Fundamental confusions of calculus

[Trippy]

But does introducing two "new" variables really say anything?

I suppose that depends on your thought processes really, where it might confuse some people, to me it emphasizes the point :shrugs:.

Since a and b both depend on y, you might get confused about it and think you now have a function of three variables?

I can imagine that some people might, sure. But if people prefer to use a and a[sup]2[/sup] that really doesn't bother me in the slightest because as long as one remembers that b=a[sup]2[/sup]=y[sup]2[/sup] it's fine, we're still talking about the same thing.

I think, personally, that index notation shouldn't be unnecessarily cluttered that way, it's not something I've seen either.

I know, it's not something that I would ordinarily use either, I was just using it to emphasize a point (or try to anyway).

 

[Tach]

The material I cited has nothing to do with how $$u$$ is defined.

Those are unambiguous errors you made that you can't dress up any other way. Live with it, Tach. This is only an issue at all because you keep trying to deflect them instead of owning up to them.

The only issue is that you failed to connect the theoretical part in post 18 with the example. This boils down to your failing to identify $$u$$. This is what I call arguing in bad faith.

 

[Trippy]

I already explained to you earlier in this thread that you can take $$u=sin^2(\theta)$$ in $$f=3 \theta +u+v$$ or $$u=sin(\theta)$$ in $$f=3 \theta +u^2+v$$ , it doesn't change ANYTHING. Remember the recent example we have gone over?

You mean the example I bought up?

That example?

You're asking me if I remember a post I made in this thread?

I'm well aware of that point, but apparently you're ignoring mine, or it went over your head.

You have on three seperate occasions:
Left u undefined explicitly.
Defined u as being sin($$\theta$$)
Defined u as being sin[sup]2[/sup]($$\theta$$)

 

[przyk]

The only issue is that you failed to connect the theoretical part in post 18 with the example. This boils down to your failing to identify $$u$$. This is what I call arguing in bad faith.

Once again, the material I cited had nothing to do with the definition of $$u$$.

You are deflecting criticism of real and unambiguous errors you made. That is bad faith.

 

[Tach]

Defined u as being sin($$\theta$$)
Defined u as being sin[sup]2[/sup]($$\theta$$)

Do you understand that there is no difference between the two cases? They produce the same partial and total derivative. Can you prove it?

 

[Tach]

Once again, the material I cited had nothing to do with the definition of $$u$$.

You are deflecting criticism of real and unambiguous errors you made. That is bad faith.

You mean the imagined errors resulting from your failure to identify $$u$$ and to apply the derivative only to $$3 \theta$$ (because $$sin^2(\theta)=u)$$. Even after being told that to be the case for about 11 times?
Let me ask you something:

$$z=3 \theta +u+v$$
$$u=sin^2(\theta)$$
$$v=x^6$$

What is $$\frac{\partial z}{\partial \theta}$$?
What is $$\frac{d z}{d \theta}$$?

 

[Trippy]

Can you prove it?

I don't have anything to prove in this thread. You do.

 

[Tach]

I don't have anything to prove in this thread. You do.

I would hope that you understand that there is no difference between the two examples. So, why do you bring up the issue?

 

[przyk]

You mean the imagined errors resulting from your failure to identify $$u$$ and to apply the derivative only to $$3 \theta$$ (because $$sin^2(\theta)=u)$$.

No. [POST=2904021]Learn to read[/POST]. I mean these very real errors, which are independent of how $$f$$ and $$u$$ are defined:

Either way , your claim that $$\frac{\partial}{\partial \theta} \Bigl( 3 \theta \,+\, \sin(\theta)^{2} \,+\, \ln(x) \Bigr) \,=\, 3 \,+\, \sin(2\theta)$$ is false since the partial derivative is always taken wrt the direct variable (in this case, $$\theta$$) and not through a function (in this case $$\sin(\theta)$$) . The moment you start using chain differentiation you cease calculating partial differentials and you are starting to calculate total differentials.

The first two words say a lot when put in the context of your entire post:

What in $$f=f(\theta,u,v)$$ is that you fail to see? Or is not clear enough to you that $$sin(\theta)$$ is not a variable but a function? It was not clear enough that $$u=u(\theta)=sin(\theta)$$ and $$v=v(x)=ln(x)$$?
Either way , your claim that $$\frac{\partial}{\partial \theta} \Bigl( 3 \theta \,+\, \sin(\theta)^{2} \,+\, \ln(x) \Bigr) \,=\, 3 \,+\, \sin(2\theta)$$ is false since the partial derivative is always taken wrt the direct variable (in this case, $$\theta$$) and not through a function (in this case $$\sin(\theta)$$) . The moment you start using chain differentiation you cease calculating partial differentials and you are starting to calculate total differentials.

You believed the part I quoted above was true in general, and independent of how $$f$$ and $$u$$ were defined. So all this stuff about how $$u$$ and $$f$$ are defined is pathetic deflection on your part.

 

[Tach]

No. [POST=2904021]Learn to read[/POST]. I mean these very real errors, which are independent of how $$f$$ and $$u$$ are defined:

So, you wouldn't have any difficulty in answering:


$$z=3 \theta +u+v$$
$$u=sin^2(\theta)$$
$$v=x^6$$

What is $$\frac{\partial z}{\partial \theta}$$?
What is $$\frac{d z}{d \theta}$$?

Very simple questions.

 

[przyk]

So, you wouldn't have any difficulty in answering:


$$z=3 \theta +u+v$$
$$u=sin^2(\theta)$$
$$v=x^6$$

What is $$\frac{\partial z}{\partial \theta}$$?
What is $$\frac{d z}{d \theta}$$?

Very simple questions.

No, I wouldn't have any difficulty answering any of these questions, and nor would anyone else participating in this thread, and you know it. What does that have to do with anything?

Didn't I just say multiple times that the errors I was pointing out had nothing to do with how $$f$$ and $$u$$ were defined?

If you'd admit you made these errors, that'd be great. If you'd rather silently acknowledge them by ceasing to reply to my posts on that subject, then I'd also consider that acceptable, as long as you don't repeat the reason I brought them up in the first place (explained [POST=2904009]here[/POST]). What's really not acceptable is deflection.

 

[Pete]

No, I wouldn't have any difficulty answering any of these questions, and nor would anyone else participating in this thread, and you know it.

Actually, I think that people would have difficulty with $$dz/d\theta$$, considering that he's asking for the total derivative of a multivariable function (assuming x is a variable, not an oddly named constant).
Trick question?
Or is Tach still confused?

 

[Trippy]

So, why do you bring up the issue?

Because of your insistence that you defined u in the first place, which you didn't, followed by your insistence that you defined u as being a particular thing, when you've defined it as being two different things.

 

[Pete]

$$z=3 \theta +u+v$$
$$u=sin^2(\theta)$$
$$v=x^6$$

Notation related question for the educated.

Regarding the above, if I wrote:

$$\frac{\partial}{\partial \theta}z(\theta, u(\theta), v(x))$$​

...would that be interpreted as equivalent to

$$\frac{\partial}{\partial \theta}z(\theta,u,v) = \frac{\partial}{\partial \theta}(3\theta + u + v)$$​

...or...

$$\frac{\partial}{\partial \theta}Z(\theta, x) = \frac{\partial}{\partial \theta}(3\theta + sin^2\theta + x^6)$$​

...or is it just poor notation?

 

[arfa brane]

$$\frac{\partial}{\partial \theta}z(\theta, u(\theta), v(x))$$​

To me, that says: "take the partial derivative of z with respect to θ, with x constant"

...would that be interpreted as equivalent to

$$\frac{\partial}{\partial \theta}z(\theta,u,v) = \frac{\partial}{\partial \theta}(3\theta + u + v)$$​

That looks like a different function that doesn't define arguments u or v as functions but variables.

...or...

$$\frac{\partial}{\partial \theta}Z(\theta, x) = \frac{\partial}{\partial \theta}(3\theta + sin^2\theta + x^6)$$​

...or is it just poor notation?

That looks like it has the same form as the first function but with u(θ) and v(x) explicitly defined. What you haven't done is write the first function as an equation.

In each case you get a different answer, though.

 

[Pete]

Thanks, I think I've got a good handle on it now.

In each case you get a different answer, though.

That's why I'm asking, and where I think Tach is confused.

Given:

$$z = Z(\theta, \, u, \, v) \\
u = U(\theta) \\
v = V(x) \\
Z_1(\theta,x) = Z(\theta, \, U(\theta), \, V(x))$$​

Then I think that Tach is misnaming $$\partial Z_1/\partial x$$ as a total derivative $$dz/d\theta$$.
In post 210, the answer he is expecting for $$dz/d\theta$$ is $$3 + \sin(2\theta)$$, which is of course the partial derivative $$\frac{\partial Z_1}{\partial \theta}$$ for $$Z_1(\theta, x) = 3\theta + \sin^2\theta + x^6$$

If that is what he's thinking, then I can certainly understand the mistake. I've learned a lot in this thread (and still learning) about the dangers of playing fast and loose with notation when playing with functions and their parameters.
(I hope that the notation in this post is sufficiently close to good practice!)

 

[temur]

Regarding the above, if I wrote:

$$\frac{\partial}{\partial \theta}z(\theta, u(\theta), v(x))$$​

...would that be interpreted as equivalent to

$$\frac{\partial}{\partial \theta}z(\theta,u,v) = \frac{\partial}{\partial \theta}(3\theta + u + v)$$​

...or...

$$\frac{\partial}{\partial \theta}Z(\theta, x) = \frac{\partial}{\partial \theta}(3\theta + sin^2\theta + x^6)$$​

...or is it just poor notation?

My answer is the last two..

 

[temur]

Given:

$$z = Z(\theta, \, u, \, v) \\
u = U(\theta) \\
v = V(x) \\
Z_1(\theta,x) = Z(\theta, \, U(\theta), \, V(x))$$​

Then I think that Tach is misnaming $$\partial Z_1/\partial x$$ as a total derivative $$dz/d\theta$$.
In post 210, the answer he is expecting for $$dz/d\theta$$ is $$3 + \sin(2\theta)$$, which is of course the partial derivative $$\frac{\partial Z_1}{\partial \theta}$$ for $$Z_1(\theta, x) = 3\theta + \sin^2\theta + x^6$$

If that is what he's thinking, then I can certainly understand the mistake.

Yes. This is what he is thinking. It doesn't help that a partial derivative is just the "total" derivative along one of the coordinate axes. In fact I would recommend forgetting altogether about "total" derivative. There are only partial derivatives for multivariate functions (more precisely, only the directional derivatives). "Total" derivative is just the ordinary derivative of a single variable function that is constructed out of a multivariate function in a certain way.

 

[Pete]

My answer is the last two.

Thought so, thanks

Yes. This is what he is thinking. It doesn't help that a partial derivative is just the "total" derivative along one of the coordinate axes. In fact I would recommend forgetting altogether about "total" derivative. There are only partial derivatives for multivariate functions. "Total" derivative is just the ordinary derivative of a single variable function that is constructed out of a multivariate function in some way.

...so rather than talking about the total derivative of a multivariable function, it's better to explicitly define the related single variable function, and talk about that derivative.

For example, in the infamous bug-on-a-hotplate video, Metzler describes functions:

$$T = f(x,y,t)$$ (the temperature of the hotplate at given coordinates at a given time)​

$$x = g(t) \\
y = h(t)$$ (the path coordinates of a bug walking on the hotplate)​

He then describes using the total derivative dT/dt to determine the rate of change in temperature felt by the bug on its walk across the plate.

But!
It would be better (would have saved much confusion!) if he had instead described a separate function for the temperature felt by the bug:

$$\begin{align}
T_{\tiny{bug}} &= F(t) \\
&= f(g(t), h(t), t)
\end{align}$$​

and talked about $$\frac{dT_{\tiny {bug}}}{dt}$$ instead.
But then again, the purpose of the video was to explain the concept of a total derivative, so perhaps not.

Questions...

• in the previous post I labelled two functions $$Z$$ and $$Z_1$$.
  Is it Ok to use a numeric subscript to simply distinguish two different but related functions, or does a subscript imply something else?
• Is it OK to use a F(t) for a path-function (or whatever you call it) through f(x,y,t), or (thinking back to high school) does F(t) imply the antiderivative of f(t)?
• If you have an expression $$y = f(x)$$, is it preferable to use $$\frac{dy}{dx}$$ or $$\frac{df}{dx}$$ or $$\frac{d}{dx} f(x)$$ to denote the derivative?
  I'm guessing it's best to leave y out altogether, and just define f(x) in the first place.

 

[temur]

Questions...

• in the previous post I labelled two functions $$Z$$ and $$Z_1$$.
  Is it Ok to use a numeric subscript to simply distinguish two different but related functions, or does a subscript imply something else?
• Is it OK to use a F(t) for a path-function (or whatever you call it) through f(x,y,t), or (thinking back to high school) does F(t) imply the antiderivative of f(t)?
• If you have an expression $$y = f(x)$$, is it preferable to use $$\frac{dy}{dx}$$ or $$\frac{df}{dx}$$ or $$\frac{d}{dx} f(x)$$ to denote the derivative?
  I'm guessing it's best to leave y out altogether, and just define f(x) in the first place.

• It is ok. Unless.. If Z is a vector, subscript may indicate its particular component.
• Ok. If you need F to be an antiderivative you explicitly say it.
• All fine. Strictly speaking, the notation $$\frac{d}{dx} f(x)$$ means that you take the derivative, and then evaluate it at point x.