\( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \)

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effective study techniques

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.

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

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. Let's watch a quick video that explains this idea in more detail.

PatrickJMT - Conservative Vector Fields - The Definition and a Few Remarks [4mins-1sec]

video by PatrickJMT

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 section 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.

\( \vec{ \nabla } \times \vec{G} \)

\( \begin{vmatrix} \vhat{i} & \vhat{j} & \vhat{k} \\ \displaystyle{\frac{\partial }{\partial x}} & \displaystyle{ \frac{\partial }{\partial y}} & \displaystyle{\frac{\partial }{\partial z}} \\ M(x,y) & N(x,y) & 0 \end{vmatrix} \)

\( \begin{vmatrix} \displaystyle{\frac{\partial }{\partial y}} & \displaystyle{\frac{\partial }{\partial z}} \\ N(x,y) & 0 \end{vmatrix} \hat{i} - \begin{vmatrix} \displaystyle{\frac{\partial }{\partial x}} & \displaystyle{\frac{\partial }{\partial z}} \\ M(x,y) & 0 \end{vmatrix} \hat{j} + \begin{vmatrix} \displaystyle{\frac{\partial }{\partial x}} & \displaystyle{\frac{\partial }{\partial y}} \\ M(x,y) & N(x,y) \end{vmatrix} \hat{k} \)

\( \left[ \displaystyle{\frac{\partial 0 }{\partial y} - \frac{\partial N(x,y)}{\partial z}} \right] \hat{i} - \displaystyle{\left[ \frac{\partial 0 }{\partial x} - \frac{\partial M(x,y)}{\partial z} \right]} \hat{j} + \displaystyle{\left[ \frac{\partial N(x,y)}{\partial x} - \frac{\partial M(x,y)}{\partial y} \right]} \hat{k} \)

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.

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.

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 \( \nabla g = \vec{G}\)

\( g(x,y,z) \)

\( \partial g / \partial x = G_i ~~~ \to ~~~ \int{G_i~dx} = g_1 + c_1(y,z) \)

\( \partial g / \partial y = G_j ~~~ \to ~~~ \int{G_j~dy} = g_2 + c_2(x,z) \)

\( \partial g / \partial z = G_k ~~~ \to ~~~ \int{G_k~dz} = g_3 + c_3(x,y) \)

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 841 shows, in detail, how this works

Practice

Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems

Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. So, Practice A01 (1) is probably the first basic practice problem, A02 (2) is probably the second basic practice problem, etc. Practice B01 is probably the first intermediate practice problem and so on.

GOT IT. THANKS!

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

Basic Problems

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

Problem Statement

Determine if the vector field \( \vec{F}(x,y,z) = 2xy\vhat{i} + (x^2+2yz)\vhat{j} + y^2 \vhat{k} \) is conservative. If it is, find the potential function.

Solution

The video does not show how to determine the potential function, so we include the details here.

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

\( \int{2xy~dx} = x^2y + c_1(y,z) \)

\( \int{x^2+2yz~dy} = x^2y + y^2z + c_2(x,z) \)

\( \int{y^2~dz} = y^2z + c_3(x,y) \)

Now we need to combine the results of the three integrals.
From the first integration, we have \(x^2y\).
From the second integration, we have \( y^2z \). We do not use \(x^2y\) since we got it from the first integration.
From the third integration, we have nothing new. The term \( y^2z \) we already have from the second integration.
So our potential function is \( f(x,y,z) = x^2y + y^2z \). Technically, we should have a constant in the equation but we usually let that be equal to zero. So we should really call this A potential function, one of many, in fact infinite, from which we could determine the conservative vector field. With a quick calculation in your head, you can verify that \( \nabla f = \vec{F} \).

841 solution video

video by PatrickJMT

close solution

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

Problem Statement

Determine if the vector field \( \vec{F} = x^2\hat{i} + y\hat{j} \) is conservative. If it is, find the potential function.

Solution

849 solution video

video by Dr Chris Tisdell

close solution

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

Problem Statement

Determine if the vector field \( \vec{F} = \langle 2x+yz, xz, xy \rangle \) is conservative. If it is, find the potential function.

Solution

850 solution video

video by PatrickJMT

close solution

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

Problem Statement

Determine if the vector field \( \vec{F} = \langle 3x+2y, 2x-3y \rangle \) is conservative. If it is, find the potential function.

Solution

852 solution video

video by MIP4U

close solution

Intermediate Problems

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

Problem Statement

Determine if the vector field \( \vec{F} = (6xy+4z^2)\hat{i} + (3x^2+3y^2)\hat{j} + 8xz\hat{k} \) is conservative. If it is, find the potential function.

Solution

851 solution video

video by PatrickJMT

close solution

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

Problem Statement

Determine if the vector field \( \vec{F} = \langle z^2+2xy, x^2+2, 2xz-1 \rangle \) is conservative. If it is, find the potential function.

Solution

853 solution video

video by MIP4U

close solution
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