Which value of k is vector field conservative

**mathsforumhelp** · October 13th 2012, 04:31 AM

$Which value of k is vector field conservative-phpkeko6g.png$

$Which value of k is vector field conservative-screen-shot-2012-10-13-7.49.37-pm.png$

Is the answer correct......

**johnsomeone** · October 13th 2012, 06:58 AM

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:

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

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

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

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

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

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.

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

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

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

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

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

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

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

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:

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

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

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

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

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

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

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

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

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

Again, you can stop here, at $k = -12$ . Stopping here, you haven't proven that $\vec{F}$ is conservative. What you've proven is that **IF** $\vec{F}$ is conservative, and its associated $\phi$ is $C^2$ , then the only possible value k could have is -12.

**mathsforumhelp** · October 13th 2012, 08:57 AM

thanks for replying.....

**TheEmptySet** · October 13th 2012, 12:26 PM

Another useful criterion for a conservative vector field is that it is curl free.

In symbols that means

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

So in your case that gives

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

If we evaluate this we get

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

So now we just have to solve

$12+k=0 \iff k=-12$

**johnsomeone** · October 13th 2012, 01:19 PM

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.

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