Can we take the partial derivatives of φ and ψ here?

  • #1
98
3

Homework Statement:

Show that the function u=φ(xy) +√xy ψ(y/x)provides the following equation.

Relevant Equations:

u=φ(xy) +√xy ψ(y/x)
I research about coordinate systems and I found the following informations about transformation.
1590492979586.png

Now, if I replace arctan (x/y) (according to the picture above) to φ, I think I can solve. But if I can do this, then what will be replaced to ψ? I mean, I know just taking partial derative about this question and I guess this symbols can look familiar someone who spend some times about the topic. I left below a clear picture of question. Thanks for now.

1590493384718.png
 

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
2019 Award
13,527
3,248
Now, if I replace arctan (x/y) (according to the picture above) to φ
Why do you think there is a relationship between the ##\phi## in the illegible picture and ##\phi(xy) ## ?
And what about ##\psi## ?

What does your textbook mean if it asks to 'provide' something ?

##\ ##

My strong impression :smile: is that you are supposed to work out the partial derivatives. Do you know how to do that ?
 
Last edited:
  • #3
98
3
Why do you think there is a relationship between the ϕϕ\phi in the illegible picture and ϕ(xy)ϕ(xy)\phi(xy) ?
And what about ψψ\psi ?
I have told you that I have had no idea about these two symbols.
What does your textbook mean if it asks to 'provide' something ?
It is expected to prove if you're talking about it. Actually I don't get what you said.
 
  • #4
98
3
My strong impression :smile: is that you are suppsoed to work out the partial derivatives. Do you know how to do that ?
Yea I know how to do but I can't take derivative if I am not sure whether one each these symbols are equations. If these are constants, then there is no problem. I wondered if these are constant or not.
 
  • #5
BvU
Science Advisor
Homework Helper
2019 Award
13,527
3,248
whether one each these symbols are equations
They are functions. Start working out ##\partial \phi\over \partial x## and post your work.
 
  • #6
BvU
Science Advisor
Homework Helper
2019 Award
13,527
3,248
Actually I don't get what you said.
A more understandable wording would be "Show that ##u = ...\ ## satisfies the equation ...."
 
  • #7
98
3
A more understandable wording would be "Show that ##u = ...\ ## satisfies the equation ...."
Yea, he don't like speaking understandably :/
 
  • #8
98
3
They are functions. Start working out ##\partial \phi\over \partial x## and post your work.
Could you want to say ##\partial \ u\over \partial x## ?
 
  • #9
BvU
Science Advisor
Homework Helper
2019 Award
13,527
3,248
Could you want to say ##\partial \ u\over \partial x## ?
No. You have to start somewhere. Function ##u = \phi\ + ... \ ##, so ##{\partial \ u\over \partial x} = {\partial \phi \over \partial x} \;+ ... ##​
 
  • #10
BvU
Science Advisor
Homework Helper
2019 Award
13,527
3,248
I was so eager to give you the easiest possible start with this assignment that I overlooked that it might still be unclear to you. What I meant may become clear when I hand you the first step as an example:

$${\partial \phi \over \partial x} = {\partial \phi(xy) \over \partial (xy)} \ {\partial (xy) \over \partial x} = \phi' \;y $$Now, this one is easy and you already see some possible cancelling come up in the second derivatives .

##\sqrt{xy}\;\psi(y/x)\ ## is a little :rolleyes: more work but the idea is the same (cross fingers...)
 
  • Like
Likes Delta2
  • #11
98
3
I would have write ∂ϕ like that;
∂ϕ = ∂(u /(xy) +√xy ψ(y/x) )
Where did
(xy)
and

∂(xy)
come from?
 
  • #12
Delta2
Homework Helper
Insights Author
Gold Member
2,990
1,046
This is quite tricky problem. The "long and hard" way for this problem would be to set ##z=xy,w=\frac{y}{x}## and then to consider the function $$u(z,w)=\phi(z)+\sqrt{z}\psi(w)$$ and then calculate the derivatives ##\frac{\partial u}{\partial z},\frac{\partial u}{\partial w}## and then calculate the partial derivatives ##\frac{\partial u}{\partial x},\frac{\partial u}{\partial y}## using the 2-variable chain rule (for example $$\frac{\partial u}{\partial x}=\frac{\partial u}{\partial z}\frac{\partial z}{\partial x}+\frac{\partial u}{\partial w}\frac{\partial w}{\partial x}$$.

But this is the long , hard and painful way, the way @BvU suggests at post #10 is a clever shortcut. According to BvU's way you just have to know how to cleverly use the 1-variable chain rule.
 
  • #13
BvU
Science Advisor
Homework Helper
2019 Award
13,527
3,248
I would have write ∂ϕ like that;
∂ϕ = ∂(u /(xy) +√xy ψ(y/x) )
That is not correct.
Where did ##(xy)## and ##\partial \over\partial(xy)## come from?
From your post #1:
1590514209480.png

##xy## is the (single) argument of the function ##\phi##. If I give you the value of ##xy##, then you can give the value of ##\phi(xy)## if you know the function (which you don't).

My writing ##{\partial\phi(xy)\over \partial(xy)}\equiv \phi'## is just a sloppy physicist's way of avoiding to introduce a new variable, e.g. ##z(x,y) = xy\ ##, but it comes down to the same thing:$${\partial \phi \over \partial x} = {\partial \phi \over \partial z} \ {\partial z \over \partial x} = \phi' \;y$$ -- the chain rule, as @Delta2 already said.

---

Yea I know how to
Can you give us an example that demonstrates that ? Not to be obnoxious, but to understand what assistance is most effective.
 
  • #14
98
3
I can't give an example because of it doesn't make sense to me.
I was so eager to give you the easiest possible start with this assignment that I overlooked that it might still be unclear to you. What I meant may become clear when I hand you the first step as an example:

$${\partial \phi \over \partial x} = {\partial \phi(xy) \over \partial (xy)} \ {\partial (xy) \over \partial x} = \phi' \;y $$Now, this one is easy and you already see some possible cancelling come up in the second derivatives .

##\sqrt{xy}\;\psi(y/x)\ ## is a little :rolleyes: more work but the idea is the same (cross fingers...)
I guess I found ## \partial Ψ##/ ## \partial \ y##. But I don't get why we are doing these calculations somehow. The question asks for ## \partial \ u##/ ## \partial \ x## etc.
1591645952108.png
 
  • #15
98
3
1591646450676.png

Why we don't study on the calculation like that?
 
  • #16
BvU
Science Advisor
Homework Helper
2019 Award
13,527
3,248
We do: ##\ u = \phi + \psi\ ##, so your first factor is a ##\;1\;##.

But in your post #15 you have completely forgotten ##\;{\partial \psi\over \partial x}\;## ?!
 
  • #17
98
3
We do: ##\ u = \phi + \psi\ ##, so your first factor is a ##\;1\;##.

But in your post #15 you have completely forgotten ##\;{\partial \psi\over \partial x}\;## ?!
I'll be hanged if I understand! Pardon but I'm just not be able to do these calculations. I'll glance at chain rule of partial derivatives. I guess we couldn't make progress.
 
  • #18
BvU
Science Advisor
Homework Helper
2019 Award
13,527
3,248
I'll be hanged if I understand! Pardon but I'm just not be able to do these calculations.
Don't give up. We'll switch gear and try again. Since $$u = \phi + \sqrt{xy}\;\psi $$ we get $$
{\partial u\over \partial x} = {\partial\phi \over \partial x} + {\partial \sqrt{xy}\;\psi \over \partial x}$$
so ##{\partial u\over \partial x} ## has two terms and we started with the first:$$
{\partial u\over \partial x} = {\partial\phi \over \partial x} + ...$$

There are a few simple rules underlying all this. Hang on a little more and you'll be allright.
Sloppy physicist's notation (consult calculus textbook for more):

Simple derivative: $$f(x)\quad\Rightarrow\quad df = {\partial f\over\partial x} \;dx $$

Two variables: $$f(x,y)\quad\Rightarrow\quad df = {\partial f\over\partial x} \;dx + {\partial f\over\partial y} \;dy $$

Change in sum = sum of changes:
$$\ u = \phi + \chi\quad
\Rightarrow \quad du = d\phi + d\chi\ $$
Clear enough, right ?

Product rule:
$$\ y = ab\quad\Rightarrow\quad dy = a\; db + b \; da$$
Also clear enough, right ?

Chain rule:
$$\ y = f\bigl (g(x)\bigr ) \quad\Rightarrow\quad dy = {\partial f\over\partial g} \;dg $$

A little more abstract! Example:
$$\ f = 1/g \quad\Rightarrow\quad df = {\partial f\over\partial g} \;dg \ = \ -{1\over g^2}\; dg$$

I'll be hanged if I understand!
Doesn't help me to help you. More useful to indicate where and how you get stuck: give an example of what you (just) understand and what about the next step you do not understand -- and why not.

(Easy to say for me, I grant you; not so easy for you -- but you are the asking party at this point :wink: )

If the above are sufficiently clear, we can go back to pick up at the point where we were before. OK ?
 
  • Like
Likes sysprog and requied
  • #19
98
3
I was studying on partial derivatives, chain rule etc uptill now. Deadline of the homework is tomorrow night. I guess I will not be able to complete the homework. If someone would give me the solution I really appreciate it. Else I will quit the question. I know that here is not a question solving site but I have to ask for it please forgive me. By the way, thanks a lot to @BvU and @Delta2. They supported me a lot.
 
  • #20
BvU
Science Advisor
Homework Helper
2019 Award
13,527
3,248
If someone would give me the solution I really appreciate it
PF isn't in favour of just giving solutions -- our rules forbid it and it isn't very helpful for the receivers either. Pity about the deadline -- perhaps you get some help on Tuesday -- but if you don't get enough from teacher we can always continue unraveling the exercise with you.
 
  • #21
WWGD
Science Advisor
Gold Member
2019 Award
5,341
3,295
A way of doing it I thought was nice was to set up a path diagram between your initial function and the variables against which you differentiate.
 
  • #22
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
34,259
5,836
No.
Think of ##\psi## as a function of z, where z happens to equal xy always.
So ##\psi## is a function of only one variable, and ##\psi'## means the (normal, not partial) derivative of ##\psi## wrt that variable.
To get the partial derivative wrt x we use the chain rule:
##\frac{\partial \psi}{\partial x}=\frac{d\psi}{dz}\frac{\partial z}{\partial x}=\psi'\frac{\partial z}{\partial x}=\psi'\frac{\partial (xy)}{\partial x}=\psi'y##

The result to be proved in this thread is interesting. The general form that satisfies it is ##u=x^ay^b\psi(x^cy^d)## where either {a=b and c=d} or {a+b=1 and c=+/-d}.
 
  • #23
BvU
Science Advisor
Homework Helper
2019 Award
13,527
3,248
I think @haruspex means ##\phi##, not ##\psi## .

We had (in post #1) $$u(x,y)\equiv \phi(xy) + \sqrt{xy} \; \psi\left (y\over x\right )$$

On the other hand I did use partials in #10: $${\partial \phi \over \partial x} = {\partial \phi(xy) \over \partial (xy)} \ {\partial (xy) \over \partial x} = \phi' \;y $$ where I should (could ?) have used ##\displaystyle {{\text d} \phi(xy) \over {\text d}xy}## as @haruspex noted. That would have been more clear.
 
  • Like
Likes haruspex

Related Threads on Can we take the partial derivatives of φ and ψ here?

  • Last Post
Replies
2
Views
2K
Replies
4
Views
1K
Replies
1
Views
766
  • Last Post
Replies
1
Views
1K
Replies
4
Views
924
  • Last Post
Replies
2
Views
881
  • Last Post
Replies
4
Views
801
  • Last Post
Replies
2
Views
518
Replies
2
Views
1K
Replies
1
Views
1K
Top