17Calculus - Potential Functions
When you have a conservative vector field, it is sometimes possible to calculate a potential function, i.e. to un-do the gradient.
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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 \( \nabla g = \vec{G}\) |
\( g(x,y,z) \) |
\( \partial g / \partial x = G_i ~~~ \to ~~~ \int{G_i~dx} = g_1 + c_1(y,z) \) | |
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\( \partial g / \partial y = G_j ~~~ \to ~~~ \int{G_j~dy} = g_2 + c_2(x,z) \) | |
\( \partial g / \partial z = G_k ~~~ \to ~~~ \int{G_k~dz} = g_3 + c_3(x,y) \) | |
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 | |
Here is a great video, with examples, showing how this works.
Dr Chris Tisdell - Potential Function Example (Gradient) [5mins-19secs]
video by Dr Chris Tisdell |
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Practice
Unless otherwise instructed, determine if the given vector field is conservative. If it is, find the potential function.
Basic |
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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.
Problem Statement |
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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 |
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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 video
video by PatrickJMT |
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Determine if the vector field \( \vec{F} = x^2\hat{i} + y\hat{j} \) is conservative. If it is, find the potential function.
Problem Statement |
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Determine if the vector field \( \vec{F} = x^2\hat{i} + y\hat{j} \) is conservative. If it is, find the potential function.
Solution |
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849 video
video by Dr Chris Tisdell |
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Determine if the vector field \( \vec{F} = \langle 2x+yz, xz, xy \rangle \) is conservative. If it is, find the potential function.
Problem Statement |
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Determine if the vector field \( \vec{F} = \langle 2x+yz, xz, xy \rangle \) is conservative. If it is, find the potential function.
Solution |
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850 video
video by PatrickJMT |
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Determine if the vector field \( \vec{F} = \langle 3x+2y, 2x-3y \rangle \) is conservative. If it is, find the potential function.
Problem Statement |
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Determine if the vector field \( \vec{F} = \langle 3x+2y, 2x-3y \rangle \) is conservative. If it is, find the potential function.
Solution |
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852 video
video by MIP4U |
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Determine if the vector field \( \nabla f = (2x+yz)\hat{i} + xz\hat{j} + xy\hat{k} \) is conservative. If it is, find the potential function.
Problem Statement |
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Determine if the vector field \( \nabla f = (2x+yz)\hat{i} + xz\hat{j} + xy\hat{k} \) is conservative. If it is, find the potential function.
Solution |
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811 video
video by PatrickJMT |
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Intermediate |
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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.
Problem Statement |
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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 |
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851 video
video by PatrickJMT |
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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.
Problem Statement |
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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 |
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853 video
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You CAN Ace Calculus
Topics You Need To Understand For This Page
Related Topics and Links
external links you may find helpful |
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Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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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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Practice Instructions
Unless otherwise instructed, determine if the given vector field is conservative. If it is, find the potential function.
