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Is the answer correct......

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- Oct 13th 2012, 04:31 AMmathsforumhelpWhich value of k is vector field conservative
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Is the answer correct...... - Oct 13th 2012, 06:58 AMjohnsomeoneRe: Which value of k is vector field conservative
Your work has a valid approach, and the correct answer. However, you seem to have overlooked that the "constants of integration" are

**FUNCTIONS**in the variables*not*integrated over. In this particular problem, that doesn't matter, but in other potential problems, you'll have trouble/get no answer/get confused/get wrong answers if you don't understand that. See Method #1:

$\displaystyle \vec{F}$ is conservative if there exists some function $\displaystyle \phi (x, y, z)$ such that $\displaystyle \vec{F} = \nabla \phi$.

For that to happen, the $\displaystyle \hat{i}, \hat{j}, \hat{k}$ components for both must be equal, as stated in the 3 equations:

1) $\displaystyle \frac{\partial \phi}{\partial x} = (14+k)x + 2y$

2) $\displaystyle \frac{\partial \phi}{\partial y} = (14+k)x + 14z$

3) $\displaystyle \frac{\partial \phi}{\partial z} = 14y$

Method #1:

From the first equation, when you integrate, you need to remember that when taking the partial deriuvative with respect to x, you were treating y and z as constant. Thus, when you perform the indefinite integral, instead of +C, you'll have a +C(y, z), which is an arbitrary function in y and z. If you take the partial w/repect to x of the result of that integration, it might make it clearer why you needed to add an arbitrary function of y and z rather than just a constant.

$\displaystyle \frac{\partial \phi}{\partial x} = (14+k)x + 2y$, so

$\displaystyle \int \frac{\partial \phi}{\partial x} \ dx = \int \ [ \ (14+k)x + 2y \ ] \ dx$, so

$\displaystyle \phi(x, y, z) = \left( \frac{14+k}{2} \right) x^2 + 2yx + C(y,z)$.

Now, plug that into the 2nd equation, $\displaystyle \frac{\partial \phi}{\partial y} = (14+k)x + 14z$, getting

$\displaystyle \frac{\partial}{\partial y} \left\{ \frac{14+k}{2}x^2 + 2yx + C(y,z) \right\} = (14+k)x + 14z$, so

$\displaystyle 2x + \frac{\partial C(y,z)}{\partial y} = (14+k)x + 14z$, so

$\displaystyle \frac{\partial C(y,z)}{\partial y} = (14+k)x -2x + 14z = (12+k)x + 14z$.

NOW - notice that C(y,z) has no dependence on x. Thus the LHS has no x dependence, so the RHS, which the LHS equals, also has no x dependence. But there's an x-term in the RHS. Thus the coefficient in front of that x-term in the RHS must be 0 (i.e. there actually is no x-term there). You could also see this by taking the partial derivative with respect to x of both sides (although that makes a differentiablity assumption). Either way, get:

$\displaystyle 12+k = 0$. Therefore $\displaystyle k = -12$.

You can stop here, at $\displaystyle k = -12$. Stopping here, you haven't proven that $\displaystyle \vec{F}$ is conservative. What you've proven is that **IF** $\displaystyle \vec{F}$ is conservative, then the only possible value k could have is -12.

Method #2:

Assuming $\displaystyle \phi$ is what's called $\displaystyle C^2$ on its domain, meaning all it's 2nd order partial derivatives exist and are continuous on its domain, then

$\displaystyle \frac{\partial^2 \phi}{\partial y \partial x} = \frac{\partial^2 \phi}{\partial x \partial y}, \ \frac{\partial^2 \phi}{\partial y \partial z} = \frac{\partial^2 \phi}{\partial z \partial y}$, and $\displaystyle \frac{\partial^2 \phi}{\partial z \partial x} = \frac{\partial^2 \phi}{\partial z \partial y}$.

We'll only use $\displaystyle \frac{\partial^2 \phi}{\partial y \partial x} = \frac{\partial^2 \phi}{\partial x \partial y}$, which means that $\displaystyle \frac{\partial}{\partial y} \left( \frac{\partial \phi}{\partial x} \right)= \frac{\partial}{\partial x} \left( \frac{\partial \phi}{\partial y} \right)$.

So if $\displaystyle \vec{F}$ is conservative with $\displaystyle \phi$, and $\displaystyle \phi$ is $\displaystyle C^2$, then using Equations 1 & 2, have:

$\displaystyle \frac{\partial}{\partial y} \left( \frac{\partial \phi}{\partial x} \right) = \frac{\partial}{\partial y} \left\{ (14+k)x + 2y \right\} = 2$, and

$\displaystyle \frac{\partial}{\partial x} \left( \frac{\partial \phi}{\partial y} \right) = \frac{\partial}{\partial x} \left\{ (14+k)x + 14z \right\} = 14+k$, so

$\displaystyle \frac{\partial}{\partial y} \left( \frac{\partial \phi}{\partial x} \right)= \frac{\partial}{\partial x} \left( \frac{\partial \phi}{\partial y} \right)$ implies that $\displaystyle 2 = 14 + k$, which implies $\displaystyle k = -12$.

Again, you can stop here, at $\displaystyle k = -12$. Stopping here, you haven't proven that $\displaystyle \vec{F}$ is conservative. What you've proven is that **IF** $\displaystyle \vec{F}$ is conservative, and its associated $\displaystyle \phi$ is $\displaystyle C^2$, then the only possible value k could have is -12. - Oct 13th 2012, 08:57 AMmathsforumhelpRe: Which value of k is vector field conservative
thanks for replying.....

- Oct 13th 2012, 12:26 PMTheEmptySetRe: Which value of k is vector field conservative
Another useful criterion for a conservative vector field is that it is curl free.

In symbols that means

$\displaystyle \nabla \times \mathbf{v}=0$

So in your case that gives

$\displaystyle \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ (14+k)x+2y & (14+k)x+14z & 14y \end{vmatirx} $

If we evaluate this we get

$\displaystyle [14-14]\mathbf{i} -[0-0]\mathbf{j}+[(14+k)-2]\mathbf{k}=0$

So now we just have to solve

$\displaystyle 12+k=0 \iff k=-12$ - Oct 13th 2012, 01:19 PMjohnsomeoneRe: Which value of k is vector field conservative
My two approaches stopped at showing that, if F is conservative, then k must be -12. Your approach has the nice advantage of showing (assuming F is defined on the entire 3-space) that k = -12 actually guarantees that F is conservative.