Solving Total Differential Homework Problem

[chemphys1]

Homework Statement

If z(x,y) = f(x/y)
show that

x(∂z/∂x)_y + y(∂z/∂y)_x = 0

Homework Equations

so I understand z(x,y)
means I can write
dz = (∂z/∂x)_ydx + (∂z/∂y)_x dyI do not understand the = f(x/y) bit though?
does that mean this?
df= (∂f/∂x)_y dx+ (∂f/∂y)_x dy
and (∂f/∂x)_y = -y/x² (∂f/∂y)_x = 1/x

although that seems wrong can't manipulate to get the answer

any help on the method or explaining f(x/y) equalling z(x,y) would be appreciated. maths is not my strongest so if possible go as basic as it comes

 

[LCKurtz]

Homework Statement

If z(x,y) = f(x/y)
show that

x(∂z/∂x)_y + y(∂z/∂y)_x = 0

That isn't true. Try it for z = f(x/y) = x/y. Generally speaking, if everything is continuous, you would have $z_{xy} = z_{yx}$ and you wouldn't expect to multiply one by x and the other by y and have them be equal with opposite signs. Did you copy the problem correctly, parentheses and all?

 

[chemphys1]

That isn't true. Try it for z = f(x/y) = x/y. Generally speaking, if everything is continuous, you would have $z_{xy} = z_{yx}$ and you wouldn't expect to multiply one by x and the other by y and have them be equal with opposite signs. Did you copy the problem correctly, parentheses and all?

complete question is attached, but the information in the original post is correctly copied as far as I can see

 

[LCKurtz]

Like I said, it is false without more context. Check my example yourself.

 

[chemphys1]

Like I said, it is false without more context. Check my example yourself.

Sorry, I really do not follow
I find it hard to understand mathematical notation, so re: z = f(x/y) = x/y
I can't see how to check the example

 

[CAF123]

Sorry, I really do not follow
I find it hard to understand mathematical notation, so re: z = f(x/y) = x/y
I can't see how to check the example

You are familiar with the notation f(x) as describing a function of x. Similarly, f(x/y) just means you have some function in the variable x/y. You can instead consider u=x/y and then you are back to the more familiar f(u).

But with LCKurtz's example of f(x/y)=x/y, I get the equation to be satisfied. I also get it to be satisfied in general. I think LCKurtz simply misread the notation, that's all.

 

[LCKurtz]

You are familiar with the notation f(x) as describing a function of x. Similarly, f(x/y) just means you have some function in the variable x/y. You can instead consider u=x/y and then you are back to the more familiar f(u).

But with LCKurtz's example of f(x/y)=x/y, I get the equation to be satisfied. I also get it to be satisfied in general. I think LCKurtz simply misread the notation, that's all.

Are you saying that $\left(\frac {\partial z}{\partial x}\right)_y$ means something other than $\frac \partial {\partial y}\left (\frac{\partial z}{\partial x}\right )$ or, as I wrote $z_{xy}$?

 

[CAF123]

Are you saying that $\left(\frac {\partial z}{\partial x}\right)_y$ means something other than $\frac \partial {\partial y}\left (\frac{\partial z}{\partial x}\right )$ or, as I wrote $z_{xy}$?

Yes, I took $\left(\frac{\partial z}{\partial x}\right)_y$ to mean, say, differentiate z wrt x, keeping y held fixed. Similarly for the other case. I actually thought that was a standard notation, although I have seen cases where they simply suppress the variable being held constant because in a sense it is obvious from the problem.

 

[LCKurtz]

Are you saying that $\left(\frac {\partial z}{\partial x}\right)_y$ means something other than $\frac \partial {\partial y}\left (\frac{\partial z}{\partial x}\right )$ or, as I wrote $z_{xy}$?

Yes, I took $\left(\frac{\partial z}{\partial x}\right)_y$ to mean, say, differentiate z wrt x, keeping y held fixed. Similarly for the other case. I actually thought that was a standard notation, although I have seen cases where they simply suppress the variable being held constant because in a sense it is obvious from the problem.

Well, that's a new one on me. In over 40 years of teaching calculus using many different texts, I never encountered that notation.