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You CAN Ace Calculus

17calculus > vector fields > conservative vector fields

### Calculus Main Topics

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free ideas to save on books - bags - supplies ATTENTION INSTRUCTORS: The new 2018 version of 17calculus will include changes to the practice problem numbering system. If you would like advance information to help you prepare for spring semester, send us an email at 2018info at 17calculus.com.

Conservative Vector Fields and Potential Functions

A conservative vector field may also be called a gradient field.

A conservative vector field is a vector field that is a gradient of some function, in this context called a potential function. The idea is that you are given a gradient and you have to 'un-gradient' it to get the original function. The only kind of vector fields that you can 'un-gradient' are conservative vector fields.

Test For Conservative Vector Field

Conservative vector fields are also called irrotational since the curl is zero. So, you now have a test to see if a vector field is conservative: calculate the curl and see if it's zero. If so, then it is conservative, otherwise it is not conservative. Okay, so you may be saying to yourself, that works for three dimensions, but what about two dimensions? Well, it works for two dimensions too but your instructor may give you a 'special' equation. The next panel discusses this 'special' equation.

### Conservative Vector Fields in Two Dimensions

In the above discussion, we said that the test for a conservative vector field is to calculate the curl to see if it is zero. However, the curl uses the idea of the cross product, which works only in three dimensions. Here is how we apply the same idea to a two dimensional situation. This frees you from having to memorize those 'special' equations.

Let's say you are given the two dimensional vector field $$\vec{F}(x,y) = M(x,y)\hat{i} + N(x,y)\hat{j}$$ and you need to determine if it is conservative.
First, we expand it to three dimensions so that we can use find the curl. So we rewrite it as $$\vec{G}(x,y) = M(x,y)\hat{i} + N(x,y)\hat{j} + 0\hat{k}$$. We renamed this new three dimensional vector as $$\vec{G}$$ just to be clear that we are now working in three dimensions. Okay, let's set up the curl and evaluate it.

$$\displaystyle{ \begin{array}{rcl} \vec{ \nabla } \times \vec{G} & = & \begin{vmatrix} \vhat{i} & \vhat{j} & \vhat{k} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ M(x,y) & N(x,y) & 0 \end{vmatrix} = \begin{vmatrix} \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ N(x,y) & 0 \end{vmatrix} \hat{i} - \begin{vmatrix} \frac{\partial }{\partial x} & \frac{\partial }{\partial z} \\ M(x,y) & 0 \end{vmatrix} \hat{j} + \begin{vmatrix} \frac{\partial }{\partial x} & \frac{\partial }{\partial y} \\ M(x,y) & N(x,y) \end{vmatrix} \hat{k} \\ \\ & = & \left[ \frac{\partial 0 }{\partial y} - \frac{\partial N(x,y)}{\partial z} \right] \hat{i} - \left[ \frac{\partial 0 }{\partial x} - \frac{\partial M(x,y)}{\partial z} \right] \hat{j} + \left[ \frac{\partial N(x,y)}{\partial x} - \frac{\partial M(x,y)}{\partial y} \right] \hat{k} \end{array} }$$

Okay, let's look at each component separately. Of course, you know that the derivative of zero is zero. That's easy. So for the $$\hat{i}$$ component we have the term $$\partial N(x,y) / \partial z$$. Notice that $$N(x,y)$$ does not contain any $$z$$'s. So this derivative is zero. The same reasoning holds for $$\partial M(x,y) / \partial z$$. This means that we have zero for both the $$\hat{i}$$ and $$\hat{j}$$ components.

For a vector field to be conservative, we need the curl to be zero. Since we have zero for both the $$\hat{i}$$ and $$\hat{j}$$ components, we just need to look at the $$\hat{k}$$ component. For it to be zero, we need

$$\displaystyle{ \frac{\partial N(x,y)}{\partial x} = \frac{\partial M(x,y)}{\partial y} }$$

And that's your 'special' equation. Cool, eh? We don't think that special equations should be memorized. We recommend that you understand the technique that applies to a wider range of problems. However, what your instructor expects comes first. So check with them before you take our advice.

Okay, let's watch a quick video that explains this idea.

 PatrickJMT - Conservative Vector Fields - The Definition and a Few Remarks

Potential Function

If you have a conservative vector field, you will probably be asked to determine the potential function. This is the function from which conservative vector field ( the gradient ) can be calculated. So you just need to set up two or three multi-variable (partial) integrals ( depending if you are working in $$\mathbb{R}^2$$ or $$\mathbb{R}^3$$ ), evaluate them and combine them to get one potential function. The equations look like this.

$$\partial g / \partial x = G_i ~~~ \to ~~~ \int{G_i~dx} = g_1 + c_1(y,z)$$ $$\partial g / \partial y = G_j ~~~ \to ~~~ \int{G_j~dy} = g_2 + c_2(x,z)$$ given conservative vector field ( a gradient ) $$\vec{G}(x,y,z) = G_i\hat{i} + G_j\hat{j} + G_k\hat{k}$$ potential function to calculate using $$\nabla g = \vec{G}$$ $$g(x,y,z)$$ combine $$g_1 + c_1(y,z)$$, $$g_2 + c_2(x,z)$$ and $$g_3 + c_3(x,y)$$ to get $$g(x,y,z)$$ practice problem A01 below shows, in detail, how this works

### Search 17Calculus

Practice Problems

Instructions - - Unless otherwise instructed, determine if the given vector field is conservative. If it is, find the potential function.

 Level A - Basic

Practice A01

$$\vec{F}(x,y,z) = 2xy\vhat{i} + (x^2+2yz)\vhat{j} + y^2 \vhat{k}$$

solution

Practice A02

$$\vec{F} = x^2\hat{i} + y\hat{j}$$

solution

Practice A03

$$\vec{F} = \langle 2x+yz, xz, xy \rangle$$

solution

Practice A04

$$\vec{F} = \langle 3x+2y, 2x-3y \rangle$$

solution

 Level B - Intermediate

Practice B01

$$\vec{F}=(6xy+4z^2)\hat{i} + (3x^2+3y^2)\hat{j} + 8xz\hat{k}$$

solution

Practice B02

$$\vec{F} = \langle z^2+2xy, x^2+2, 2xz-1 \rangle$$

solution