anyway if I distribute the right hand side I can easily see that the units match on the right ... but where $\frac{q}{mc}$ is at, that's where I can't see the units matching.. energy absorbed per time, q, is calories/time

@usukidoll long time no see :3

teadawg !

Hello @Ted!

m is mass... that's I'm F*********ing dying in this equation @teadawg1337

m is a unit of time, but that's also grams

shakes head

shakes Ted's head

c is unknown how do I find out whatttt ittttt isssss

besides the fact that I have calories/time and since mass is a unit of time... with grams attached to its butt I have calories/grams... supposedly it's calories/grams x degree whereeee? Is that because that part of the equation is missing a degree so I have to multiply by degree just askingz

@usukidoll I'd help if I could :(

@usukidoll You wanna know the unit of $c$ ?

YES!

how do I aim for that?

Although, if I plan on double majoring in math and physics, I need to step it up

@usukidoll Give me a sec

ok yes I know now that q energy absorbed in time that's indeed calories/time ( I looked it up :P) and yes mass is grams and it's also a unit of time. AND yeah I am missing degrees in that area... so how do I make my unit of c something with degrees using the fact that q/mc

@usukidoll $[c]=\dfrac{L^2}{t^2T}$

That should be it

@usukidoll, you still need help with "Prove that for any two points A and B.."?

@Marco yeah I got stuck midway ..

@Hippalectryon I've seen those .. but my prof. doesn't use them... what the heck is that?

@usukidoll WHat part ?

@usukidoll L is length, t is time, T is temperature

$[c]=\dfrac{L^2}{t^2T}$ I've read of them

but the problem is that my prof doesn't use any of that in his lectures

$[c]$ is the dimension of c

?!

Is something the matter ?

wait..... I have read about those.. I was looking for a list.. it was in M L T format and it had all the units

@usukidoll It's like the unit

yeah

@usukidoll That means that the unit of $c$ is $m^2/(s^2K^2)$

but how do we figure it out... especially the T part .. is that T part theta in the equation?

m is meters, s is seconds, K is Kelvins (or Celcius)

Theta is a temperature, so its dimension is T indeed

how is mass = time ?

@TedShifrin ??

@Hippa: See this.

Olol that is indeed messed up

wait slow down... how do we get c if we know that q - energy absorbed per time and m is mass which is also gramz

That's why I shook my head when you shook it for me, @Hippa.

@usukidoll Ok let's do this step by step

YAY!

gramz?

gramz?

:( this is not English class -_-

$d\theta/dt$ has the same unit as $q/(mc)$ right ?

Have you learned something good this weekend, teadawg?

I tried to tell @Mike that I = me when it's a direct object, but he got snippy.

I know that the units on the left must equal the units on the right

which means if I have degrees/time on the left I should have degrees/time all over the right

Hence the unit of $c$ is the unit of $\dfrac{dt}{d\theta}\dfrac{q}m$

Ok so far ?

did you distribute the $\frac{d \theta}{dt}$

No

$\frac{d \theta}{dt} = \frac{q}{mc}-\frac{k}{mc}(\theta-T)$ is my equation

@teadawg1337 Here, a Grammar Godwin Point

I'll match you bit for bit, @teadawg.

@usukidoll I think you didn't get my first point, let's start over.

where -k/mc onwards it's easy to see that degrees/time yay but for q/mc it's hard

yes please x(

$\frac{d \theta}{dt} = \frac{q}{mc}-\frac{k}{mc}(\theta-T)$ hence the unit of $d\theta/dt$ is the unit of $q/(mc)$. Is that ok ?

?

The unit on the left must be the same as the unit on the right

yes :)

And you can only add two expressions which have same units

add expressions?

unless there's cancellation because the second expression is really two terms? :D

$\frac{q}{mc}-\frac{k}{mc}(\theta-T)$

That means that the two parts have same units

Add/subtract

huh

?

why does q/mc have the same unit?!

Because if you write $a+b$, $a$ and $b$ must have the same units. Otherwise it makes no sense.

do we know $\theta$ and $T$ have the same units?

You can't add apples and bananas.

@TedShifrin Of course since they're subtracted :D

I can add apples and (bananas + apples - bananas).

Plz no

LOL

bashes Ted - gets counter bashed

lol

big T is strictly degrees.. theta is degrees and time

You realize I haven't got you back for all your evil deeds, @Hippa.

degrees and time, huh?

@usukidoll Then it's impossible

Your formula is wrong.

?! what the heck

You can't add two things that don't have the same units

it is exactly as I copied it down from the book

If $T$ is in K and $\theta$ is in K/s then it's wrong

@user159870: Go back to beginning high school algebra. $(x-r_1)(x-r_2)(x-r_3) = $?

@usukidoll Theta is usually just a temperature, not a temp/seconds

I'll scan the page for ya

Yeah, send the page.

Books aren't allowed to make mistakes, and certainly readers never make mistakes.

@Ted so is that what people were saying before we replaced it with the phrase "everything on the internet is true"?

That was only that ad, @JM :P

Oh, its been said long before the advert

sarcastically of course

well, one poor chatter did have a typo in his physics solution in the text yesterday, @JM. So I'm right.

So, @JM, have you gone to play bridge yet?

Does $f(x)=x^3+ax+b$ always have $3$ roots? @TedShifrin

in $\Bbb C$, of course.

someone there who know about probability theory?

@Hippalectryon hold on a sec I'll get it

@Hippalectryon there's the page

@usukidoll It clearly says that $\theta$ is in degree

@TedShifrin Hmm... I've learned that I'm stuck on proving that $\displaystyle \sqrt{\frac7{12}-\frac{7^{\frac23}(1-i\sqrt{3})}{12(2^{\frac23})\sqrt[3]{-1+3i\s‌​qrt{3}}}-\frac{1+i\sqrt{3}}{24}\sqrt[3]{\frac72(-1+3i\sqrt{3})}}\gt\sqrt{\frac7{1‌​2}-\frac{7^{\frac23}(1+i\sqrt{3})}{12(2^{\frac23})\sqrt[3]{-1+3i\sqrt{3}}}-\frac{‌​1-i\sqrt{3}}{24}\sqrt[3]{\frac72(-1+3i\sqrt{3})}}=\sin\left({\frac{\pi}7}\right)$

@Hippa: I tell my students that reading is a prerequisite for my courses.

@TedShifrin Haha

@teadawg: What the **** are you doing writing inequalities with complex numbers?

And why are you doing anything like this?

Because it came up when I was solving for $\sin\left({\frac{\pi}7}\right)$

Why????

@teadawg1337 I think the question is: what do you mean by $\leq$ over $\Bbb C$?

Well, I thought I'd asked that. But I wasn't clear. Then I asked why he was doing this ... :P

I agree with Ted. I don't do large computations.

I don't even know, to be honest....

Ted is proud of being able to do elaborate computations, but there needs to be method to his madness.

I had no idea it would end up being so daunting....

I think $\sin(\pi/7)$ is a perfectly fine name for that number.

I make my algebra students figure out the fifth root of unity, @teadawg, but I've never tried to do $\sin(\pi/7)$, never.

Well, I do know why I'm doing it. I was using the multiple angle formulas to see if I could find a pattern or something

no patterns ...

who the hell did that long latex have some consideration?

LOL, @usukidoll, you've done plenty of spam here.

@usukidoll So, is the formula ok now ?

@TedShifrin *throws a beanie baby

ducks and cackles loudly

@Hippalectryon the formula is from the book. exactly as scanned no sabatoge

so how did they get c

@usukidoll I know that. But you told me that theta was in temp/time

They clearly say that theta is in temp only

you mean theta know

and she also said that mass was units of mass or time

*theta knot yeah I know it's degrees only, but I'm not focused on nondimensionalizing the ic

@usukidoll That's not what you told me -__-

Anyway soo

@usukidoll ^

it says and I quote that the "[...] variables theta and t have dimensions (degrees and time)"

@usukidoll Which means that theta in in degree, and t in time

it's badly written; they mean "respectively." $\theta$ is degrees and $t$ is time.

Idiot book writers.

the unit of d theta/dt is the unit of q/mc??????????? how... erm

Wolfram Alpha shows that $\sin\left(\frac{\pi}7\right)$ is a root of $64x^6-112x^4+56x^2-7$, but I ended up with the huge monstrosities above

@TedShifrin amen I'm not the only one who thinks this book sucks

@usukidoll What don't you understand ? The unit of the LHS must be the one of the RHS

So it can be solved explicitly, using the cubic formula, @teadawg, but who cares?

Presumably, two of the roots are real and four are not.

:(

All six are complex expressions, though...

Enough about that, I'm just making a fool out of myself lol

@Hippalectryon I don't understand how c ... since it wasn't explained in the book... is calories/grams x degree

Yes, it's a theorem that even when your cubic has all real roots, there's no expression for them without going through horrid complex roots.

I'm not calling you a fool, @teadawg. I'm telling you it's ok to quit and do something else :P

@usukidoll That's what I'm trying to explain, so if you would please answer when I'm asking you if you're following so far ...

k

now @Hippa has some empathy for his teachers :P

So. Have you understood the line above ?

This one ^

hmm, super bowl or math or keyboard?

hmm...

@usukidoll Is that supposed to mean yes or no ?

no :/

Ok. So what don't you understand ?

Do you get that the unit of $\frac{d \theta}{dt}$ is the unit of $\frac{q}{mc}-\frac{k}{mc}(\theta-T)$ ?

You don't get what I mean.

If you have an equation $a=b$ then $a$ and $b$ must have the same unit

Do you get that ?

yeah

My computer crashed ._.

So if we have $\frac{d \theta}{dt} = \frac{q}{mc}-\frac{k}{mc}(\theta-T)$ then both sides have the same unit, ok ?

oh now I get it XD yeahhh it should have the same unit

Ok so $\frac{d \theta}{dt}$ and $\frac{q}{mc}$ must have the same unit, ok ?

yes

Great

That means that $c$ has the unit of $\dfrac{dt}{d\theta}\dfrac{q}{m}$, right ?

why dt/dtheta ? OH wait are you dividing by dtheta/dt?

Kind of. If $[a]$ is the unit of $a$, then we showed that $[\frac{d \theta}{dt}]=[\frac{q}{mc}]$ hence $[c]=[\dfrac{dt}{d\theta}\dfrac{q}{m}]$

Ok so far ?

yeah so you made c by itself but dividing 1/c on both sides too

Ok so now what's the unit of $q$ ?

which mean I need q by itself

<---- goes to cook dinner and abandons @Hippa to his purgatory

@TedShifrin q_q I see why you wanna retire now :D

@TedShifrin throws a muffin

ROFL

divide 1/m on both sides

I've been doing this for 40+ years, @Hippa.

@TedShifrin Oh that's why you're nearly as mad as me ? :P

You're bad enough as it is, @Hippa.

$mc= \frac{dt}{d \theta} q$

then divide dt/dtheta over

You have school in the morning, @Hippa.

@usukidoll Not what I mean

dsfl;aa

@usukidoll What do they tell you the unit of $q$ is

True @TedShifrin

energy per time

energy absorbed per time... what the hell those two labels are different -_____-

@usukidoll so $dt\cdot q$ is in .. ?

$mc d \theta = dt q$ ?!

-___-

way off I know I'm sorry D:

If $a$ is in bladibla per time, then $a*t$ is in .. ?

a bladiblia per time x time = a bladiblia per time squareD?

Ugh no

per time means divided by time

Like in 1 mile per hour

oh ._.

So, bladibla per time * time = ?

ugh I see this through math eyes... like (a)(a) = a^2

-__- Anyway, (bladibla per time)*time = bladibla, right ?

like bladibla/time x time -> bladibla..

Exactly

So, $q\cdot dt$ is in what unit ?

q per dt?

No

per = divided by

But $q$ in in energy per time

So $q\cdot dt$ is in .. ?

?!!!!!!!!!

What's not clear ?

The unit of $q$ is Energy/time. You said it yourself.

what unit is q x dt... what are you talking about

oh yeah

So the unit of (q*time) = ?

energy /time x dt

energy / time x d(time)

(dt)'s unit is time

the 'd' just means that's it's an infinitesimal

AH!

energy / time x time = ENERGYYYYYYYYYYYYYYYYYYYYYYYY

Yee

And calories are in .. ? Energy :D

So you got $c$'s unit

calories/(mass*temp)

$[c]=[\dfrac{dt}{d\theta}\dfrac{q}{m}]$, remember.

ok.. so err can we repeat this process again ? step by step.. I'm starting to get it but just wanna be sure

oh I'm starting to see what's going on ^_^... but can we please repeat ittttt pleaseee...

Try to repeat it yourself here, I'll stop you if you make a mistake

ok

where the left hand side is in degrees/time

and the right hand side must also match degrees/time

Urm wait

What do you mean in that last sentence

x!

like if I do something like -k/mc theta + k/mc T

I'm not saying it's wrong, I just wanna make sure it's right.

and since T is strictly degrees and k/mc is time

then it has the fraction degrees/time?

It's simpler than that.

really?

you can only add things with the same units, so if $a=b+c$ then $b,c$ have the same unit. That way, $a$ has the unit of $b$ or $c$, you can chose the one that's the easiest.

Here, it's easiest to use $q/(mc)$

...

?

Let me recap

We have that the unit of $\frac{d \theta}{dt}$ is the unit of $\frac{q}{mc}-\frac{k}{mc}(\theta-T)$. What is more, the unit of $\frac{q}{mc}$ must be the unit of $\frac{k}{mc}(\theta-T)$. Hence, the unit of $\frac{d \theta}{dt}$ is that of $\frac{q}{mc}$. Ok ?

so d theta/dt is the unit of the entire right hand side

$d\theta/dt$ has the same unit as the RHS indeed

and that q/mc from that right side but also be the unit of the k/mc (theta-T)

Indeed, since they must have the same unit.

so if we were to solve for c, we just need $\frac{d \theta}{dt} = \frac{q}{mc}$

Exactly.

curious question... what if we choose the k/mc blob it would be messy to solve for c right?

It would be messier, but the end result would be the same.

ah ok let's leave that alone T_T

and then divide by $\frac{d \theta}{dt}$

$ c = \frac{q}{m}\frac{dt}{d \theta}$

hmmmmmmmmm.

so the time labels cancel out and we have energy

Guh ?

$c = \frac{q}{m}\frac{dt}{d \theta}$

You've figured out that $q\cdot dt$ was an energy, i.e. the unit of calories

yeah but towards the end I couldn't figure out the label for c besides that I solve it and got an energy label

The $(m\cdot d\theta)^{-1}$ remains, don't forget it

oh ffffffffffml I see the problem omg

yeah I was looking at m dtheta OMG I SEE IT!

:D

woohoo

the non-dimensionalization is easy... if there's a theta* and a t* we just have to find something in degrees and time to cancel out the parameters in the equation and then it will be dimensionless ^_^

Ye

yay!

because t* is still a time unit so we can only pick the variables with the same time unit right?

There is only one time unit

time

and k/mc has it... so if I flip it to make it t* = mc/k then the k/mc should cancel

I guess so

I think I have it on my notebook

one sec

ignore the animals problem that one had units given and it was real easy to see # of animals/days throughout the question

oh how I wish I got more problems like that one

What do we do in dekstra's algorithm if we have 2 equal edges?