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Partial derivative

  1. Jan 13, 2014 #1
    Hi everyone!

    I'm not sure if this is the right forum to post my question. If i'm wrong, let me know it.

    The question:

    Let us consider the functions [itex]\theta=\theta(x,y)[/itex], and [itex]M=M(\theta)[/itex], where M is a operator, but i doesn't relevant to the problem. I need to know the derivative [tex]\frac{\partial M}{\partial \theta}[/tex] in terms of x and y. Additional information: x and y are the longitud and latitude of a sphere, thus, every arc of sphere, θ, can be descomposed in two arcs, one associated to longitud and other associated with the lattitude. This is the aim of achieve the ∂M/∂θ in terms of x and y.

    It's a trivial question, but I'm stuck...

    This is my try...

    [tex] dM=\frac{\partial M}{\partial x}dx+\frac{\partial M}{\partial y}=\frac{\partial M}{\partial \theta}d\theta=\frac{\partial M}{\partial \theta}\frac{\partial \theta}{\partial x}dx+\frac{\partial M}{\partial \theta}\frac{\partial \theta}{\partial y}dy[/tex]

    Therefore, I get, [tex]\frac{\partial M}{\partial x}=\frac{\partial M}{\partial \theta}\frac{\partial \theta}{\partial x}[/tex] and[tex]\frac{\partial M}{\partial y}=\frac{\partial M}{\partial \theta}\frac{\partial \theta}{\partial y}[/tex]

    But, if this calculus are right, them ∂M/∂θ has two differents expressions, due to I get two equations. I'm stuck

    Thanks in advance

    Regards
     
    Last edited by a moderator: Jan 16, 2014
  2. jcsd
  3. Jan 13, 2014 #2

    pasmith

    User Avatar
    Homework Helper

    This is straightforward: if you know [itex]M(\theta)[/itex] and you know [itex]\theta(x,y)[/itex] then you can calculate [itex]dM/d\theta = M'(\theta)[/itex], and then [itex]M'(\theta(x,y))[/itex] will give you [itex]dM/d\theta[/itex] in terms of [itex]x[/itex] and [itex]y[/itex].

    It follows that the two expressions must be equal where defined, so that if [itex]\partial \theta/\partial x \neq 0[/itex] and [itex]\partial \theta/\partial y \neq 0[/itex] at a point then you will have
    [tex]
    \frac{dM}{d\theta} = \left. \frac{\partial M}{\partial x} \right/ \frac{\partial \theta}{\partial x} = \left. \frac{\partial M}{\partial y} \right/ \frac{\partial \theta}{\partial y}
    [/tex]
    but if either [itex]\partial \theta/\partial x = 0[/itex] or [itex]\partial \theta/\partial y = 0[/itex] at a point then the corresponding expression will be an indeterminate form [itex]0/0[/itex] and cannot be used to compute [itex]dM/d\theta[/itex] at that point. But you can always use direct substitution into [itex]M'(\theta)[/itex].
     
    Last edited by a moderator: Jan 16, 2014
  4. Jan 14, 2014 #3
    Thanks for your reply.

    I have a few doubts. Accordingly with you, my calculus are right, perfect!. I'm not a complete foolish XD.

    In order to check the result. I consider a function: θ=xy^2, then

    [tex]\frac{dM}{d\theta} =\frac{\partial M}{\partial x}\frac{1}{y^2} = \frac{\partial M}{\partial y}\frac{1}{2xy}[/tex]

    And therefore,

    [tex]\frac{\partial M}{\partial x}= \frac{\partial M}{\partial y}\frac{y}{2x}[/tex]

    It does mean, the partical derivative in the variable x, is related with the partial derivative of the variabley. In other words, the surface θ=θ(x,y), represent a constraint, and thereby, the partial derivatives respect x and y, are dependient each other. Thus, the total derivative respect θ can be expressed using ∂M/∂x or ∂M/∂y alternatively. Is this the right reasoning?

    Thanks in advance again!

    Regards
     
  5. Jan 14, 2014 #4

    Mark44

    Staff: Mentor

    Since M is a function of θ alone, the derivative with respect to θ is dM/dθ. Now, since θ is a function of x and y, the derivatives of M with respect to x or y are partial derivatives, and are given by the chain rule.

    $$\frac{\partial M}{\partial x} = \frac{dM}{dθ} \frac{\partial θ}{\partial x} $$
    $$\frac{\partial M}{\partial y} = \frac{dM}{dθ} \frac{\partial θ}{\partial y} $$
     
  6. Jan 16, 2014 #5
    Thanks Mark44 for casting more light on the problem. Now I'm completely sure of my calculus. My last question in this issue was if the reason is this, I quote: "the partical derivative in the variable x, is related with the partial derivative of the variabley. In other words, the surface θ=θ(x,y), represent a constraint, and thereby, the partial derivatives respect x and y, are dependient each other. Right?

    Thanks
     
  7. Jan 16, 2014 #6

    Mark44

    Staff: Mentor

    This doesn't make much sense to me. I don't understand how θ being a function of x and y represents a constraint, unless θ is held constant.

    By "partial derivatives respect x and y" do you mean the partial of M with respect to x and the partial of M with respect to y? And what do you mean by "dependent on each other"?
     
  8. Jan 17, 2014 #7
    I mean, the equations obtained, implies:

    [tex]\frac{\partial M}{\partial x}\frac{\partial \theta}{\partial y} = \frac{\partial M}{\partial y}\frac{\partial \theta}{\partial x}[/tex]

    This relation is due to M is a function of a single variable, θ, and θ=θ(x,y). And the variable x, and y are relate via the surface θ=θ(x,y). This is what I was trying to tell, when I said constrain.

    Regards!
     
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