Partial Derivative of Sphere in Terms of x and y

[Grufey]

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 \theta=\theta(x,y), and M=M(\theta), where M is a operator, but i doesn't relevant to the problem. I need to know the derivative \frac{\partial M}{\partial \theta} 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...

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

Therefore, I get, \frac{\partial M}{\partial x}=\frac{\partial M}{\partial \theta}\frac{\partial \theta}{\partial x} and\frac{\partial M}{\partial y}=\frac{\partial M}{\partial \theta}\frac{\partial \theta}{\partial y}

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

 

[pasmith]

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 \theta=\theta(x,y), and M=M(\theta), where M is a operator, but i doesn't relevant to the problem. I need to know the derivative \frac{\partial M}{\partial \theta} 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.

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

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

Therefore, I get, \frac{\partial M}{\partial x}=\frac{\partial M}{\partial \theta}\frac{\partial \theta}{\partial x} and\frac{\partial M}{\partial y}=\frac{\partial M}{\partial \theta}\frac{\partial \theta}{\partial y}

But, if this calculus are right, them ∂M/∂θ has two differents expressions

It follows that the two expressions must be equal where defined, so that if \partial \theta/\partial x \neq 0 and \partial \theta/\partial y \neq 0 at a point then you will have
<br /> \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}<br />
but if either \partial \theta/\partial x = 0 or \partial \theta/\partial y = 0 at a point then the corresponding expression will be an indeterminate form 0/0 and cannot be used to compute dM/d\theta at that point. But you can always use direct substitution into M&#039;(\theta).

 

[Grufey]

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

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

And therefore,

\frac{\partial M}{\partial x}= \frac{\partial M}{\partial y}\frac{y}{2x}

It does mean, the particle 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

 

[Mark44]

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.

$$\frac{\partial M}{\partial x} = \frac{dM}{dθ} \frac{\partial θ}{\partial x} $$
$$\frac{\partial M}{\partial y} = \frac{dM}{dθ} \frac{\partial θ}{\partial y} $$

 

[Grufey]

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 particle 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

 

[Mark44]

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 particle 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?

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"?

 

[Grufey]

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"?

I mean, the equations obtained, implies:

\frac{\partial M}{\partial x}\frac{\partial \theta}{\partial y} = \frac{\partial M}{\partial y}\frac{\partial \theta}{\partial x}

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!