Finding the Potential Function for a Conservative Vector Field

[fredrick08]

Homework Statement

F(x,y,z)=(3(x^2)*yz+y+5, (x^3)*z+x-z,(x^3)y-y+7) show that F is conservative and find the potential function V(x,y,z) with V(0,0,0)=10 which gives rise to F

Homework Equations

F x del=0
F=-delV

The Attempt at a Solution

ok i found that F was conservative, but can someone please tell me how to find the potential V(x,y,z), i thought if i could intergrate -F=V but then when I intergrate i get intergration constants, and i have no idea how to find them constants, so that V(0,0,0)=10

thanks for any help

 

[fredrick08]

idk i thought i could do a line intergral or something,V= - intergral F(r).dr, but i just keep confusing myself please someone help

 

[fredrick08]

sorry if anyone doesn't understand, we call del the gradient function.

 

[Cyosis]

2. Homework Equations
F x del=0
F=-delV

The first equation is wrong it should be \nabla \times \vec{F}=0

The potential function will be a function of the form \phi(x,y,z) and you know that \vec{F}=-\nabla \phi(x,y,z). Therefore the first component of \vec{F} is given by -\partial_x \phi(x,y,z). It is not hard to see what the general form of \phi has to be to at least satisfy the x component of F. Namely just integrating the first component of \vec{F} with respect to x. So from this you can conclude that -\phi(x,y,z)=x^3yz+xy+5x+f(y,z)+constant. Keep in mind that for any function f(y,z) you still satisfy the x component of F, because f(y,z) is constant with respect to x. Now try to continue this line of thought for the second and third component.

 

[fredrick08]

lol you srry... ok yes, i think i memba from vector calculus thankyou

 

[fredrick08]

ok if i follow on from what u did, i got

-V(x,y,z)=(x^3)yz+xy+5x+f(y,z)
Fy=(x^3)z+x-z=>f(y,z)=(x^3)yz+xy-yz+g(z)
-V(x,y,z)=2(x^3)yz+2xy-yz+5x+g(z)
dy/dz=(x^3)y-y=(x^3)y-y+7+dg/dz=>dg/dz=-7=>g(z)=-7z
-V(x,y,z)=2(x^3)yz+2xy-yz+5x-7z+(A+B+C, i think)

where A+B+C=10?

im pretty sure i done something wrong, i dotn know what to do with the constants... because there should be 3? i think

 

[fredrick08]

does that mean that those constants add up to 10?? and there is no way of finding that out?

 

[Cyosis]

-V(x,y,z)=(x^3)yz+xy+5x+f(y,z)

You assumed the constant to be 0 here. The general expression for -V(x,y,z) is -V(x,y,z)=x^3yz+xy+5x+f(y,z)+constant.

Fy=(x^3)z+x-z=>f(y,z)=(x^3)yz+xy-yz+g(z)

The first equation is correct (it's given to be that) the second equation is wrong. You could have easily checked that by checking if it still satisfies Fx. Using f(y,z)=x^3yz+xy-yz+g(z) we obtain -V(x,y,z)=2x^3yz+2xy-yz+5x+g(z). If we take the derivative with respect to x we get \partial_x (-V(x,y,z))=6x^2yz+2y+5 \neq 3x^2yz+y+5=F_x. On top of that it doesn't satisfy the y-component either. Always check your answer!

The correct way of doing it is to solve the equation F_y=\partial_y (-V(x,y,z)) for f(y,z). Don't forget the constant!

 

[fredrick08]

im confused now, because when i intergrate Fy to find f(y,z) i get f(y,z)=(x^3)yz+xy-yz+g(z)+B is this wrong? because don't i have to add it to the first equation? of course -V(x,y,z) doesn't equal Fx becasue it is added to Fy, so its Fx+Fy.. is this wrong?

 

[fredrick08]

therefore i think -V(x,y,z)=2(x^3)yz+2xy-yz+5x-7z+10?? its only thing i can think of doing, becasue the extra constants have to add up to 10?

 

[Cyosis]

I am confused now, because when i intergrate Fy to find f(y,z) i get f(y,z)=(x^3)yz+xy-yz+g(z)+B is this wrong? because don't i have to add it to the first equation? of course -V(x,y,z) doesn't equal Fx becasue it is added to Fy, so its Fx+Fy.. is this wrong?

Yes this is wrong. If you take the negative derivative of the potential function with respect to x you should get the x-component of F. If you take the negative derivative of the potential function with respect to y you should get the y-component of F. Besides how can a function f(y,z) that only depends on variables y and z turn out to be f(y,z)=(x^3)yz+xy-yz+g(z)+B, a function that clearly depends on x,y and z?

therefore i think -V(x,y,z)=2(x^3)yz+2xy-yz+5x-7z+10?? its only thing i can think of doing, becasue the extra constants have to add up to 10?

Before you worry about the constant, which is the last step, worry about the errors you made leading up to that point first, as explained in post #8.

I would like you to take the gradient of the potential function you have found and show me the calculations. You will see that this does not equal F and therefore cannot be the right potential.

 

[fredrick08]

ok, sorry i really struggling to remember.

the grad would be (3(x^2)yz+2y+5,2(x^3)z+2x-z,2(x^3)y-y-7) you doesn't equal F.

ok then how do i do this then, because i thought u had to intergarte Fx with respect to x, plus interate Fy with with respect to y then plus intergrate Fz with respect to z??

 

[fredrick08]

from this i can clearly see the answer should be (x^3)yz+yx+5x-zy+7z, lol can't memba how to prove it

 

[Cyosis]

ok, sorry i really struggling to remember.

the grad would be (3(x^2)yz+2y+5,2(x^3)z+2x-z,2(x^3)y-y-7) you doesn't equal F.

You have difficulty remembering what a grad is? Before you even attempt a question you first familiarize yourself with the definition of the objects present in said question.

While the conclusion is correct, you took the gradient wrong. Whats the derivative of 2x^3? According to you its 3x^2 which is incorrect.

I've seen you use 'memba' twice now, does this mean remember? I am not familiar with words like that and I would appreciate it if you kept it to English.

from this i can clearly see the answer should be (x^3)yz+yx+5x-zy+7z, lol can't memba how to prove it

You forgot the constant but other than that the answer is correct.

I have explained to you how to obtain this result without guessing in post #4 and #8. In post #4 I calculated the x-component for you and in post #8 I listed the equation you need to solve to get the correct y-component. I suggest you read both posts thoroughly again and then show me your solution, including the z-component.

 

[fredrick08]

yes i know the grad function, sorry its quite late my head is not thinking too well. ok, now to find f(y,z) u said in post#8 to intergrate Fy with respect to y, i did this and it has x's in it... which you said was wrong, this is how i understood what you said, so I am confused. because part i intergrated Fx with respect to x, and this is correct from what you said. so how can i get f(y,z)? do i have to subtract from the first equation or something?

 

[Cyosis]

In part one we integrated the x-component to find a general -V(x,y,z) that satisfies the x-component. The solution to this problem entails a potential that of course when differentiated with respect to x yields us the correct x-component of F. Now that we have this general potential we have to fiddle around with it so that when differentiated with respect to y yields Fy and when differentiated with respect to z yields Fz.

To obtain the correct potential, that if differentiated with respect to y, yields the correct Fy we are only allowed to fiddle with the function f(y,z) and the constant in the general potential. If we change the other terms it will not satisfy the x-component anymore. Do you understand this?

All you need to do now is to find a general form of f(y,z) such that F_y=-\partial_y V(x,y,z). Write this equation out explicitly on both sides and solve for f(y,z).

 

[fredrick08]

ok i think i finally get it now, i was getting mixed up with what i was differentiating.

start from start.

-V(x,y,z)=(x^3)yz+xy+5x+f(y,z)+A
(x^3)z+x=Fy+df/dy=f(y,z)=-yz+g(z)+B
y=Fz+dg/dz=>g(z)=7z+C

-V(x,y,z)=(x^3)yz+yx+5x-zy+7z+10?? is that how to do it, or is my process still wrong

 

[fredrick08]

no that doesn't make sense... ok let me write it down clearly...

 

[Cyosis]

Great then I don't have to decipher it.:P

 

[fredrick08]

omg i can be stupid sometimes... (all the time)

-V(x,y,z)=(x^3)yz+yx+5x+f(y,z)+A
(x^3)z+x-z=(x^3)z+x+df/dy=>df/dy=-z=>f(y,z)=-zy+B+g(z)
-V(x,y,z)=(x^3)yz+yx+5x-zy+A+B+g(z)
(x^3)y-y+7=(x^3)y-y+dg/dz=>7=>g(z)=7z+C
-V(x,y,z)=(x^3)yz+yx+5x-zy+7z+10=>V(x,y,z)=-(x^3)yz-yx-5x+zy-7z-10

please say that is correct

 

[Cyosis]

(x^3)z+x-z=(x^3)z+x+df/dy=>df/dy=-z=>f(y,z)=-zy+B

Almost, don't forget that differentiating any function that depends on z with respect to y will yield 0. Therefore the general solution is f(y,z)=-zy+B+h(z).

In the next line the g(z) suddenly appears so the rest of the computation is correct!

 

[fredrick08]

sorry i was trying to work from example i had in my book... tipical it turned out that my examples were wrong lol, that's why i was getting confused.

sorry bad typing... i need to learn how to do that fancy maths writing, i just more confused when i look at what i have written sometimes...

 

[Cyosis]

It's unlikely that multiple examples in your book were wrong. It could be that they used a different method, but this is one way of doing things. You can check your answer by taking the gradient.

 

[fredrick08]

thanks for all your help, your the most persistent and helpful helper on here. you deserve much credit = ) (especially when dealing with idiots like me = p)

im not completely stupid, i just lack the knowledge required to completely understand a problem, your right though i have to stop rushing into things and understand what i have to do before i do it.

 

[Cyosis]

I think you should also practice a few more problems of this type so it becomes 'second nature'. Math is learned by just doing it and preferably doing it a lot.