does this mean that the vector field is not conservative? Is it a ~if and only if~ condition?

 

[mathmari]

Or does this only mean that we cannot conclude by using this formula that the vector field is conservative? So we don't know if it is conservative?

 

[I like Serena]

Hey!

Determine if the vector field $\overrightarrow{F}=y\hat{i}+(x+z)\hat{j}-y\hat{k}$ is conservative or not.The vector field $\overrightarrow{F}=M\hat{i}+N\hat{j}+P\hat{k}$ is conservative if $$\frac{\partial{M}}{\partial{y}}=\frac{\partial{N}}{\partial{x}}, \frac{\partial{M}}{\partial{z}}=\frac{\partial{P}}{\partial{x}}, \frac{\partial{N}}{\partial{z}}=\frac{\partial{P}}{\partial{y}}$$

In this case:
$$\frac{\partial{M}}{\partial{y}}=1=\frac{\partial{N}}{\partial{x}}, \frac{\partial{M}}{\partial{z}}=0=\frac{\partial{P}}{\partial{x}}, \frac{\partial{N}}{\partial{z}}=1 \neq \frac{\partial{P}}{\partial{y}}=-1$$

Does this mean that the vector field is not conservative? Is it a ~if and only if~ condition?

Hi! :DFirst note that your condition is equivalent to $\nabla\times\overrightarrow{F} = \mathbf{0}$.From wikipedia:

A vector field $\mathbf{v}$ is said to be conservative if there exists a scalar field $\varphi$ such that
$$\mathbf{v}=\nabla\varphi$$
A vector field $\mathbf{v}$ is said to be irrotational if its curl is zero. That is, if
$$\nabla\times\mathbf{v} = \mathbf{0}$$
Therefore every conservative vector field is also an irrotational vector field.

Suppose that $S\subseteq\mathbb{R}^3$ is a region of three-dimensional space where $\mathbf{v}$ is defined.
Provided that $S$ is a simply connected region, the converse of this is true: every irrotational vector field is also a conservative vector field.
The above statement is not true if $S$ is not simply connected.
​

Can you deduce an answer from that?

 

[mathmari]

Hi! :DFirst note that your condition is equivalent to $\nabla\times\overrightarrow{F} = \mathbf{0}$.From wikipedia:

A vector field $\mathbf{v}$ is said to be conservative if there exists a scalar field $\varphi$ such that
$$\mathbf{v}=\nabla\varphi$$
A vector field $\mathbf{v}$ is said to be irrotational if its curl is zero. That is, if
$$\nabla\times\mathbf{v} = \mathbf{0}$$
Therefore every conservative vector field is also an irrotational vector field.

Suppose that $S\subseteq\mathbb{R}^3$ is a region of three-dimensional space where $\mathbf{v}$ is defined.
Provided that $S$ is a simply connected region, the converse of this is true: every irrotational vector field is also a conservative vector field.
The above statement is not true if $S$ is not simply connected.
​

Can you deduce an answer from that?

So the vectorfield $\overrightarrow{F}$ is not irrotational since $\nabla \times \overrightarrow{F} \neq 0$, is it?

The other condition, to find a $f$ such that $\overrightarrow{F}=\nabla f$ is as followed:
$$\overrightarrow{F}=\nabla f=\frac{\partial{f}}{\partial{x}}\hat{i}+\frac{ \partial{f}}{\partial{y}}\hat{j}+\frac{\partial{f}}{\partial{z}}\hat{k}$$
$$\frac{\partial{f}}{\partial{x}}=y \Rightarrow f=yx+g(y,z)$$
$$\frac{\partial{f}}{\partial{y}}=x+\frac{\partial{g}}{\partial{y}}=x+z \Rightarrow \frac{\partial{g}}{\partial{y}}=z \Rightarrow g(y,z)=yz+h(z)$$
$$f=yx+yz+h(z)$$
$$\frac{\partial{f}}{\partial{z}}=y+h'(z)=-y \Rightarrow h'(z)=-2y \Rightarrow h(z)-2yz+c$$
So $f=yx+yz-2yz+c \Rightarrow f=yx-yz+c$.

So since there exists such a $f$, does this mean that the vectorfield $\overrightarrow{F}$ is conservative?

 

[I like Serena]

Let me first respond on your original post.

The vector field $\overrightarrow{F}=M\hat{i}+N\hat{j}+P\hat{k}$ is conservative if $$\frac{\partial{M}}{\partial{y}}=\frac{\partial{N}}{\partial{x}}, \frac{\partial{M}}{\partial{z}}=\frac{\partial{P}}{\partial{x}}, \frac{\partial{N}}{\partial{z}}=\frac{\partial{P}}{\partial{y}}$$

Does this mean that the vector field is not conservative? Is it a ~if and only if~ condition?

From the wikipedia definition, it should be clear that this is not the definition of a conservative field. (I am wondering if it was intended to be...?) It is similar, but there are special cases that are not covered by this statement.
However, if we're only looking at vector fields that are differentiable on $\mathbb R^3$, the definitions are equivalent. Within that context it is an if-and-only-if.

So indeed, no, the vector field is not conservative.

So the vectorfield $\overrightarrow{F}$ is not irrotational since $\nabla \times \overrightarrow{F} \neq 0$, is it?

The other condition, to find a $f$ such that $\overrightarrow{F}=\nabla f$ is as followed:
$$\overrightarrow{F}=\nabla f=\frac{\partial{f}}{\partial{x}}\hat{i}+\frac{ \partial{f}}{\partial{y}}\hat{j}+\frac{\partial{f}}{\partial{z}}\hat{k}$$
$$\frac{\partial{f}}{\partial{x}}=y \Rightarrow f=yx+g(y,z)$$
$$\frac{\partial{f}}{\partial{y}}=x+\frac{\partial{g}}{\partial{y}}=x+z \Rightarrow \frac{\partial{g}}{\partial{y}}=z \Rightarrow g(y,z)=yz+h(z)$$
$$f=yx+yz+h(z)$$
$$\frac{\partial{f}}{\partial{z}}=y+h'(z)=-y \Rightarrow h'(z)=-2y \Rightarrow h(z)-2yz+c$$
So $f=yx+yz-2yz+c \Rightarrow f=yx-yz+c$.

So since there exists such a $f$, does this mean that the vectorfield $\overrightarrow{F}$ is conservative?

Looks like a good approach! ;)
But... let's verify.

With $f=yx-yz+c$ we have:
$$\nabla f = y \hat\imath + (x-z)\hat \jmath -y \hat k$$
Hey! That is not equal to the original $\overrightarrow F$ we had!

 

[mathmari]

From the wikipedia definition, it should be clear that this is not the definition of a conservative field. (I am wondering if it was intended to be...?) It is similar, but there are special cases that are not covered by this statement.

There is a theorem:
Let $M(x,y,z), N(x,y,z), P(x,y,z)$ and their first partial derivatives be continuous functions, then a necessary condition so that the vectorfield $\overrightarrow{F}$ is conservative is that the following equations are satisfied:
$$\frac{\partial{M}}{\partial{y}}=\frac{\partial{N}}{\partial{x}}, \frac{\partial{M}}{\partial{z}}=\frac{\partial{P}}{\partial{x}}, \frac{\partial{N}}{\partial{z}}=\frac{\partial{P}}{\partial{y}}$$

That means that if these equations are satisfied, the vectorfield is conservative.
If these equations are not satisfied, as in this case, we cannot say that the vectorfield is conservative or not.

So we use the other condition, right?
$$\overrightarrow{F} \text{ conservative } \Leftrightarrow \overrightarrow{F}=\nabla f$$

Looks like a good approach! ;)
But... let's verify.

With $f=yx-yz+c$ we have:
$$\nabla f = y \hat\imath + (x-z)\hat \jmath -y \hat k$$
Hey! That is not equal to the original $\overrightarrow F$ we had!

A ok!

 

[I like Serena]

There is a theorem:
Let $M(x,y,z), N(x,y,z), P(x,y,z)$ and their first partial derivatives be continuous functions, then a necessary condition so that the vectorfield $\overrightarrow{F}$ is conservative is that the following equations are satisfied:
$$\frac{\partial{M}}{\partial{y}}=\frac{\partial{N}}{\partial{x}}, \frac{\partial{M}}{\partial{z}}=\frac{\partial{P}}{\partial{x}}, \frac{\partial{N}}{\partial{z}}=\frac{\partial{P}}{\partial{y}}$$

That means that if these equations are satisfied, the vectorfield is conservative.
If these equations are not satisfied, as in this case, we cannot say that the vectorfield is conservative or not.

So we use the other condition, right?
$$\overrightarrow{F} \text{ conservative } \Leftrightarrow \overrightarrow{F}=\nabla f$$

Ah okay. So it was "only" a theorem (with additional conditions involved) and not a definition.

There is also another theorem that says that if a field is conservative, it has to be irrotational (the converse).
Or in other words, if it is it not irrotational, it is not conservative.

Anyway, yeah, going back to the definition is always a good way to go! ;)

 

[mathmari]

Ah okay. So it was "only" a theorem (with additional conditions involved) and not a definition.

There is also another theorem that says that if a field is conservative, it has to be irrotational (the converse).
Or in other words, if it is it not irrotational, it is not conservative.

Anyway, yeah, going back to the definition is always a good way to go! ;)

Ok! Thank you very much for your help! (Smile)