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. |
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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.
video by PatrickJMT |
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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.
Okay, so you may be wondering, what's the point? Why do we care? Well, conservative vectors fields are unique in that, if we assume the vector field we have is a the result of taking the gradient of another function, it is sometimes possible to undo the gradient to find the other function, which we call a potential function.
You CAN Ace Calculus
external links you may find helpful |
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The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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