# Slice of X

Tags:
1. Aug 1, 2015

### Stephanus

Dear PF Forum,
I'm studying Relativity,
And there's some term in math that I don't know. And I am even not an English native.
in $\frac{dX}{dY}$ What does "d" mean?
I know that dX is the tiny slice of X. What do we say it in English?

2. Aug 1, 2015

3. Aug 1, 2015

### Stephanus

I know I should have asked this in "Language forum"?
So, "infinitesimal" is the antonym of "infinite"?

4. Aug 1, 2015

### Dr. Courtney

https://en.wikipedia.org/wiki/Infinitesimal

The words "opposite" and "antonym" are too imprecise in english to be of much use in mathematics.

What is the opposite of x? Is it -x? Or is it 1/x?

What is the opposite of INF? Is it -INF? Or is it 1/INF? Something else?

5. Aug 1, 2015

### Stephanus

THAT'S RIGHT. Another good point for me. Thanks.
So dX is if you slice X again and again and again,... to infinity.
If you challenge me, "How do you slice X? In half or 1/3 or 1/4 or...?
Supposed we divide the biggest part of X with n = 1/3
1. There is X -> A: 1/3 and B: 2/3
2. Divide the biggest part; B: 2/3 -> C: 2/9 and D: 4/9
3. Divide the biggest part; D: 4/9 -> E: 2/27 and F: 4/27
4. Divide the biggest part: A: 1/3 -> G: 1/9 and H: 2/9
5. Divide the biggest part; A or C: 2/9 -> etc...
No matter what method you choose for n, dX still infinitesimal]

6. Aug 1, 2015

### HallsofIvy

Staff Emeritus
Your mistake is in thinking that "d" has a separate meaning! "dy/dx" is defined as a single entity, the "derivative of function y with respect to variable x. That is defined early in an introductory Calculus course. A little later in such a course, we define the quantities "dy" and "dx", separately, in such a way that "dy" divided by "dx" is equal to the single quantity "dy/dx" originally defined. But we do NOT define "d" as a separate quantity.

(You can calculate an approximation to "dy/dx" by using "small changes in y divided by small changes in x" giving the idea that "d" means "small changes in" but that is just a "mnemonic".)

7. Aug 1, 2015

### paisiello2

Isn't d/dy considered as an operator on x?

8. Aug 1, 2015

### Stephanus

Yes, HallsofIvy.
If you divide dX by dY in Y = X3 + n, you'll get the gradient, or if you touch a ruler to that point. Y = 3X2
If you divide dX by dY in C2 = x2+y2, you'll get the tangent.
But as a non English speaker, I want to know what is the "word" for d.
"Derivative" is a good one.
I just can't keep saying for dX/dY like this:
"You should take the very very small part of X and divide it by the very very small part of Y and ..."
This one is better
"Take the derivative of X/Y and..."
Dr. Courtney choosen of word is good "Infinitesimal", and yours too. "Derivative".
Perhaps I would use both of them. Derivative seems good.

Now, I can go back to Special Relativity (SR) and ask this question
"In $U^\mu = \frac{dX^\mu}{d\tau}$ why do we have to use the derivative of X instead of X only". Seems good sentence?

And it just hits me with this SR thing. Am I being fooled by SR. In the past 3 months I was just doing math, and math, and math. Not a single thing about physics??

9. Aug 6, 2015

### Staff: Mentor

No, you don't divide dX by dY. If Y is a function of X, as above, you can find the derivative of Y with respect to X, or dY/dX (usually read as "dee Y dee X"). This new function gives you the slope of the tangent line on the original function at an arbitrary point.
Here you can use implicit differentiation to find either dX/dY or dY/dX. The latter, dY/dX, gives the slope of the tangent line at an arbitrary point on the circle that the equation represents. It's called implicit differentiation because the equation doesn't give either X or Y as a function of the other variable.

Since mathematicians normally use lower case letters, I'm going to switch to the more-used form, dy/dx etc.
No, it isn't. If you are given that, say, y = x2, for example, the differential of y, denoted dy, is often defined as $dy = \frac{dy}{dx} \cdot dx = 2x dx$. Here dy is the differential of y and dx is the differential of x.
I'm not familiar with this equation, but it seems to be defining $U^{\mu}$. On the right side is the derivative of $x^{\mu}$ (not the derivative of x) with respect to $\tau$.

10. Aug 7, 2015

### MrAnchovy

For dy/dx we say "dee wye bye dee eks" or sometimes "dee bye dee eks ov wye". We can also say "the derivative of y with respect to x", or if it is obvious from the context what we are differentiating by just "the derivative of y". I think Dr Courtney misunderstood what you were asking because we would not normally use "infinitesimal" in this way.

11. Aug 7, 2015

### Staff: Mentor

This might be a cultural thing. In my experience of the past 50 years, in the US we say for dy/dx, "dee wye dee eks" without the "by" or "of."

12. Aug 7, 2015

### Dr. Courtney

I have found it useful to focus on d/dx as an operator that differentiates what it acts on with respect to x. dy/dx = (d/dx) y

13. Aug 7, 2015

### MrAnchovy

My experience of calculus in the UK only goes back 35 years, and "dee wye dee eks" is certainly something that would always have been well understood, and possibly the prefererence now.

14. Aug 7, 2015

### aikismos

I worked as a math teacher in high school up to two years ago and "dee wye dee eks" was the norm in the department.

15. Aug 7, 2015

### Dr. Courtney

Being aware of how students can get the wrong impression from certain notations, I try and mix in a fair amount of:

d/dt [x(t)]

and

d/dx [f(x)]

and other forms where the letter specifying the function and the input variable are frequently different from y and x.

Students need to understand calculus with a wide variety of functions and input variables. Sticking to religiously to y as the output and x as the input works against this understanding.

16. Aug 7, 2015

### Stephanus

Yes, in voice . How do you type it in written? Because I was trying to understand SR, and there's something like that. $U^{\mu} = \frac{dX^{\mu'}}{d\tau}$. And I wanted to express my question in 'English'.
Something like: "Why we should divide the dee eks(distance) by the dee tau (time)?"
But I read many useful math concepts from this thread beside the question that I asked here.

17. Aug 7, 2015

### Dr. Courtney

See: http://www.sjsu.edu/faculty/watkins/infincalc.htm

My answer was intentional. The word "infinitesimal" probably should be used more often in the teaching and learning of Calculus. Newton and Leibniz did intend the derivative to be understood as an [(infinitesimal of the input variable) divided by an (infinitesimal of the output variable) ].

Thus, I believe understanding "d" as an infinitesimal is correct and what I intended.

18. Aug 7, 2015

### Stephanus

Ahh, I put it upside down. After 30 years from high school, I should have joined "Are you smarter than fifth grader"
Yes, I remember. The gradient of vertical line is infinity right. And horizontal line is zero.
Should have written $\frac{dY}{dX}$
Okay, okay.

Ok.

Yes! In math language.
But what is in 'English'
My problem is...
From $\frac{dx^{\mu'}}{d\tau}$ It's the (a)tiny/(b)slice/(c)derivative distance divided by (a)tiny/(b)slice/(c)derivative time so it's velocity at that particular time. It's the English that I want to know before I go back to SR forum again and ask my question in a better sentence.
I'm not an English speaker, I wasn't definitely went to English school before. So I don't know the word for this dX.
I'm aware if you have a curve, say y = x2. Then you touch a ruler to that curve in some arbitrary point, say in x=5, so the gradient of your ruler is $\frac{dY}{dX}$, right. = 6, correction: 10
Gradient = $\frac{dx}{dy}$

[Add: I know the number, I know the result, I don't know the 'English']

19. Aug 7, 2015

### Staff: Mentor

No, it's the rate of change of $x^{\mu}$ with respect to $\tau$ or the derivative of $x^{\mu}$ with respect to $\tau$. I don't see how this would be a velocity at all, even if $\tau$ happens to represent time.
I can't say it any better than I did just above without knowing what $x^{\mu}$ represents.
As I said already, dx is the differential of x.
Right, the slope of the tangent line to y = x2 at x = 5 is 10. In other words, $\frac{dy}{dx} |_{x = 5} = 10$.
No, this would normally be dy/dx, not dx/dy.

20. Aug 8, 2015

### Stephanus

This equation is from this: http://www.lecture-notes.co.uk/susskind/special-relativity/lecture-6/relativistic-kinematics/

And I wonder, why Velocity is $U^{\mu'} = \frac{dX^{\mu'}}{d\tau} \rightarrow (1,0,0,0)$?It's not the equation that I ask, it's the English.
the $\frac{sin(dx)}{dx} = cos(x)$ It's just the 'English' that I don't know.