For more information, please see full course syllabus of Multivariable Calculus

For more information, please see full course syllabus of Multivariable Calculus

### Potential Functions, Continued

_{F}for the vector field F(x,y) = (xy

^{2},x

^{2}y) over an open plane S.

- We first verify that a potential function exists by showing that [df/dy] = [dg/dx] where F(x,y) = (f(x,y),g(x,y)).
- Since f(x,y) = xy
^{2}and g(x,y) = x^{2}y we obtain [df/dy] = 2xy = 2xy = [dg/dx], and so a potential function P_{F}exists. - To find the potential function P
_{F}recall that by definition ∇P_{F}= F(x,y) = ( [(dP_{F})/dx],[(dP_{F})/dy] ). - So ∇P
_{F}= (xy^{2},x^{2}y) and [(dP_{F})/dx] = xy^{2}. Integrating this partial derivative in terms of x yields ∫[(dP_{F})/dx]dx = ∫xy^{2}dx or equivalently P_{F}= [1/2]x^{2}y^{2}+ h(y). - We proceed by computing [d/dy]P
_{F}= [d/dy]( [1/2]x^{2}y^{2}+ h(y) ) or equivalently [(dP_{F})/dy] = x^{2}y + h′(y). Note that this expression must equal the same value as the y - component of ∇P_{F}. - Then x
^{2}y + h′(y) = x^{2}y and thus conclude that h′(y) = 0. It follows that h(y) is a constant. (Recall that the derivative of a constant is zero).

_{F}= [1/2]x

^{2}y

^{2}+ C, if we take C = 0 we have our potential function P

_{F}= [1/2]x

^{2}y

^{2}.

_{F}for the vector field F(x,y) = (x

^{2}+ 2xy

^{2},y

^{2}+ 2x

^{2}y) over an open plane S.

- We first verify that a potential function exists by showing that [df/dy] = [dg/dx] where F(x,y) = (f(x,y),g(x,y)).
- Since f(x,y) = x
^{2}+ 2xy^{2}and g(x,y) = y^{2}+ 2x^{2}y we obtain [df/dy] = 4xy = 4xy = [dg/dx], and so a potential function P_{F}exists. - To find the potential function P
_{F}recall that by definition ∇P_{F}= F(x,y) = ( [(dP_{F})/dx],[(dP_{F})/dy] ). - So ∇P
_{F}= (x^{2}+ 2xy^{2},y^{2}+ 2x^{2}y) and [(dP_{F})/dy] = y^{2}+ 2x^{2}y. Integrating this partial derivative in terms of y yields ∫[(dP_{F})/dy]dy = ∫( y^{2}+ 2x^{2}y )dy or equivalently P_{F}= [1/3]y^{3}+ x^{2}y^{2}+ h(x). - We proceed by computing [d/dx]P
_{F}= [d/dx]( [1/3]y^{3}+ x^{2}y^{2}+ h(x) ) or equivalently [(dP_{F})/dy] = 2xy^{2}+ h′(x). Note that this expression must equal the same value as the x - component of ∇P_{F}. - Then 2xy
^{2}+ h′(x) = x^{2}+ 2xy^{2}and thus conclude that h′(x) = x^{2}. Computing ∫h′(x)dx = ∫x^{2}dx yields h(x) = [1/3]x^{3}+ C

_{F}= [1/3]y

^{3}+ x

^{2}y

^{2}+ [1/3]x

^{3}+ C, if we take C = 0 we have our potential function P

_{F}= [1/3]y

^{3}+ x

^{2}y

^{2}+ [1/3]x

^{3}.

_{F}for the vector field F(x,y) = ( xy

^{2}+ [1/3]y

^{3},x

^{2}y + xy

^{2}) over an open plane S.

- We first verify that a potential function exists by showing that [df/dy] = [dg/dx] where F(x,y) = (f(x,y),g(x,y)).
- Since f(x,y) = xy
^{2}+ [1/3]y^{3}and g(x,y) = x^{2}y + xy^{2}we obtain [df/dy] = 2xy + y^{2}= 2xy + y^{2}= [dg/dx], and so a potential function P_{F}exists. - To find the potential function P
_{F}recall that by definition ∇P_{F}= F(x,y) = ( [(dP_{F})/dx],[(dP_{F})/dy] ). - So ∇P
_{F}= ( xy^{2}+ [1/3]y^{3},x^{2}y + xy^{2}) and [(dP_{F})/dx] = xy^{2}+ [1/3]y^{3}. Integrating this partial derivative in terms of x yields ∫[(dP_{F})/dx]dx = ∫( xy^{2}+ [1/3]y^{3})dx or equivalently P_{F}= [1/2]x^{2}y^{2}+ [1/3]xy^{3}+ h(y). - We proceed by computing [d/dy]P
_{F}= [d/dy]( [1/2]x^{2}y^{2}+ [1/3]xy^{3}+ h(y) ) or equivalently [(dP_{F})/dy] = x^{2}y + xy^{2}+ h′(y). Note that this expression must equal the same value as the y - component of ∇P_{F}. - Then x
^{2}y + xy^{2}+ h′(y) = x^{2}y + xy^{2}and thus conclude that h′(y) = 0. It follows that h(y) is a constant. (Recall that the derivative of a constant is zero).

_{F}= [1/2]x

^{2}y

^{2}+ [1/3]xy

^{3}+ C, if we take C = 0 we have our potential function P

_{F}= [1/2]x

^{2}y

^{2}+ [1/3]xy

^{3}.

_{F}for the vector field F(x,y) = ( xln(y),[(x

^{2})/2y] ) over an open rectangle S.

- Since f(x,y) = xln(y) and g(x,y) = [(x
^{2})/2y] we obtain [df/dy] = [x/y] = [x/y] = [dg/dx], and so a potential function P_{F}exists. - To find the potential function P
_{F}recall that by definition ∇P_{F}= F(x,y) = ( [(dP_{F})/dx],[(dP_{F})/dy] ). - So ∇P
_{F}= ( xln(y),[(x^{2})/2y] ) and [(dP_{F})/dx] = xln(y). Integrating this partial derivative in terms of x yields ∫[(dP_{F})/dx]dx = ∫( xln(y) )dx or equivalently P_{F}= [1/2]x^{2}ln(y) + h(y). - We proceed by computing [d/dy]P
_{F}= [d/dy]( [1/2]x^{2}ln(y) + h(y) ) or equivalently [(dP_{F})/dy] = [(x^{2})/2y] + h′(y). Note that this expression must equal the same value as the y - component of ∇P_{F}. - Then [(x
^{2})/2y] + h′(y) = [(x^{2})/2y] and thus conclude that h′(y) = 0. It follows that h(y) is a constant. (Recall that the derivative of a constant is zero).

_{F}= [1/2]x

^{2}ln(y) + C, if we take C = 0 we have our potential function P

_{F}= [1/2]x

^{2}ln(y).

_{F}for the vector field F(x,y) = ( e

^{ − xy},e

^{ − xy}) over an open plane S.

^{ − xy}and g(x,y) = e

^{ − xy}we obtain [df/dy] = − xe

^{ − xy}− ye

^{ − xy}= [dg/dx], and so a potential function P

_{F}does not exists.

_{F}for the vector field F(x,y) = ( xe

^{ − y}, − [1/2]x

^{2}e

^{ − y}) over an open plane S.

- Since f(x,y) = xe
^{ − y}and g(x,y) = − [1/2]x^{2}e^{ − y}we obtain [df/dy] = − xe^{ − y}= − xe^{ − y}= [dg/dx], and so a potential function P_{F}exists. - To find the potential function P
_{F}recall that by definition ∇P_{F}= F(x,y) = ( [(dP_{F})/dx],[(dP_{F})/dy] ). - So ∇P
_{F}= ( xe^{ − y}, − [1/2]x^{2}e^{ − y}) and [(dP_{F})/dx] = xe^{ − y}. Integrating this partial derivative in terms of x yields ∫[(dP_{F})/dx]dx = ∫xe^{ − y}dx or equivalently P_{F}= [1/2]x^{2}e^{ − y}+ h(y). - We proceed by computing [d/dy]P
_{F}= [d/dy]( [1/2]x^{2}e^{ − y}+ h(y) ) or equivalently [(dP_{F})/dy] = − [1/2]x^{2}e^{ − y}+ h′(y). Note that this expression must equal the same value as the y - component of ∇P_{F}. - Then − [1/2]x
^{2}e^{ − y}+ h′(y) = − [1/2]x^{2}e^{ − y}and thus conclude that h′(y) = 0. It follows that h(y) is a constant. (Recall that the derivative of a constant is zero).

_{F}= [1/2]x

^{2}e

^{ − y}+ C, if we take C = 0 we have our potential function P

_{F}= [1/2]x

^{2}e

^{ − y}.

_{F}for the vector field F(x,y) = (ycos(x),sin(x)) over an open plane S.

- Since f(x,y) = ycos(x) and g(x,y) = sin(x) we obtain [df/dy] = cos(x) = cos(x) = [dg/dx], and so a potential function P
_{F}exists. - To find the potential function P
_{F}recall that by definition ∇P_{F}= F(x,y) = ( [(dP_{F})/dx],[(dP_{F})/dy] ). - So ∇P
_{F}= (ycos(x),sin(x)) and [(dP_{F})/dx] = ycos(x). Integrating this partial derivative in terms of x yields ∫[(dP_{F})/dx]dx = ∫ycos(x)dx or equivalently P_{F}= ysin(x) + h(y). - We proceed by computing [d/dy]P
_{F}= [d/dy]( ysin(x) + h(y) ) or equivalently [(dP_{F})/dy] = sin(x) + h′(y). Note that this expression must equal the same value as the y - component of ∇P_{F}. - Then sin(x) + h′(y) = sin(x) and thus conclude that h′(y) = 0. It follows that h(y) is a constant. (Recall that the derivative of a constant is zero).

_{F}= ysin(x) + C, if we take C = 0 we have our potential function P

_{F}= ysin(x).

_{F}for the vector field F(x,y) = (4x

^{2}+ e

^{y},y

^{2}+ xe

^{y}) over an open plane S.

- Since f(x,y) = 4x
^{2}+ e^{y}and g(x,y) = y^{2}+ xe^{y}we obtain [df/dy] = e^{y}= e^{y}= [dg/dx], and so a potential function P_{F}exists. - To find the potential function P
_{F}recall that by definition ∇P_{F}= F(x,y) = ( [(dP_{F})/dx],[(dP_{F})/dy] ). - So ∇P
_{F}= (4x^{2}+ e^{y},y^{2}+ xe^{y}) and [(dP_{F})/dx] = 4x^{2}+ e^{y}. Integrating this partial derivative in terms of x yields ∫[(dP_{F})/dx]dx = ∫( 4x^{2}+ e^{y})dx or equivalently P_{F}= [4/3]x^{3}+ xe^{y}+ h(y). - We proceed by computing [d/dy]P
_{F}= [d/dy]( [4/3]x^{3}+ xe^{y}+ h(y) ) or equivalently [(dP_{F})/dy] = xe^{y}+ h′(y). Note that this expression must equal the same value as the y - component of ∇P_{F}. - Then xe
^{y}+ h′(y) = y^{2}+ xe^{y}and thus conclude that h′(y) = y^{2}. Computing ∫h′(y)dy = ∫y^{2}dx yields h(y) = [1/3]y^{3}+ C.

_{F}= [4/3]x

^{3}+ xe

^{y}+ [1/3]y

^{3}+ C, if we take C = 0 we have our potential function P

_{F}= [4/3]x

^{3}+ xe

^{y}+ [1/3]y

^{3}.

_{F}for the vector field F(x,y,z) = (y + z,x + z,x + y) over an open connected region S.

- We first verify that a potential function exists by showing that [dh/dy] = [dg/dz], [dh/dx] = [df/dz] and [dg/dx] = [df/dy] where F(x,y,z) = (f(x,y),g(x,y),h(x,y)).
- Since f(x,y) = y + z, g(x,y) = x + z and h(x,y) = x + y we obtain [dh/dy] = 1 = [dg/dz], [dh/dx] = 1 = [df/dz] and [dg/dx] = 1 = [df/dy], and so a potential function P
_{F}exists. - To find the potential function P
_{F}recall that by definition ∇P_{F}= F(x,y) = ( [(dP_{F})/dx],[(dP_{F})/dy],[(dP_{F})/dz] ). - So ∇P
_{F}= (y + z,x + z,x + y) and [(dP_{F})/dx] = y + z. Integrating this partial derivative in terms of x yields ∫[(dP_{F})/dx]dx = ∫( y + z )dx or equivalently P_{F}= xy + xz + k(y,z). - We proceed by computing [d/dy]P
_{F}= [d/dy]( xy + xz + k(y,z) ) or equivalently [(dP_{F})/dy] = x + [dk/dy]. Note that this expression must equal the same value as the y - component of ∇P_{F}. - Then x + [dk/dy] = x + z and thus conclude that [dk/dy] = z. Taking the partial derivative in terms of y yields ∫[dk/dy]dy = ∫( z )dy or equivalently k(y,z) = yz + l(z).
- So P
_{F}= xy + xz + k(y,z) = xy + xz + yz + l(z). Computing [d/dz]P_{F}= [d/dz]( xy + xz + yz + l(z) ) gives [(dP_{F})/dz] = x + y + l′(z). Again, this expression must equal the same value as the z - component of ∇P_{F}. - Then x + y + l′(z) = x + y and thus conclude that l′(z) = 0. It follows that l(z) is a constant. (Recall that the derivative of a constant is zero).

_{F}= xy + xz + yz + C, if we take C = 0 we have our potential function P

_{F}= xy + xz + yz.

_{F}for the vector field F(x,y,z) = (yz,xz,xy) over an open connected region S.

- We first verify that a potential function exists by showing that [dh/dy] = [dg/dz], [dh/dx] = [df/dz] and [dg/dx] = [df/dy] where F(x,y,z) = (f(x,y),g(x,y),h(x,y)).
- Since f(x,y) = yz, g(x,y) = xz and h(x,y) = xy we obtain [dh/dy] = x = [dg/dz], [dh/dx] = y = [df/dz] and [dg/dx] = z = [df/dy], and so a potential function P
_{F}exists. - To find the potential function P
_{F}recall that by definition ∇P_{F}= F(x,y) = ( [(dP_{F})/dx],[(dP_{F})/dy],[(dP_{F})/dz] ). - So ∇P
_{F}= (yz,xz,xy) and [(dP_{F})/dx] = yz. Integrating this partial derivative in terms of x yields ∫[(dP_{F})/dx]dx = ∫( yz )dx or equivalently P_{F}= xyz + k(y,z). - We proceed by computing [d/dy]P
_{F}= [d/dy]( xyz + k(y,z) ) or equivalently [(dP_{F})/dy] = xz + [dk/dy]. Note that this expression must equal the same value as the y - component of ∇P_{F}. - Then xz + [dk/dy] = xz and thus conclude that [dk/dy] = 0 and k(y,z) is a constant or a function of z alone. With out loss of generality let k(y,z) = l(z).
- So P
_{F}= xyz + k(y,z) = xyz + l(z). Computing [d/dz]P_{F}= [d/dz]( xyz + l(z) ) gives [(dP_{F})/dz] = xy + l′(z). Again, this expression must equal the same value as the z - component of ∇P_{F}. - Then xy + l′(z) = xy and thus conclude that l′(z) = 0. It follows that l(z) is a constant. (Recall that the derivative of a constant is zero).

_{F}= xyz + C, if we take C = 0 we have our potential function P

_{F}= xyz.

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Potential Functions, Continued

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Potential Functions 0:52
- Theorem 1
- Example 1
- Theorem in 3-Space
- Example 2
- Example 3

### Multivariable Calculus

### Transcription: Potential Functions, Continued

*Hello and welcome back to educator.com and multivariable calculus.*0000

*Today we are going to continue our discussion of potential functions.*0004

*In our last lesson, we gave a condition under which we can actually test to see whether a potential function exists for a given vector field, this thing about checking the quality or nonequality, as it turns out, of the partial derivatives of the 2 coordinate functions.*0008

*In this lesson, so, the converse turned out not to be true. Just because df/dy equals dg/dx, we cannot necessarily conclude that a potential function exists.*0025

*In this lesson, we are actually going to add onto, so add some hypotheses. *0040

*We are going to talk about conditions under which we can conclude that a potential function exists, so let us just get started.*0046

*Let us start off with a particular theorem. So, theorem, now, let f and g be differentiable on an open set s -- and of course f and g are the coordinate functions of f.*0055

*If s is the entire plane or a rectangle, if the partials of f and g exist and are continuous and if df/dy = dg/dx, which is equivalent to -- I will just write or -- d2f1 equals d1f2, then there does exist a potential function.*0102

*p _{f} for the vector field f coordinate functions f and g. So, again, in our last lesson, we said that just because df/dy = dg/dx, we cannot necessarily conclude that a potential function exists, without certain other things being in place.*0177

*This theorem lays out those other things that need to be there if the df/dy = dg/dx.*0199

*So, what has to happen is this. Not only does df/dy have to equal dg/dx, but we have to make sure that f and g are defined over the entire plane or the particular set that we are talking about is either the entire plane or a rectangle. *0207

*It is very, very important that it is a rectangle. It cannot just be some random set. It has to be a rectangle. If these hypotheses are satisfied, then we can conclude that there exists a potential function. *0227

*Let us just do an example. Example 1. So, let f = 4xy and 2x ^{2}, so this is f and this is g. First and second coordinate functions.*0241

*So, the first set of hypotheses is definitely satisfied. As far as this vector field is differentiable over the entire plane, so that is taken care of. The entire plane is taken care of.*0264

*Now we want to see if these are satisfied, df/dy, so, df/dy is going to equal 4x, and we are going to take dg/dx. dg/dx = 4x.*0282

*Those two are equal. So, all of the hypotheses of the theorem are satisfied. df/dy = dg/dx, therefore we can conclude, so there does exist a potential function for f. Now, let us see if we can find it.*0300

*Let us see if we can find it. This is the important part, well, existence is important, but from a practical standpoint, actually constructing the function is what we want.*0319

*So, here is what we do. So, let us make sure that we know what we want here.*0335

*We want some potential function p _{f} of x and y, such that d(p_{f}) dx = f -- we can be more specific that that, we know what f is -- = 4xy, and the derivative of this potential function with respect to y is equal to 2x^{2}.*0341

*That is what we want. You want to find this p _{f}. We want to find that. Okay, let me go ahead and move on over to blue and move over to the next page. *0376

*The first thing that we are going to do -- and this is a general procedure that you can always run through, this is what is really nice, we can actually construct this function by integrating one variable at a time.*0385

*Okay. First thing that we want to is we are going to integrate f with respect to x, so we are going to differentiate the first coordinate function with respect to x, the first variable. Okay.*0396

*Now, the d _{x} = 4xy.*0413

*I am just going to work formally here. In other words, I am going to move symbols around, and I am just going to move symbols around... you should be able to follow the mathematics. *0419

*I am going to move the dx. I cannot really do this, I mean -- really -- because this partial derivative is not the same as what we were working with in single variable calculus. *0429

*Remember when we had something like dy/dx, this actually represent a division, and you can sort of move things around. Remember we said dy = f'(dx), things like that. *0438

*Well, I am going to do the same sort of thing here, I am just working formally. I am just manipulating symbols, but the underlying mathematics is valid.*0452

*This is the situation, this dp _{f} = 4xy. Well, I am going to go ahead and move this over here, so this is going to be -- oops -- of p_{f} = 4xy dx.*0460

*Now I am just going to integrate both sides. Essentially what I have done is since I am just integrating with respect to one variable, this is just your normal dx.*0487

*I am just working formally so that you see where this comes from. Since I know that the potential function exists, the derivative when I take the derivative with respect to x of that potential function, I get my f which is 4xy, my first coordinate function.*0496

*I am just going to integrate, I am just going to work backwards. I am going to integrate with respect to x and recover a function.*0510

*Okay. When I integrate, I get that the potential function is equal to -- so when you integrate 4xy with respect to x, you are going to get 4x ^{2}/2, right?*0518

*Then this becomes y + some function of y, so remember when you integrate, this 4xy is what we are integrating. But, we are integrating only with respect to the x. *0538

*This becomes 4x ^{2}/2, the y stays because the y is just a constant.*0560

*This is going to be some function of y, because when you differentiate this function of y which is only y and differentiate with respect to x because it is only a function of y, it is going to go to 0.*0565

*In other words, the x derivative is 0. So, I actually have to include that. It is not just a constant that I am adding, it is an actual function of y, because when I differentiate this with respect to x it goes to 0, because it is just a function of y, partial derivatives but y is constant.*0579

*So I have this. Now, I am going to go ahead and take the d dy of this.*0598

*So, d of p _{f} dy = ... oh wait, let me simplify this a little bit, this is going to be 2x^{2} + h(y)... so when I differentiate that, this is equal to 4x -- wait a second here, what have I got, I have got 2x^{2} + h(y), so when I take the d dy of that, I am going to end up with... not this is 2x^{2}y.*0612

*Now, when I take the d dy of this thing, this is going to be 2x ^{2} + dh dy.*0653

*Set this equal to g, because that is exactly what it is. When I am taking the derivative of the potential function with respect to y, that is the same as the second coordinate function g.*0670

*What I have done is I have recovered the potential function in this form. If I differentiate it, I get this, but this happens to be equal to g because by definition the derivative of the potential function with respect to y is g, the second coordinate function.*0684

*So, let us see. So, 2x ^{2} + dh dy = 2x^{2} + 0. This comes from that, differentiating the potential function, this comes from, well, just the function g.*0707

*dh dy is equal to 0, which implies that h(y) is a constant.*0739

*Because it is a constant, so our potential function for f happens to be 2x ^{2}y + c, and we can always take c = to 0.*0753

*So, our potential function for this vector field is equal to 2x ^{2}y.*0785

*Then you can confirm this by taking the partial derivative with respect to x to get the f function, and you can take the partial derivative with respect to y to get the g function.*0793

*Let us stop and take a look at what it is that we have done. We have taken the f, the first coordinate function, we have integrated it with respect to x. We came up with this thing right here.*0804

*Then, what we did is we differentiated this with respect to y to get this thing, and then we set it equal to g, which it is by definition, and we just matched things up. We ended up with this = this, and now we integrated this with respect to y.*0816

*That is really what we are doing. We are integrating first with respect to x, finding out what dh dy is, since it is 0, well, when you integrate this, you are going to end up with a constant. We are going to keep doing that.*0835

*Let us do some more examples. Let me go ahead and write the theorem in 3-space, and then we will go ahead and do some 3-space examples. Okay. Theorem in 3-space.*0848

*Okay. Let f = f1, f2, f3 -- you know what, I need to write these a little bit clearer here -- okay, let f = f1, f2, f3 be a continuously differentiable vector field on an open set s in R3, because we are talking about 3-space, so, in R3.*0872

*So, again, if f -- if s -- is all of R3 or a rectangular box in R3, that is the important part, is a rectangular box and if d2f1 = d1f2, d2f3 = d3f2, d1f3 = d3f1, then f has a potential function.*0930

*Again, if these partials are equal and if it happens to be defined, if our domain happens to be this connected set s which happens to be a rectangle or the entire space, then we can conclude that f has a potential function.*0976

*Okay. Let me go ahead and write out the other version of this thing right here, so, again, this is if f is written as f(x,y,z) = f1(x,y,z), f2(x,y,z), f3(x,y,z), so f1, f2, f3, and our three variables, first, second, third, are x, y, z, then this condition right here in the other notation looks as follows: df1dy = df2dx, df3dy = df2dz, and df3dx = df1dz. This is just the equivalent of that.*0998

*Okay. Let us do another example here. Example 2.*1073

*Let f (x,y,z) = 2x, 3y, 4z. So, this is f1, this is f2, and this is f3. Find the potential function for f... let me do this in red.*1085

*Just to go ahead and... we need to check to see that the partials are actually equal... let us go ahead and just do that first, so, just to confirm.*1112

*So, df1dy = 0, df2dx = 0, df1dz = 0, df3dx = 0, so those are equal.*1125

*Now, df2dz, that equals 0, and df3dy, that also equals 0. So, 0, 0, 0, 0, 0, the equality is there.*1151

*This particular vector field is defined over the entire space and it is continuously differentiable over the entire space, so all of the hypotheses of the theorem are satisfied, so we know a potential function exists. Let us construct that potential function using our process.*1165

*We are going to integrate the first function with respect to the first variable. So, again, working formally, I know that the partial derivative of this potential function f with respect to x is equal to 2x.*1180

*Therefore, I can go ahead and say dp _{f} = 2x dx. I can integrate both sides and I can end up with this potential function for f is equal to x^{2} + some function h(y and z). *1200

*Now because I am working with 3 variables, x, y, and z. I differentiated -- sorry, I integrated with respect to the variable x, so the constant of integration is some function of y and z, because if I differentiate some function of y and z, it goes to 0.*1224

*This is not just a constant of c, it is some function of the other 2 variables.*1243

*So, now, let me go ahead and take the... I am going to take d dy of x ^{2} + h, I will just go ahead and write it that way.*1250

*That is equal to the derivative with respect to y of x ^{2} is 0, so what I just get is dh dy. Okay.*1266

*That happens to equal, the d of the potential function with respect to y happens to equal the second coordinate function, so it equals 3y.*1276

*Now I can go ahead and integrate this, right? dh = 3y dy, and again, this is probably not the best way of doing this, as far as working formally, but you know what is happening. This is just the normal dy. *1289

*I am integrating this function with respect to dy, and when I do that, I end up with the following.*1305

*I end up with h = 3y ^{2}/2 + some function, now of z, right?*1312

*I am doing one variable at a time, I integrated with respect to x and I ended up adding this constant, which happens to be a function of y and z. *1325

*I differentiated that with respect to y, I set that equal to 3y, and now I integrate it and I ended up with now this plus a constant which happens to be a function of z alone.*1334

*So, my potential function now is equal to x ^{2} + this h, which happens to be 3y^{2} + k, which is a function of z. *1348

*Now, if I go ahead and differentiate this with respect to z. I am going to set it equal to that because that is what it is by definition.*1364

*So d(p _{f}) dz, the derivative of this = 0, the derivative of this with respect to z = 0, so I get d(k) dz, well this happens to equal the 4z.*1373

*When I integrate that, I get k is equal to 2z ^{2} + c, now my c is just a normal constant because I am integrating just with respect to z. I have taken care of everything else.*1389

*So, my final potential function is... x ^{2} + 3y^{2}/2 + 2z^{2} + some constant, but of course we can take the constant equal to 0, so this is my potential function.*1409

*You can confirm this by taking the partial with respect to x, partial with respect to y, partial with respect to z.*1436

*In other words, take the gradient of this thing and you are going to recover the vector field.*1441

*Okay, let us do one more example. Let us do this one in black. Example 3.*1449

*y ^{3}z + y, 3xy^{2}z + x + z, and xy^{3} + y.*1465

*Okay, so that is our vector field, let us recover the potential function. Let us go ahead and make sure that all of the partials are equal.*1491

*So, we will do df... so this is f1, this right here is f2, so df1dy is going to equal... df1dy is going to be 3y ^{2}z + 1.*1497

*If I take df2dx, I get 3y ^{2}z + 1, so that checks out.*1517

*Now I will take df1dz = y ^{2} and df3dx, that is equal to y^{3}, so that checks out. *1529

*If I take df2 with respect to z, I end up with 3xy ^{2} + 1, and if I take df3 with respect to y, I get the same thing, 3xy^{2} + 1, and I hope that you are confirming this for me, I hope that I have done it right. *1543

*Okay, so the first thing that I am going to do is I am going to integrate with respect... I am to integrate this first function with respect to the first variable, x.*1568

*So, I am going to... this time I am going to write it straight... y ^{3}z + y, I am going to integrate with respect to x and I end up with xy^{3}z + xy + some function h which is a function of y and z. *1576

*So this is our potential function. Now, I am going to take this thing and I am going to differentiate it with respect to y and I am going to set it equal to the second coordinate function, which his exactly what it is. *1605

*So if I differentiate this with respect to y, I am going to get 3xy ^{2}z + x + dh dy. That equals 3xy^{2}z + x + z.*1623

*Therefore, that is equal to that, because this is this, this is this, and so therefore we have these two equal to each other. So, dh dy = z.*1652

*So, I have got dh dy = z, so now I am going to integrate this with respect to y, because it is y here. *1668

*I am going to integrate, so h = the integral of z with respect to y, and that is going to equal z ^{2}/2, right? -- no, not z^{2}/2 see I made the same mistake, this is where you have to be careful, I am actually integrating with respect to y, so z is a constant.*1682

*This is actually going to be zy, zy + some function of z, so be very careful. As you can see, I made the mistake.*1709

*So, now, our potential function is equal to xy ^{3}z + xy + yz + kz.*1723

*Now I am going to go ahead and take d dz of this and I get xy ^{3} + y, because this is 0, so + y + dk dz, and that is going to equal our third coordinate function, which is xy^{3} + y + 0.*1744

*Therefore, dk dz = 0, which implies that k = some constant, which we can always take to be 0.*1780

*So, our potential function for this particular vector field is... well... it is this thing but k is equal to a constant which we can take as 0, so what we are left with is xy ^{3}z + xy + yz. There you go, that is our potential function.*1793

*So, once we know the certain hypotheses are satisfied, we can go ahead and just start to construct this potential function.*1824

*We do it by integrating one variable at a time. We take the first coordinate function integrate with respect to x. We get our potential function, we differentiate that with respect to the next variable which is y, and we say equal to the second coordinate function.*1831

*We match corresponding terms, and then we integrate with respect to y. Along the way, we recover an extra constant function. We differentiate that with respect to the next variable z, we set it equal to the third coordinate function.*1847

*We match corresponding terms, and then we integrate with respect to z. Then we put it all together in the end to come up with our potential function. *1865

*That is it. It is really that simple. Really, the only thing that you have to watch out for here is pretty much the mistake that I made myself.*1872

*Up here, this here we are integrating z with respect to y. z is a constant, so this is just zy, or yz. That is it, you just have to make sure that you are keeping track of the variable with respect to which you are integrating and any of the other variables inside.*1881

*Thank you for joining us here at educator.com, we will see you next time for the conclusion and summary of potential functions. Take care, bye-bye.*1899

1 answer

Last reply by: Professor Hovasapian

Wed Nov 25, 2015 7:11 PM

Post by Hen McGibbons on November 25, 2015

potential function lectures have been great. was beyond confused with what was going on in class and then watched these lectures. everything clicked

1 answer

Last reply by: Professor Hovasapian

Mon May 13, 2013 2:12 AM

Post by Josh Winfield on May 11, 2013

Example 1: Why is it the partial of h wrt partial y when h only depends on one variable y

2 answers

Last reply by: mateusz marciniak

Sat Jan 19, 2013 11:07 AM

Post by mateusz marciniak on January 16, 2013

hello professor at the end of the lecture you explained the step by step way we solved the problem. you mentioned we integrate with respect to z but i don't recall us doing that. once we matched up the corresponding terms with the second function, we then took the potential and differentiated with respect to z and matched up the terms with the third function, we found the constant to be 0 and that was it, we never integrated with respect to z as far as i can tell. i'm a little confused