Why Does the Scalar k Equal 4/9 in the Second Partial Derivatives Problem?

[cooev769]

Homework Statement

Suppose z=ψ(2x-3y), Show that the second partial derivative of z with respect to x, is equal to the second partial derivative with respect to y multiplied by a scalar k.

Homework Equations

The Attempt at a Solution

I thought this was too simple to be correct, and apparently it was. I just use the chain rule and found the first partial derivative with respect to x as 2ψ and with respect to y as -3ψ, hence both the second partial derivatives will be 0, and hence the scalar k is just 0. But apparently this is incorrect, apparently the scalar k is 4/9. I'm not exactly sure why, but our maths teacher told us to redefine z in terms of u, and then make u depend on x and y. Any help would be appreciated. Cheers.

 

[cooev769]

Is it because we can't assume that psi is not dependent on x and hence we have to redefine a function u which is only dependent on x and y, and use the chain rule to solve that zx=zu*ux?

 

[Chestermiller]

Your first partial involves ψ. So, when you take the second partial with respect to x, you need to take the partial of the ψ in the first partial with respect to x again. Same for y.

Chet

 

[cooev769]

But don't you treat all the other variables as constant when you are doing partial derivatives?

 

[haruspex]

But don't you treat all the other variables as constant when you are doing partial derivatives?

You treat all other independent variables as constant. x and y are independent, but ψ depends on x and y.

 

[cooev769]

How do you know that, they're just a bunch of undefined variables. Are we just supposed to make the assumption?

 

[cooev769]

And I commented above I thought this may be the case, but it seems stupid that they've left the variables undefined not showing what they're dependent on, and just making you assume, without knowing if psi is dependent on x and y.

 

[cooev769]

Nope I can't figure it out. The answer says find psi', but it has evaluated psi' for both the x partial and y partial, saying they're the same thing, but on is psi with respect to x and one is psi with respect to y goddamit!

 

[cooev769]

Like I've sampled the answer by chucking psi=x+y and evaluating it and it doesn't work, stupid question is pissing me off now.

 

[Chestermiller]

Nope I can't figure it out. The answer says find psi', but it has evaluated psi' for both the x partial and y partial, saying they're the same thing, but on is psi with respect to x and one is psi with respect to y goddamit!

Try thinking about it by making the substitution η=2x-3y, so that z = ψ(η). So, by the chain rule,

\frac{∂ψ}{∂x}=\frac{dψ}{dη}\frac{∂η}{∂x}=ψ'(2)

Similarly,

\frac{∂^2ψ}{∂x^2}=2\frac{d^2ψ}{dη^2}\frac{∂η}{∂x}=4ψ''

Hope this helps.

Chet

 

[cooev769]

Yes that definitely does help, but it seems really weird haha. Thanks.

 

[cooev769]

Wait wait wait, but seeing as psi and eta are both dependent on x, doesn't this mean that z partial with x would require the product rule and it still doesn't work.

 

[cooev769]

Also how can you assume that psi is only dependent on eta, what if psi isn't a function that is a multiple of eta, then psi is a function of x and y as well.

 

[cooev769]

All we've been taught is the chain rule, and if i just draw z as a function of variables psi and eta, then eta is a function of x and y, and psi is a function of eta, but it can't just be a function of eta, because if it's not a scalar multiple of eta then you have to take on the difference in terms of x and y, so psi is of eta and x or y, and eta is a function of x and y so you end up with a stupid tree which doesn't even give me the right answer.

 

[cooev769]

Here's the file, question 6. The answer just makes no sense to me.

 

[cooev769]

If you put psi=x, and you solve for zxx and zyy, it is clearly not a scalar multiple of their suggested k out. Stupid question!

 

[haruspex]

All we've been taught is the chain rule, and if i just draw z as a function of variables psi and eta,

No, ψ is a function specifying how the variable z depends on η. z is not "a function of ψ".

then eta is a function of x and y, and psi is a function of eta, but it can't just be a function of eta,

Yes it is. For any given value of η, the function ψ applied to that value tells you the value of z.

because if it's not a scalar multiple of eta then you have to take on the difference in terms of x and y,

sorry, you've lost me there.

so psi is of eta and x or y, and eta is a function of x and y.

ψ is a function, not a function of a specific variable. z is a dependent variable, a function of η.
E.g. you could have another dependent variable w = ψ(y). Same function, different variables.

 

[cooev769]

So is it just analogous to:

y=f(x)?

Not z=psi*(2x-3y)=psi2x-psi3y?

 

[cooev769]

If so why the heck didn't they just write z=f(2x-3y)!

 

[Chestermiller]

Suppose, for example, ψ(η)=η²+2η+1=(2x-3y)²+2(2x-3y)+1

Then \frac{∂ψ}{∂x}=\frac{dψ}{dη}\frac{∂η}{∂x}

\frac{dψ}{dη}=ψ'=2η+2=2(2x-3y)+2
\frac{∂η}{∂x}=2
So,
\frac{∂ψ}{∂x}=4η+4=4(2x-3y)+4

See what you get when you evaluate it by taking the partial derivative of (2x-3y)²+2(2x-3y)+1 with respect to x.

Chet

 

[cooev769]

Yeah i just need to clarify, that the question stating z=psi(2x-3y) isn't saying psi multiply by the brackets, it is actually analogous to:

z is equal to a function of (2x-3y), where psi is the function? If so then this problem is no problem for me mathematically I just had no idea that was the case because it's so ambitious, at least to me it is.

 

[Chestermiller]

If so why the heck didn't they just write z=f(2x-3y)!

They did. They just used the symbol ψ instead of the symbol f.

 

[cooev769]

Why that's so bloody ambiguous. Why didn't they just write z(2x-36) jesus tap dancing Christmas.

 

[Chestermiller]

Yeah i just need to clarify, that the question stating z=psi(2x-3y) isn't saying psi multiply by the brackets, it is actually analogous to:

z is equal to a function of (2x-3y), where psi is the function?

Yes. That's exactly what the problem is saying.

If so then this problem is no problem for me mathematically I just had no idea that was the case because it's so ambitious, at least to me it is.

Excellent. You'll soon get used to using other symbols besides f to represent functions.

Chet

 

[cooev769]

But how do we then differentiate between a function multiplied by brackets and the function of variables in the brackets, mathematics you weird.

 

[Chestermiller]

But how do we then differentiate between a function multiplied by brackets and the function of variables in the brackets, mathematics you weird.

That's a great question. It has to come from proper specification of the problem. When I'm writing something like that, I try to make it very clear which I am talking about. Your problem statement just assumed that you would be able to read their minds.

Chet

 

[haruspex]

But how do we then differentiate between a function multiplied by brackets and the function of variables in the brackets, mathematics you weird.

Yes, it can be ambiguous. Generally people rely on customary usage. You mentioned using f, as is very common for functions, but g, h, k, \phi, \psi, \theta... also get used. In the present case, I would also apply the principle that the mere multiplicative constant interpretation makes the problem too trivial.

You also mentioned z(2x-3y). I assume you mean as in z = z(2x-3y). This can also be confusing since it is a 'pun'. The z on the left is a dependent variable, while that on the right is a function. It gets really awkward if you want to express another pair of variables related by the same function. E.g. if I write y = f(x) and z = f(t) then you understand the two functions to be the same but the variables different. I can't express that with the y = y(x) notation.

You also asked at one point how you know which are dependent and which independent variables. That too can be a puzzle. What if I have z = f(x, y), but also y = g(x)? The answer is that ∂f/∂x should be thought of as treating x and y as independent. So we can write dz/dx = ∂f/∂x + ∂f/∂y.g'. Not sure if a meaning can be assigned to ∂z/∂x here.

 

[cooev769]

Thank you all very much for your responses. They are all extremely helpful. I actually went and saw my professor and he disagreed with me. He thought the idea of psi being a variable itself was perverse and there's no need to include the explicit statement that it is a function. I guess as a physicist we take shortcuts and are probably prone to using incorrect notation.

 

[Fredrik]

But how do we then differentiate between a function multiplied by brackets and the function of variables in the brackets, mathematics you weird.

It took me a while to understand what you're asking here. I think you're asking how we know if (for example) x(yz) is the product of x and yz or the value of the function x at yz. There's no other way than to keep track of what your variables represent, in particular which ones represent functions and which ones represent numbers. If you have defined them yourself, this should be easy. If someone else have defined them for you, they should have told you what the symbols mean. If they haven't, you're going to have to guess. The symbols used are usually a hint. If they're close together in the alphabet, they often represent elements of the same set.

I would think about the equality $z=\psi(2x-3y)$ in the following way. x,y,z are variables that represent real numbers. $\psi$ is a variable that represents a function. (If it had represented a real number, I would have expected to see a symbol like w instead). The equation also defines a relationship between the values of the variables x,y,z. This relationship is such that the value of z is completely determined by the values of x and y. So indirectly, the equation defines the function that takes (x,y) to z. This function can be denoted by a new symbol, like $\phi$. We would then have $\phi(x,y)=\psi(2x-3y)$ for all $x,y\in\mathbb R$ such that 2x-3y are in the domain of $\psi$.

But it can be annoying to have to introduce new symbols every time we do something like this. So you may prefer to denote the new function by z. Now the symbol z has two different meanings, and you must be careful not to confuse them.