@maks That logic would work if the third vertex were at the origin.

components of $r'(t) = \frac{dx}{dt}\vec{x} + \frac{dx_2}{dt}\vec{x_2} + ...$ then $\nabla f \cdot r'(t) \to \frac{\partial f(r(t))}{\partial x}\frac{dx}{dt}\vec{x} + \frac{\partial f(r(t))}{\partial x_2} \frac{dx_2}{dt}\vec{x_2} + ...$?

@Maks the fact that the vector associated to the points are orthogonal does not imply that they form a rectangle triangle. One simple way to see this is that the fact that the triangle is rectangle depends on the third point, whereas your argument does not take it into account

As a simple example, consider the points (1,0,0), (0,1,0), (0,0,1)

is that last expression correct @semiC ?

No.

for one, r'(t) would be $r'(t)= x'(t)\hat{x}+y'(t)\hat{y}+z'(t)\hat{z}$

oh so I need $\frac{dx(t)}{dt}$

@Semiclassical Then how would I solve it ??

Make one of the points your origin by shifting your coordinate system.

@DHMO it must be soooo much work to animate all that shit

@Semiclassical Another way ? We havent seen that technique

alright.

@NaCl agreed

in that case, note that the points given point correspond to vectors pointing from the origin to your vertices, not from vertex to vertex

to get a vector pointing from the first vertex to the second, you'd compute (3,1,1)-(-1,2,1)=(4,-1,0)

that's the displacement vector of the second vertex relative to the first.

How does $\frac{\partial f(r(t))}{\partial x} \frac{dx(t)}{dt} + \frac{\partial f(r(t))}{\partial x_2} \frac{dx_2(t)}{dt} + ...$ get integrated to return $f(x(b)) - f(x(a)) + f(x_2(b)) - f(x_2(a)) + ...$ for some bounds of integration?

@obliv you haven't taken dot products correctly.

once you take the dot product, you can't have any vectors left over.

Oh my god.. you're right.

and the length of the vector <4,-1,0> is sqrt(17)

@obliv take a look at the first term---does it remind you of anything?

from 1-variable calculus

@Maks Do you know scalar product ?

@maks There's two basic approaches here: Vector algebra, and law of cosines.

(they're actually the same thing, under the surface, but the mindset is different)

@Semiclassical I'm trying the one you just said

@Astyx yes I do

You can compute the norm of the sides of the triangle, and dividing the scalar product by the product of the norms gives you the cosine of the angle between the vectors

$\vec{a}\cdot \vec{b}=ab\cos \theta$

Yes, but I need to get the vectors from the points right ? Like @Semiclassical said

Yes

Well, suppose you want the angle ABC

@semiC I forgot to include $dt$ with which the integral is being integrated w.r.t . When you include that, you can change $r'(t)$ to $dx(t)\vec{x} + dx_2(t)\vec{x_2} + ...$

eh, you could use that for $r'(t)\,dt$, but I don't think it's too useful here.

taking integral w.r.t. the infinitesimal components of the vector that $f$ is a function of makes it more simple I think

focus more on this: What does $\frac{\partial f}{\partial x}\frac{dx}{dt}$ remind you of from 1-variable calculus?

@maks What defines the angle ABC are the edges AB and BC

So we should take a look at the vectors which point from B to A and from B to C

if we write the three position vectors as $\vec{a},\vec{b},\vec{c}$, then the displacement vectors are $\vec{a}-\vec{b}$ and $\vec{c}-\vec{b}$

We can now use the formula Astyx and I gave above to find the cosine of the angle between those two vectors

it just looks like the total derivative of a scalar function @semiC I can't recall anything from 1-var calculus that helps me here D:

Well, that's where I was going :)

I was getting at its similarity to the 1-variable chain rule

Just a simple doubt, If I get a vector from the difference of two points, that vector should go through both those points??

i.e. if $f$ only depended on $x$, you'd have $\frac{df}{dx}\frac{dx}{dt}=\frac{df}{dt}$

@Maks No.

@Semiclassical then what are those vectors ?

However, if you add $\vec{a}-\vec{b}$ to $\vec{b}$, you'll get $\vec{a}$.

image

with $\vec{p}$ and $\vec{a}$ in this case

So, now I have to calculate the difference between those vectors I just got

And that gives me the vectors that goes through the points ?

Well, it looks like you already did that

No. You don't want vectors that go through those points.

You want vectors that point from one vertex to the other.

this is basically just vector addition: mathworld.wolfram.com/images/eps-gif/ParallelogramLaw_1000.gif

@obliv So therefore $\nabla f\cdot r'(t)=?$

@Semiclassical isnt that the same ? I'm sorry I can't catch it

maybe we're talking past each other.

@Semiclassical I'll tell you what I had understood from my theory

okay. then the difference is definitely not of that form

the difference is the $k \vec{a}$ part.

i.e. $V$ corresponds to point b, and $p$ to point a

@semiC it's $\nabla f(r'(t))$?

that'd be a vector still, and a dot product is a scalar quantity.

you said it yourself earlier...

So, the difference of two vectors is another vector, that goes from the tail of one vector to the tail of the other one ?

right.

Ok got it

@semiC The total derivative of the function $f(r(t))$

?

right. it's just $\frac{d}{dt}f(r(t))$

So, If I do $ A - B $ , $ B - C $ and $ A - C $

I'll get all the sides of the triangle

So your integral becomes? @Obliv

Being A,B,C the vectors I just got from the difference of the points

@Maks Right.

And then you can use the dot product formula to find the angles involved.

Now, if any $ (A-B) (B-C) (A-C)$ are perpendicular there I can say that the triangle is a rectangle triangle

Or I can calculate the angles using the dot product formula

right.

@semiC $\int_a^b \frac{d}{dt}f(r(t)) dt \to f(r(b)) - f(r(a))$

Thanks, now I got it

@obliv yup

That's really the point of the gradient theorem: the dot product of the gradient and r'(t) is just a total derivative

That makes sense but just to clarify for doing it in components, does $\int_a^b \frac{\partial f(r(t))}{\partial x} dx \to f(x(b)) - f(x(a))$?

Hmm.

Yes, it does.

Actually, wait.

if it's $\frac{\partial f}{\partial x}$, yes.

The expression $\frac{\partial}{\partial x} f(r(t))$, though, would probably not be right because that allows y,z to change as well.

So I'd be suspicious of that, though my brain isn't coming up with a counter-example at the top of my head.

consider $r(t) = \langle x(t),x_2(t), x_3(t),...,\rangle$ then you have $\frac{\partial f(r(t))}{\partial x}$ to be something of the form $ax^l + ax_2^m x^n + ...$ and if you take integral w.r.t x you have the original function f(x(t)) I think.

I'm pretty sure what you just wrote doesn't make a lot of sense.

oops.

You'd need to further assume that $f(\vec{r})$ is itself a polynomial.

I'd suggest focusing on trying some examples, though.

Main thing to note is that the gradient theorem is a statement about the entire dot product, not some particular portion of it.

Okay i'll try some examples and get back to you

Mmkay.

@Semiclassical I'm sorry I'm back haha, but when I first calculate the difference between the two points, shouldnt that already give me the vector that is one side of the triangle ??

Why do I need to go one step further?

Not sure what you're getting at. My point was that, if you start from the position vectors A,B,C corresponding to the coordinates you specified, then the differences are precisely the AB,BC,AC vectors you wrote before.

@semiC Okay I checked it for a function $f(r(t)) = x^2 + 3xx_3 - 2x_2 + x_4 - 6$ took the partials and took integrals with respect to the same variable and it worked.

hmm

actually, here's an objection:

err actually what did I say it would return originally, f(x(t))?

what is the integration variable in $\int_a^b \frac{\partial f(r(t))}{\partial x} dx$?

that doesn't make sense..

You write $x$, but the limits would suggest $t$.

(3,1,1)-(-1,2,1) is enough

that'll be A-B.

well, you also need to compute B-C and A-C

but those won't be the difference between A-B and C

@semiC when we say $\int_a^b \nabla f \cdot r'(t)~dt = f(r(b)) - f(r(a))$ are we doing $r(b) = \langle x(b) , x_2(b) , x_3(b), ..., \rangle$ ?

sure, and then we're plugging those values into $f(\vec{r})$

(I insist on writing $f(\vec{r})$ since $f(r)$ is used in the context of spherical coordinates to denote a function depending only on $|\vec{r}|$)

so you could have $f(\vec{r})=x_1+x_2^2+x_3^3$

and if all three coordinates ended up going from $a$ to $b$, that'd be $(b+b^2+b^3)-(a+a^2+a^3)$

@Maks do you know distance between two points formula?

@Ramanujan $ || a || - || b || $ ?

No

Distance

@maks $|\vec{a}-\vec{b}|$

but we're getting messy. you gave your AB,BC,AC earlier

@Maks here

What would the lengths of AB,AC,BC be?

@Ramanujan that side should be $ (12,14) - (4,8) $

@Semiclassical || AB || ?

@Maks can you see any triangle?

i'm talking about your triangle, btw, not Ramanujan's

As "length" I understand the number that tells me what is its size

what are the three side lengths based on what you wrote earlier about AB, BC, AC?

hi guys

?

@Hey-men-whatsup hi

those are the displacement vectors. what are their lengths?

@Semiclassical And the length should be the absolute value of that

Hi

Length of AB = $ \sqrt{1^2 + 3^2 + (-4)^2} = \sqrt{26}$

oh hi dudes

is that a magnitude?

yeah it seems.

@Maks similarly find length of BC and CA

Then apply Pythagoras theorem

back, woops

those are correct, yeah.

now, if that was a right triangle, how should those be related?

@Ramanujan Pythagoras doesnt work because it's a rectangle triangle

Whoah!

@Semiclassical $ \sqrt{41}^2 = \sqrt{17}^2 + \sqrt{26}^2 $

right. but 17+26=43>41

So, yeah.

Vectors fuck with my head .-

okay. now lets do the angle ABC

You should make drawings to understand them betters

@Semiclassical btw what is rectangle triangle?

for that, let's take the vectors A-B and C-B. (I want those directions to avoid any sign issues later on)

I think he meant that it's not a right triangle.

And be sure to fully understand vector in 2D before attempting to understand them in 3D

@Ramanujan A right triangle, sorry, not an english speaker

just to keep things honest, what are A-B and C-B?

Vectors!

sure. which specific vectors are they, though?

@Maks you can edit it within 2 minutes

i.e. repeat the expressions.

Hey Astyx

As we called them

Right.

do you still remember the question I asked a few days ago @Astyx ?

CB would be -BC based on what you gave above. It's that minus sign I was after.

@Hori hi

Which one ?

@Semiclassical Yes, I follow you

Okay. So now we do the dot product. What's (A-B).(C-A)?

i.e. crunch it out.

I think I managed to solve the problem, but I am not sure if it's the right answer, wanna join a private room to discuss it @Astyx?

Ok fine

Where are you from @maks? I'm just curious because they're rectangle triangles in Italian too

@semiC I think $\int_a^b \frac{\partial f}{\partial x} dx$ gives a different answer like you said. $\frac{\partial f(r(t))}{\partial x} = \frac{\partial}{\partial x}\left[ x^2 + x_2x_3 - x_4 + 6\right]$ then taking integral gives $x^2 + x_2x_3$ But, the integral of the total derivative would yield something like $f(r(b)) - f(r(a)) = x(b)^2 + x_2(b)x_3(b) - x_4(b) + 6 - [...]$ right?

@Alessandro Are they not rectangle triangle in english ? They ar ein french ar well

sounds right. i'm not paying as much attention at the moment.

It's 'right triangle' in english.

Oh yes of course

which I guess you could say as 'rectriangle'

I like this one

$ (1,3,-4) x (-3,4,-4) = 1*(-3) + 3*4 + (-4)*(-4) = 25 $ ?

right.

@Alessandro Argentina, we call them "triangulos rectangulos"

on the other hand, the dot product formula had a*b = |a||b| cos(C)

and you already found that the corresponding lengths of AB,CB were sqrt(26), sqrt(41)

Yes, so the cos(c) = a*b / |a||b|

right.

It's the same (triangoli rettangoli) in Italian and apparently in French too, must be a latin languages thing

from that you can get the value of theta, which should be acute since the dot product is positive.

@Ramanujan, how's your school days?

at this point, it's numbers.

for the other angles, you'd want to do (B-A) * (C-A) and (A-C) * (B-C)

tan right

@Hey-men-whatsup currently passing time in this chat room,its 12:35AM now (again had to wake up at 6)

is it?

@maks I'm going to have to head out, but I think you can see the rest of this.

right.

Thank you man :)

:)

well, AB.CB, but yeah.

later

@Ramanujan, you may don't sleep at all for some nights once you entered an university

@Hey-men-whatsup why so?

Projects?

yeah...

If (a,b) is a vector then (-b,a) should be parallel ?

No, calculate the dot product

@Hey-men-whatsup imagining assignments in computer science…very enjoy full

nah, not really

@semiC are you leaving? Thanks for your help :)

@Hey-men-whatsup If you are not able to get a full nights sleep due to attending university then that is your own poor planning

@Alessandro it was (b,-a)

@Tobias Tell it to Ramanujan not me, I'm pointing this to him so he will pay attention

@maks, still, calculate the dot product

@Hey-men-whatsup OK,going for sleep now…!

yeap that's my point lol, save energy while you have

and good night

@Hey-men-whatsup Btw did you made any game(or any app) till now?by your own skills learned in classes

Yes, but mostly web for now

But I have sudden interest in mathematic

In which topic?of mathematics

Games takes vector, a bit physic, some geometry.

Do we need to solve problems or computer will do it?

computer vision takes function, transformation, statistic, calculus etc.

you're good at math, then everything will be easier

What is a programmer favorite place?…foo bar

What does it mean?

I don't know

There are lot of jokes on internet saying only programmers will understand…

@robjohn ty! this is precisely what I was looking for, and I learned something too

foo bar abit racist tough, lol

but you'll get it someday

racist ?

@Ramanujan foo and bar ar typical names given to functions/programs that are not part of a bigger project and only exist to demonstrate a programming concept in the instant

yeah, something like that, perhaps you know it better

google.co.in/…

What's the difference?

Just different identation styles

Yup, and some people do not tolerate the identation style they do not usually use

the left side is written by programmer, while the right is copy paste

@Alessandro Not quite just the indentation

@Semiclassic: I used to say Taylor, but I learned CORDIC was correct about 10 years ago and used to forward people a handout I had on it.

@Hey-men-whatsup lol

hi @Alessandro @Tobias @Astyx

@TedShifrin Hi

@TedShifrin Hi ! How are you ?

Ça va, @Astyx, merci bien.

@Tobias yes but there's no actual difference from the compiler's perspective, maybe coding style fits bettee than identation there

Good evening @Ted

@Alessandro I just mean that it is also about whether you put a line break before beginning the {} (which is technically not indentation)

I found out why I could not see avatars

@Astyx Oh?

(my code would probably make most programmers turn away in disgust)

That makes 2 of us @Tobias

@BalarkaSen I made a slight error in the post. It has been corrected.

A plugin I have on my browser blocked Gravatar - which was considered a threat to my privacy - and gravatar is what makes non-personnalised image show

Howdy @MikeM

Oh, interesting, @Astyx.

Hi.

So if any of you encouter the same problem, you'll know :p

my internet provider blocks it also

Would one of you happen to know Knuth's algorithm for solving the mastermind ?

@MikeMiller Got it. Thanks for letting me know.

Hi @Ted.

Hi @Balarka. Yes, I'm home.

Anything good going on?

Hah, nice.

Well, my exam is in 3 weeks or so, been busy with that. Have been reading isolated things for procrastinating.

It's been pretty hectic, politically and economically, around this part of the world too.

Still procrastinating. Don't forget to sleep and stay healthy

Yeah, I think the world is globally a total mess.

I'm working on two letters of recommendation for gifted former students who want to go to law school.

Yeah. Did you hear what happened here, then?

Nice. Interesting choice of career, too.

Procrastination is a plague

I've done nothing in the last 5 hours ...

As long as you can do controlled procrastination, it's not. It's been a life-saver for me these few weeks.

@Astyx see it as gathering yourselves

What is controlled procrastination if I may ask ?

Hi chat.

Hi

I've been gathering myself for far too long then

Fix the time you want to spend procrastinating.

Hi @Fargle

I only get motivated to do my math homework, I can't be bothered to write the rest

better procrastination then pro castration i guess?

Huh

@Null O_o

just saying

It's his way of saying "hi"