Question

Partial derivatives question; not sure how to go about this, posted below in a moment.

Answer 1

My confusion is this: I've drawn a variable tree, but I've never dealt with a problem structured in this manner, and just conceptually am not sure how to do this (attaching file in a moment).

Answer 2

How the heck do I get from one tree to the other, in terms of expressing one function through the derivatives of its equivalent (but different) function?

Answer 3

I'm not sure if this answers your question or not, but are you able to see: \[r=\sqrt{x^2+y^2}\]\[\theta = \tan^{-1}(\frac{y}{x})\] so that you can extend your "g" tree.

Answer 4

I was thinking of that, but wasn't sure; let me take a shot from here.

Answer 5

Ok, cool. I never really learned by the "tree" method, so I'm a little unfamiliar with it. I know how partial derivatives work though. =P

Answer 6

I-yeah, I guess my question is, I don't even know-in terms of notation-what the second partial derivative of g with respect to theta would equal. I just don't know my rules well enough to know what assumptions I can make. That does "extend" my tree, but I still don't know what to do, because I don't know what is allowed, or why.

Answer 7

(Expressed in partial derivatives of f.)

Answer 8

Also, are you comfortable with this notation: \[\frac{\partial f}{\partial \theta}=f_\theta\] because I prefer to use the subscripts for partials, but I won't if it makes it more confusing for you.

Answer 9

Yep, familiar with it! I just prefer the liebniz stuff because it makes me feel "safer", lol.

Answer 10

All good, so I'll help you get started then. We'll start with what we know and take the derivative, turning it it into something that makes sense step by step hopefully!

\[g=f\]\[g_\theta = f_\theta\] So now you should note that f is a function of x and y, so we should show that! \[g_\theta = f_xx_\theta + f_yy_\theta\]
Now the next step will be similar. Try it out. You can plug in values now or later, but I prefer later as it's messier otherwise.

Answer 11

Alright, one sec.

Answer 12

I still just don't get.....it's just the way the question is phrased, I don't understand how I'm supposed to do this. Also, I'm not sure how there *is* a next step that avoids plugging in values.

Answer 13

It's not going to look incredibly pretty, but it'll be somewhat symmetric in an interesting kind of way, at least in my opinion.

Answer 14

(Thank you for your patience, btw.)

Answer 15

Oh, ok all good. I'll show you what the next step looks like, and maybe that'll make it a little more obvious. I'm sort of leaving this open-ended on purpose to sort of promote you to think about it a little more. I'm also not sure what all you know, so I'm kinda figuring it out by poking you. haha sorry ok so here we go:

\[g_\theta = f_xx_\theta + f_yy_\theta\]

Actually before we move on, does this equation make perfect sense to you or not? There's no point in moving forward otherwise, and I can explain whatever is necessary.

Answer 16

Yes, chain rule, makes sense.

Answer 17

I guess I just previously wasn't aware that if some given function of some variables equals another given function of some variables, their partial derivatives with respect to some same variable are equal, e.g. (picture in a sec)

Answer 18

Is this statement valid?

Answer 19

(And why, and how can I prove it, etc etc etc.)

Answer 20

I'm just looking for like, a theorem, postulate, or worded way to describe whatever principle is at work that allows us to do this.

Answer 21

Ok so continuing along, you seem pretty good: \[g_\theta = f_xx_\theta + f_yy_\theta\]\[g_{\theta \theta}= (f_xx_\theta + f_yy_\theta)_\theta\]
Essentially this is what we're going for. The second derivative of g with respect to theta. So now we just use more calculus tricks to get only derivatives of f wrt x & y.

So now we know differentiation is a linear operator. If that word sounds crazy, it's nothing you don't already use so don't worry, it just lets us do this trick where the derivative of a sum of things is the sum of their derivatives, \[g_{\theta \theta}= (f_xx_\theta)_\theta + (f_yy_\theta)_\theta\]
Now we invoke my favorite, the product rule:
\[g_{\theta \theta}= (f_{x \theta} x_\theta+f_xx_{\theta \theta}) + (f_{y \theta}y_\theta+f_yy_{\theta \theta})\]
Now we have 4 terms up there, 2 of them are fine, but two of them aren't. Specifically the f_{x theta} and f_{y theta} So see if you can take it further.

I know I didn't plug in anything, but you can see how much different and possibly cleaner it looks this way. You can do it any way you like though.

Answer 22

Yes, you're right in your example you made up.

Answer 23

While you try to understand what I've written I'll come up with an example to show you.

Answer 24

EXAMPLE:

\[p(x,y)=x^2+y^2 \\ x=r \cos \theta \\y=r \sin \theta \\ p(x,y)=q(r, \theta) \\ q(r, \theta) = r^2\]So here I've just said hey, this is a function p(x,y) and here's how to convert it to r and theta, so p=q. Just plug in values if you don't believe it. But we'll continue on since it should be made clear with a concrete example.
\[p=q \\ p_r=q_r\]
So
\[q_r= q_xx_r+q_yy_r\]
now we need to take some derivatives.
\[q_r = 2r\\q_x=2x \\ x_r=\cos \theta \\ q_y=2y \\ y_r =\sin \theta\]
Plugging in we get: \[2r = 2x \cos \theta + 2y \sin \theta\] Don't forget, x and y can be converted to polar coordinates, so let's do that \[2r = 2(r \cos \theta) \cos \theta + 2(r \sin \theta) \sin \theta\\2r = 2r \cos^2 \theta + 2 r \sin^2 \theta\\2r=2r(\cos^2 \theta + \sin^2 \theta)\]
So if that doesn't make it believeable then try out another example and as long as you do it correctly you'll get a "magically" correct answer. =P

Answer 25

Ok, but now you need to turn these \[f_{x \theta}=f_{xx}x_\theta+f_{xy}y_\theta\] That way you're only taking x or y derivatives of f, never a theta or r.

Answer 26

Alright, I really think (other than that thing of functions equalling each other you just taught me) this was such an issue of phrasing. The "...in terms of partial derivatives of f" kind of made no sense to me because f's partial derivatives *could be* described in terms of variables of g! IDK, I just think the question was poorly written. Lemme take a shot at doing with w.r.t. x and y instead.

Answer 27

Maybe it's helpful to look at it like this: \[(f_x)_\theta=(f_x)_xx_\theta + (f_x)_yy_\theta\] That way it seems more clear.

Answer 28

Well maybe I misread it as well, let me check lol.

Answer 29

I hope this doesn't seem like a bother (and don't expect you to lift a finger unless it's convenient), but I think I'm gonna go to bed and look at this later. You've helped me so much with this that I think I may need to just get some rest and look at this tomorrow morning, and then I can just come back with an answer.

Answer 30

Well it seems to be about right since these are all partial derivatives of f that I've been doing.

Answer 31

Yeah, no problem. I'll probably just post the rest of the solution if you want me to right now just so it's there to look at. I don't think you're wasting my time so I think you'll spend the time to understand it unlike some people on this site haha.

Answer 32

\[f_{xx}x_\theta^2+f_{yy}y_\theta^2+2f_{xy}x_\theta y_\theta+f_xx_{\theta \theta}+f_yy_{\theta \theta}\]\[x_\theta = -r \sin \theta \\ x_ {\theta \theta}=-r \cos \theta \\ y_\theta = r \cos \theta \\ y_{\theta \theta} = -r \sin \theta \] \[f_{xx}r^2\sin^2 \theta+f_{yy} r^2 \cos ^2 \theta-2f_{xy}r^2 \sin \theta \cos \theta-f_xr \cos \theta - f_y r \sin \theta\]A few minor adjustments\[r^2[f_{xx}\sin^2 \theta+f_{yy} \cos ^2 \theta-f_{xy} \sin(2 \theta)] -(f_xr \cos \theta + f_y r \sin \theta)\] There are some nice simplifications if certain things are true about f. If you have more questions just ask, I'll probably be around later.

Answer 33

I don't understand how you got this statement.

Answer 34

Yeah, I guess I just don't get how you're getting f_x_theta, I just don't know how that would be done, seeing as there isn't an equation for f itself, just equations for intermediate variables.

If I was in any situation where I had to take the second partial derivative of a function, where the function wasn't actually given as an expression and was just stated to have so-and-so variables...I wouldn't know how to do it.