# Partial derivative sphere

1. Jan 13, 2014

### 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

Regards

Last edited by a moderator: Jan 16, 2014
2. Jan 13, 2014

### pasmith

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 $y$.

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
$$\frac{dM}{d\theta} = \left. \frac{\partial M}{\partial x} \right/ \frac{\partial \theta}{\partial x} = \left. \frac{\partial M}{\partial y} \right/ \frac{\partial \theta}{\partial y}$$
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 $M'(\theta)$.

Last edited by a moderator: Jan 16, 2014
3. Jan 14, 2014

### Grufey

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 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?

Regards

4. Jan 14, 2014

### 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.

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

5. Jan 16, 2014

### 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 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

6. Jan 16, 2014

### 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"?

7. Jan 17, 2014

### Grufey

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!