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Lecture Comments (7)

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

Potential Functions, Continued

Find (if any) the potential function PF for the vector field F(x,y) = (xy2,x2y) 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) = xy2 and g(x,y) = x2y we obtain [df/dy] = 2xy = 2xy = [dg/dx], and so a potential function PF exists.
  • To find the potential function PF recall that by definition ∇PF = F(x,y) = ( [(dPF)/dx],[(dPF)/dy] ).
  • So ∇PF = (xy2,x2y) and [(dPF)/dx] = xy2. Integrating this partial derivative in terms of x yields ∫[(dPF)/dx]dx = ∫xy2dx or equivalently PF = [1/2]x2y2 + h(y).
  • We proceed by computing [d/dy]PF = [d/dy]( [1/2]x2y2 + h(y) ) or equivalently [(dPF)/dy] = x2y + h′(y). Note that this expression must equal the same value as the y - component of ∇PF.
  • Then x2y + h′(y) = x2y and thus conclude that h′(y) = 0. It follows that h(y) is a constant. (Recall that the derivative of a constant is zero).
We have PF = [1/2]x2y2 + C, if we take C = 0 we have our potential function PF = [1/2]x2y2.
Find (if any) the potential function PF for the vector field F(x,y) = (x2 + 2xy2,y2 + 2x2y) 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) = x2 + 2xy2 and g(x,y) = y2 + 2x2y we obtain [df/dy] = 4xy = 4xy = [dg/dx], and so a potential function PF exists.
  • To find the potential function PF recall that by definition ∇PF = F(x,y) = ( [(dPF)/dx],[(dPF)/dy] ).
  • So ∇PF = (x2 + 2xy2,y2 + 2x2y) and [(dPF)/dy] = y2 + 2x2y. Integrating this partial derivative in terms of y yields ∫[(dPF)/dy]dy = ∫( y2 + 2x2y )dy or equivalently PF = [1/3]y3 + x2y2 + h(x).
  • We proceed by computing [d/dx]PF = [d/dx]( [1/3]y3 + x2y2 + h(x) ) or equivalently [(dPF)/dy] = 2xy2 + h′(x). Note that this expression must equal the same value as the x - component of ∇PF.
  • Then 2xy2 + h′(x) = x2 + 2xy2 and thus conclude that h′(x) = x2. Computing ∫h′(x)dx = ∫x2dx yields h(x) = [1/3]x3 + C
We have PF = [1/3]y3 + x2y2 + [1/3]x3 + C, if we take C = 0 we have our potential function PF = [1/3]y3 + x2y2 + [1/3]x3.
Find (if any) the potential function PF for the vector field F(x,y) = ( xy2 + [1/3]y3,x2y + xy2 ) 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) = xy2 + [1/3]y3 and g(x,y) = x2y + xy2 we obtain [df/dy] = 2xy + y2 = 2xy + y2 = [dg/dx], and so a potential function PF exists.
  • To find the potential function PF recall that by definition ∇PF = F(x,y) = ( [(dPF)/dx],[(dPF)/dy] ).
  • So ∇PF = ( xy2 + [1/3]y3,x2y + xy2 ) and [(dPF)/dx] = xy2 + [1/3]y3. Integrating this partial derivative in terms of x yields ∫[(dPF)/dx]dx = ∫( xy2 + [1/3]y3 )dx or equivalently PF = [1/2]x2y2 + [1/3]xy3 + h(y).
  • We proceed by computing [d/dy]PF = [d/dy]( [1/2]x2y2 + [1/3]xy3 + h(y) ) or equivalently [(dPF)/dy] = x2y + xy2 + h′(y). Note that this expression must equal the same value as the y - component of ∇PF.
  • Then x2y + xy2 + h′(y) = x2y + xy2 and thus conclude that h′(y) = 0. It follows that h(y) is a constant. (Recall that the derivative of a constant is zero).
We have PF = [1/2]x2y2 + [1/3]xy3 + C, if we take C = 0 we have our potential function PF = [1/2]x2y2 + [1/3]xy3.
Find (if any) the potential function PF for the vector field F(x,y) = ( xln(y),[(x2)/2y] ) over an open rectangle 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) = xln(y) and g(x,y) = [(x2)/2y] we obtain [df/dy] = [x/y] = [x/y] = [dg/dx], and so a potential function PF exists.
  • To find the potential function PF recall that by definition ∇PF = F(x,y) = ( [(dPF)/dx],[(dPF)/dy] ).
  • So ∇PF = ( xln(y),[(x2)/2y] ) and [(dPF)/dx] = xln(y). Integrating this partial derivative in terms of x yields ∫[(dPF)/dx]dx = ∫( xln(y) )dx or equivalently PF = [1/2]x2ln(y) + h(y).
  • We proceed by computing [d/dy]PF = [d/dy]( [1/2]x2ln(y) + h(y) ) or equivalently [(dPF)/dy] = [(x2)/2y] + h′(y). Note that this expression must equal the same value as the y - component of ∇PF.
  • Then [(x2)/2y] + h′(y) = [(x2)/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).
We have PF = [1/2]x2ln(y) + C, if we take C = 0 we have our potential function PF = [1/2]x2ln(y).
Find (if any) the potential function PF for the vector field F(x,y) = ( e − xy,e − xy ) 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) = e − xy and g(x,y) = e − xy we obtain [df/dy] = − xe − xy − ye − xy = [dg/dx], and so a potential function PF does not exists.
Find (if any) the potential function PF for the vector field F(x,y) = ( xe − y, − [1/2]x2e − 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) = xe − y and g(x,y) = − [1/2]x2e − y we obtain [df/dy] = − xe − y = − xe − y = [dg/dx], and so a potential function PF exists.
  • To find the potential function PF recall that by definition ∇PF = F(x,y) = ( [(dPF)/dx],[(dPF)/dy] ).
  • So ∇PF = ( xe − y, − [1/2]x2e − y ) and [(dPF)/dx] = xe − y. Integrating this partial derivative in terms of x yields ∫[(dPF)/dx]dx = ∫xe − ydx or equivalently PF = [1/2]x2e − y + h(y).
  • We proceed by computing [d/dy]PF = [d/dy]( [1/2]x2e − y + h(y) ) or equivalently [(dPF)/dy] = − [1/2]x2e − y + h′(y). Note that this expression must equal the same value as the y - component of ∇PF.
  • Then − [1/2]x2e − y + h′(y) = − [1/2]x2e − 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).
We have PF = [1/2]x2e − y + C, if we take C = 0 we have our potential function PF = [1/2]x2e − y.
Find (if any) the potential function PF for the vector field F(x,y) = (ycos(x),sin(x)) 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) = ycos(x) and g(x,y) = sin(x) we obtain [df/dy] = cos(x) = cos(x) = [dg/dx], and so a potential function PF exists.
  • To find the potential function PF recall that by definition ∇PF = F(x,y) = ( [(dPF)/dx],[(dPF)/dy] ).
  • So ∇PF = (ycos(x),sin(x)) and [(dPF)/dx] = ycos(x). Integrating this partial derivative in terms of x yields ∫[(dPF)/dx]dx = ∫ycos(x)dx or equivalently PF = ysin(x) + h(y).
  • We proceed by computing [d/dy]PF = [d/dy]( ysin(x) + h(y) ) or equivalently [(dPF)/dy] = sin(x) + h′(y). Note that this expression must equal the same value as the y - component of ∇PF.
  • 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).
We have PF = ysin(x) + C, if we take C = 0 we have our potential function PF = ysin(x).
Find (if any) the potential function PF for the vector field F(x,y) = (4x2 + ey,y2 + xey) 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) = 4x2 + ey and g(x,y) = y2 + xey we obtain [df/dy] = ey = ey = [dg/dx], and so a potential function PF exists.
  • To find the potential function PF recall that by definition ∇PF = F(x,y) = ( [(dPF)/dx],[(dPF)/dy] ).
  • So ∇PF = (4x2 + ey,y2 + xey) and [(dPF)/dx] = 4x2 + ey. Integrating this partial derivative in terms of x yields ∫[(dPF)/dx]dx = ∫( 4x2 + ey )dx or equivalently PF = [4/3]x3 + xey + h(y).
  • We proceed by computing [d/dy]PF = [d/dy]( [4/3]x3 + xey + h(y) ) or equivalently [(dPF)/dy] = xey + h′(y). Note that this expression must equal the same value as the y - component of ∇PF.
  • Then xey + h′(y) = y2 + xey and thus conclude that h′(y) = y2. Computing ∫h′(y)dy = ∫y2dx yields h(y) = [1/3]y3 + C.
We have PF = [4/3]x3 + xey + [1/3]y3 + C, if we take C = 0 we have our potential function PF = [4/3]x3 + xey + [1/3]y3.
Find (if any) the potential function PF 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 PF exists.
  • To find the potential function PF recall that by definition ∇PF = F(x,y) = ( [(dPF)/dx],[(dPF)/dy],[(dPF)/dz] ).
  • So ∇PF = (y + z,x + z,x + y) and [(dPF)/dx] = y + z. Integrating this partial derivative in terms of x yields ∫[(dPF)/dx]dx = ∫( y + z )dx or equivalently PF = xy + xz + k(y,z).
  • We proceed by computing [d/dy]PF = [d/dy]( xy + xz + k(y,z) ) or equivalently [(dPF)/dy] = x + [dk/dy]. Note that this expression must equal the same value as the y - component of ∇PF.
  • 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 PF = xy + xz + k(y,z) = xy + xz + yz + l(z). Computing [d/dz]PF = [d/dz]( xy + xz + yz + l(z) ) gives [(dPF)/dz] = x + y + l′(z). Again, this expression must equal the same value as the z - component of ∇PF.
  • 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).
We have PF = xy + xz + yz + C, if we take C = 0 we have our potential function PF = xy + xz + yz.
Find (if any) the potential function PF 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 PF exists.
  • To find the potential function PF recall that by definition ∇PF = F(x,y) = ( [(dPF)/dx],[(dPF)/dy],[(dPF)/dz] ).
  • So ∇PF = (yz,xz,xy) and [(dPF)/dx] = yz. Integrating this partial derivative in terms of x yields ∫[(dPF)/dx]dx = ∫( yz )dx or equivalently PF = xyz + k(y,z).
  • We proceed by computing [d/dy]PF = [d/dy]( xyz + k(y,z) ) or equivalently [(dPF)/dy] = xz + [dk/dy]. Note that this expression must equal the same value as the y - component of ∇PF.
  • 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 PF = xyz + k(y,z) = xyz + l(z). Computing [d/dz]PF = [d/dz]( xyz + l(z) ) gives [(dPF)/dz] = xy + l′(z). Again, this expression must equal the same value as the z - component of ∇PF.
  • 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).
We have PF = xyz + C, if we take C = 0 we have our potential function PF = 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

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

pf 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 2x2, 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 pf of x and y, such that d(pf) 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 2x2.0341

That is what we want. You want to find this pf. 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 dx = 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 dpf = 4xy. Well, I am going to go ahead and move this over here, so this is going to be -- oops -- of pf = 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 4x2/2, 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 4x2/2, 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 pf dy = ... oh wait, let me simplify this a little bit, this is going to be 2x2 + h(y)... so when I differentiate that, this is equal to 4x -- wait a second here, what have I got, I have got 2x2 + h(y), so when I take the d dy of that, I am going to end up with... not this is 2x2y.0612

Now, when I take the d dy of this thing, this is going to be 2x2 + 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, 2x2 + dh dy = 2x2 + 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 2x2y + c, and we can always take c = to 0.0753

So, our potential function for this vector field is equal to 2x2y.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 dpf = 2x dx. I can integrate both sides and I can end up with this potential function for f is equal to x2 + 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 x2 + h, I will just go ahead and write it that way.1250

That is equal to the derivative with respect to y of x2 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 = 3y2/2 + 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 x2 + this h, which happens to be 3y2 + 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(pf) 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 2z2 + 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... x2 + 3y2/2 + 2z2 + 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

y3z + y, 3xy2z + x + z, and xy3 + 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 3y2z + 1.1497

If I take df2dx, I get 3y2z + 1, so that checks out.1517

Now I will take df1dz = y2 and df3dx, that is equal to y3, so that checks out. 1529

If I take df2 with respect to z, I end up with 3xy2 + 1, and if I take df3 with respect to y, I get the same thing, 3xy2 + 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... y3z + y, I am going to integrate with respect to x and I end up with xy3z + 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 3xy2z + x + dh dy. That equals 3xy2z + 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 z2/2, right? -- no, not z2/2 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 xy3z + xy + yz + kz.1723

Now I am going to go ahead and take d dz of this and I get xy3 + y, because this is 0, so + y + dk dz, and that is going to equal our third coordinate function, which is xy3 + 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 xy3z + 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