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Force and potential problem

  1. Aug 19, 2009 #1
    1. The problem statement, all variables and given/known data
    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

    2. Relevant equations
    F x del=0

    3. 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
    Last edited: Aug 19, 2009
  2. jcsd
  3. Aug 19, 2009 #2
    idk i thought i could do a line intergral or something,V= - intergral F(r).dr, but i just keep confusing myself plz someone help
  4. Aug 19, 2009 #3
    sorry if anyone doesnt understand, we call del the gradient function.
  5. Aug 19, 2009 #4


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

    The potential function will be a function of the form [itex]\phi(x,y,z)[/itex] and you know that [itex]\vec{F}=-\nabla \phi(x,y,z)[/itex]. Therefore the first component of [itex]\vec{F}[/itex] is given by [itex]-\partial_x \phi(x,y,z)[/itex]. It is not hard to see what the general form of [itex]\phi[/itex] has to be to at least satisfy the x component of F. Namely just integrating the first component of [itex]\vec{F}[/itex] with respect to x. So from this you can conclude that [itex]-\phi(x,y,z)=x^3yz+xy+5x+f(y,z)+constant[/itex]. 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.
    Last edited: Aug 19, 2009
  6. Aug 19, 2009 #5
    lol ya srry... ok yes, i think i memba from vector calculus thankyou
  7. Aug 19, 2009 #6
    ok if i follow on from what u did, i got

    -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
    Last edited: Aug 19, 2009
  8. Aug 19, 2009 #7
    does that mean that those constants add up to 10?? and there is no way of finding that out?
  9. Aug 19, 2009 #8


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

    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 [itex]f(y,z)=x^3yz+xy-yz+g(z)[/itex] we obtain [itex]-V(x,y,z)=2x^3yz+2xy-yz+5x+g(z)[/itex]. If we take the derivative with respect to x we get [itex]\partial_x (-V(x,y,z))=6x^2yz+2y+5 \neq 3x^2yz+y+5=F_x[/itex]. 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 [itex]F_y=\partial_y (-V(x,y,z))[/itex] for f(y,z). Don't forget the constant!
    Last edited: Aug 19, 2009
  10. Aug 19, 2009 #9
    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 dont i have to add it to the first equation? of course -V(x,y,z) doesnt equal Fx becasue it is added to Fy, so its Fx+Fy.. is this wrong?
  11. Aug 19, 2009 #10
    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?
  12. Aug 19, 2009 #11


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    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 [itex] f(y,z)=(x^3)yz+xy-yz+g(z)+B[/itex], a function that clearly depends on x,y and z?

    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.
  13. Aug 19, 2009 #12
    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) ya doesnt 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??
  14. Aug 19, 2009 #13
    from this i can clearly see the answer should be (x^3)yz+yx+5x-zy+7z, lol cant memba how to prove it
  15. Aug 19, 2009 #14


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    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.

    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.
  16. Aug 19, 2009 #15
    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 im 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?
  17. Aug 19, 2009 #16


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    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 [itex]F_y=-\partial_y V(x,y,z)[/itex]. Write this equation out explicitly on both sides and solve for f(y,z).
  18. Aug 19, 2009 #17
    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+yx+5x-zy+7z+10?? is that how to do it, or is my process still wrong
  19. Aug 19, 2009 #18
    no that doesnt make sense... ok let me write it down clearly...
  20. Aug 19, 2009 #19


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    Great then I don't have to decipher it.:P
  21. Aug 19, 2009 #20
    omg i can be stupid sometimes... (all the time)


    plz say that is correct
    Last edited: Aug 19, 2009
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