The h Bar - 2017-01-10 (page 1 of 2)

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12:51 AM

@BernardoMeurer notice the power of the $(x-2)^2$ decreases in the denominator and the sign of the integration flips (the bounds flip), it looks like he divided out the -(2-x) in the numerator. I dont think there is any property that allows for you to extract like terms like that from integrals just willy nilly and he skipped a few simplifying steps in the solution

Bernardo Meurer

@Skyler I have been rekd by my exam by now man :P

oh wait, actually since the integral is with respect to t and not x he can just pull that out of the integral

Bernardo Meurer

But thanks anyway

yw, I know that feeling all too well

1 hour later…

user228700

1:55 AM

Hi, everyone :-)

user228700

I have a very quick question about the definition of periodic functions. My textbook says that the period of any function, $T$ must always be greater than zero while other sources that I have found say that it is sufficient that $T \ne 0$. The latter seems to be correct but I wanted to make sure; so which is it?

user228700

2:09 AM

^ Nvm.

2:56 AM

Negative periods would be weird :-P

Super graphene! MIT reveals revolutionary ultralight material that's TEN TIMES stronger than steel / dailymail

user228700

3:57 AM

@DavidZ Would they really? .__.

4:17 AM

@Kaumudi.H well, only in the sense that one would wonder why you're using negative periods when positive periods work just as well and are identical

4:32 AM

@Kaumudi.H The point is $f(x - T) = f(x) = f(x + T)$, so it doesn't matter.

Like DavidZ said.

user228700

5:00 AM

@DavidZ Oh, of course, OK.

user228700

5:57 AM

@JohnR: Morning :-)

Morning :-)

Maths today?

user228700

Both. Physics now...

user228700

This song (sans the angst) sums up my feelings about "the system":

user228700

This Isn't Hogwarts! A Harry Potter Song

user228700

6:18 AM

I have a bit of a homework-tsy question about the electric field intensity due to an infinite wire with uniform charge density $\lambda$. The system is as shown below:

user228700

user image

user228700

The electric field at any point at a perpendicular distance $r$ from an infinite wire is given by: $(2k\lambda/r) \hat{i}$

user228700

Using this, it is clear that the electric field intensity at point $C$ is $(2k \lambda / \sqrt{3^2+4^2}) \hat{OC}$

Missing }

user228700

Thanks :-)

user228700

6:23 AM

Which is ($2k\lambda /5) \hat{OC}$

user228700

This matches the answer in my textbook but it took me awhile to get here so just to be sure, what I've done, it's correct, no?

Yes, that all looks fine

user228700

Phew. Thanks!

For a moment I was worried you were going to ask me how to derive the field for an infinitely charged wire, and I have no idea. Gauss' thereom?

user228700

:-P Dyou remember this:

user228700

6:28 AM

Jan 1 at 6:28, by Kaumudi. H

user image

Oh yes, that was to do with a slightly weird way of measuring angles IIRC

user228700

Yep. We got an expression for $\vec{E}$ in terms of $\theta_1$ and $\theta_2$. In case the wire's infinite in length, $\theta_1=\theta_2=\pi/2$, remember? :-)

Ah, yes, OK. Though I think if you use Gauss' theorem the answer falls into your lap

user228700

My textbook hasn't introduced Gauss' theorem to me yet--they're having me derive all this crap from scratch.

Probably useful practice ...

user228700

6:30 AM

:-P Optimistic as always.

user228700

Ooh, infinitely large uniformly charged sheet. Such fun!

Presumably you take the result for a line and integrate it up?

user228700

Exactly that, but it's infinite and all, so. Gimme a minute, I'm figuring this out...

In case I miss the chat session, someone please mention the best of 2016 meta post.

user228700

Yaaas. Phew.

7:19 AM

Do anybody know what happen to mafia

user228700

@AnkitSharma Oh, wow, he's gone .__.

user228700

@JohnR: Are u busy?

No ...

user228700

I thought that I had figured out the infinitely large uniformly charged sheet of charge but I'm a little confused. Here's the diagram:

user228700

7:34 AM

user image

user228700

I think that it's pretty clear what I did; took those strips and integrated from one end to the other.

user228700

The field at $P$ due to one such strip is just $2K\lambda / r$ where $r$ is the perpendicular distance b/w point $P$ and the strip.

user228700

Now $\lambda = \sigma dx$

user228700

...right?

Gosh that's a confusing diagram ...

user228700

7:42 AM

No kidding. Hang on...

I take it $r$ is the perpendicular distance from the sheet?

Before we go any further, what result did you get?

user228700

Hang on, please be my rubber-duck for one second...

OK

Yes, $\lambda = \sigma dx$

@Kaumudi.H that's why I asked. I though someone from here might know. First he changed his name tho user**** then his account looks deleted

user228700

If we take an infinitesimal rectangular strip on this sheet with length $dy$ and breadth $dx$, we have $dA=dy\times dx$. We know that the charge $dQ$ carried by this rectangular element is given by $dQ=\sigma \times dA= \sigma \times dx \times dy$.

7:46 AM

hi everyone

@JohnRennie can u pls help with my black hole question

@Kaumudi.H OK, though it isn't clear to me why you're using an element dx dy

user228700

One second, I'm still figuring this out...

@Kenshin Link?

i mean one I ask here

I've heard that when you watch someone fall towards a black hole, tyou see them approach the event horizen but never reach it

Is this just an optical effect/illusion or does the person really never reach it from my frame?

user228700

@JohnRennie I'm trying to figure out why $\lambda = \sigma \times dx$.

user228700

7:50 AM

@Kenshin Hey :-) @AnkitSharma: I see. I have no clue @Sir: Dyou know about MAFIA?

Halo @Kaumudi.H

@Kaumudi.H Ah OK. These charge densites are per unit length and per unit area respectively. Yes?

user228700

Yeah.

So lets take a unit length of the line. Then the charge in this unit length is $1 \times \lambda = \lambda$.

user228700

Gimme one second (sorry :-|) ...

7:53 AM

@JohnRennie did u see my question

On our sheet the "line" is actually a thin strip of unit length and width dx so it has area $1 \times dx = dx$.

@Kenshin a moment ...

user228700

@JohnRennie The infinite strip has unit length?

@Kaumudi.H A unit length of our "line" is actually a thin strip of unit length and width dx

user228700

OK..?

You seem doubtful?

user228700

8:00 AM

No no, do go on...oh, u were done. OK, gimme a minute...

OK. The charge on unit length of the line is $1 \times \lambda = \lambda$. Yes?

user228700

Yeah...

On the sheet our "line" is actually a thin strip of unit length and width $dx$ so its area is $1 \times dx = dx$. Still OK?

user228700

Yeah...

user228700

Wait, what was our "line" originally?

8:03 AM

"line segment" I should probably say. i.e. a unit length of the infinite line.

user228700

Right, OK...

So the area of the strip on the sheet corresponding to the "line segment" is $dx$. Yes?

user228700

Yeah...

user228700

I can see where u're going with this.

And its charge is just areal charge density times area $\sigma dx$

So $\lambda = \sigma dx$

user228700

8:05 AM

Gimme one minute, please...

user228700

I'm not thinking straight. Dyou mind if we revisit this after awhile if u're not busy then? (I'm really sorry :-|)

Yes, just ping me when you want to have another go.

@JohnRennie pls

user228700

@JohnRennie Thank you :-)

@Kenshin Suppose you are watching a spaceship accelerating away from you.

8:08 AM

yeh

As the spaceship gets faster and faster its time dilation gets bigger and bigger so from your perspective the clock on the ship gets slower and slower.

when you say the ship gets slower, you mean it's accelerationg decreases?

it doesn't decelerate does it?

Well the spaceship cannot reach $c$, so as observed by you it's acceleration must decrease. Otherwise it would reach than pass $c$.

agreed

In fact in your frame the acceleration you observer, $a'$, is $$a' = \frac{a}{\gamma^3} $$ where $a$ is the acceleration measured on the ship and $\gamma$ is the Lorentz factor.

8:11 AM

yeah ok

Anyhow, the spaceship approaches $c$ asymptotically, and as it does so its clock asymptotically approaches a complete stop.

yep

If you watched for an infinite time you'd see the spaceship reach $c$ and its clock stop completely. But only if you watch for an infinite time.

yeah i understand that

Something similar but subtly different happens if you're watching the ship falling into a black hole.

8:14 AM

ok hang on

ni the spaceship example

not only does it "appear" the ship's acceleration slows down

but in my frame of reference, the ship's acceleration really is slowing down

it's not an optical effect

it's a real effect due to the nature of spacetime

By the ship really is slowing down did you mean the ship's acceleration really is slowing down or the ship's time really is slowing down

yeah

Correct

I mean the ship's acceleration in my frame really is slowing down

so in a black hole though

is the object really assymptotically approaching the event horizen in my frame

or does the object only appear to be assymptotically approaching the event horizon optically?

It's very rare we deal with how things look i.e. optical effects. Generally speaking when discussing GR you can take the verb see to mean observe i.e. assign spacetime coordinates to.

8:17 AM

ok cool

sometimes it gets confusing because people say "appear" etc very loosely

and "looks"

So when when we say the ship approaches the horizon asymptotically we mean in our coordinates the trajectory of the ship approaches the horizon asymptotically.

ok good to know

thanks @JohnRennie

I also have a follow up quesiton

Does this mean if I'm always far from a black holes, I'll never see one form?

Yes

WOW

that's amazing and weird

129

Q: Why does Stephen Hawking say black holes don't exist?

Recently, I read in the journal Nature that Stephen Hawking wrote a paper claiming that black holes do not exist. How is this possible? Please explain it to me because I didn't understand what he said. References: Article in Nature News: Stephen Hawking: 'There are no black holes' (Zeeya Mera...

black-holes quantum-gravity hawking-radiation event-horizon

8:21 AM

too hard

@JohnRennie suppose I trill a hole in the Earth, and drop an object down

OK

the object will reach the centre of the earth (where there is 0 gravity) and then the momentum will keep it going

it will then reach the other side of the earth, and be pulled back to the centre

and will osciliate correctg?

Yes

Now suppose I'm moving into a black hole

and I'm a dot (i.e. ignore tidal forces that would distort my shape)

When I reach the centre of the black hole, will I pass through it and move to the other side

and osccilate like in the Earthe xample?

With the Earth we write the equation of motion for the falling object, then we integrate it. At on point given the position and speed the next incremental movement $dr$ is given by the equation of motion.

With a black hole the same is true, but there's a problem. At the moment you reach the singularity the equation of motion becomes singular. It predicts the next step $dr$ is infinite.

So the trajectory for the infalling is only defined up to an infinitesimal distance from the singularity. We can't tell what happens to the object after that.

8:26 AM

i see

so most likely our theory has broken down

or a glitch in the computer that built the universe

Yes, the usual view is that as we approach the singularity we reach the bounds where GR is a good model of the real world and some more complicated theory is needed.

why is there a singularity

And most think that more complicated theory will be some form of quantum gravity.

I mean, I've heard black holes are just really dense groups of matter

so I thought gravity is 0 in the centre of even a dense ball of matter

@Kenshin careful!

In the real universe it takes an infinite time for a black hole to form.

So when we say black hole we actually mean a really dense object on its way to becoming a black hole

8:28 AM

so black holes don't exist?

so we don't have to worry about the singularity?

129

Q: Why does Stephen Hawking say black holes don't exist?

Recently, I read in the journal Nature that Stephen Hawking wrote a paper claiming that black holes do not exist. How is this possible? Please explain it to me because I didn't understand what he said. References: Article in Nature News: Stephen Hawking: 'There are no black holes' (Zeeya Mera...

black-holes quantum-gravity hawking-radiation event-horizon

@JohnRennie the answers there say that it does exist

@Kenshin by changing what we mean by black hole i.e. changing the term to mean a really dense object on its way to becoming a black hole

lol i see

so under the new terminology, there is no singularity in a black hole?

the "black hole" is approaching a structure that would have a singularity

but it would never become such a structure

Correct

8:31 AM

thus the singularity is purely mathematical interest, not a physical thing

But again be careful

For us outside observers it takes an infinite time for the singularity to form.

But if you're the one falling in then by your clock the singularity forms in a finite time

but not for observers close to the black hole

yes

But this means black holes could exist?

That depends what you mean by exist

depending on how close u r to it

It doesn't matter how close you are to the black hole. As long as you remain outsdie the Schwarzshild radius you will never observe it to form.

8:33 AM

how close do you have to be for the black hole to form in finite time?

oh k

You have to get inside the Schwarzschild radius to see the black hole form. And once you're inside the Schwarzschild radius you cannot remain at a fixed position. You're doomed to fall into the singularity.

yes

interesting

So you'll see the singularity form at the moment it kills you :-)

ty @JohnRennie this has cleared up many htings

8:50 AM

This is so weird, it's as if the universe has two histories that are glued together at the event horizon: Outside, it acts like there is never a singularity, inside the schwartchild radius, the singularity can form in finite time.

user228700

Does anybody know a proper definition for the composition of two functions? The one in my textbook that I shared the other day sucks so...

user228700

(This is it:

yes f(g(x))

@Secret the black hole can never form

user228700

Dude -_- Hang on, I'm searching for the definition in my textbook...

@Secret there is no "inside the schwarzchild radius" because there is no black hole

"black holes" as used by scientists means structures approaching the theoretical black holes

8:56 AM

21 mins ago, by John Rennie

You have to get inside the Schwarzschild radius to see the black hole form. And once you're inside the Schwarzschild radius you cannot remain at a fixed position. You're doomed to fall into the singularity.

yes

John is saying if black holes were real

but they're not real

because they can never form

see John's answer here:

user228700

Jan 3 at 1:35, by Kaumudi. H

> "Let $f: X \rightarrow Y_1$ and $g: Y_2 \rightarrow Z$ be two functions and $D$ is the set of values of $x$ such that if $x \in X$, then $f(x) \in Y_2$. If $D \ne \phi$, then the function $h$ defined on $D$ by $h(x) = g[f(x)]$ is called composite function of $g$ and $f$ and is denoted by $g \circ f$"

"However the Schwarzschild metric is time independent so it would only describe a real black hole if that black hole had existed for an infinite time and would continue to exist for an infinite time. Both of these are not possible in the real universe. So the Schwarzschild metric is only an approximate description of a real black hole, though we expect it to be a very good approximation. "

129

Q: Why does Stephen Hawking say black holes don't exist?

Recently, I read in the journal Nature that Stephen Hawking wrote a paper claiming that black holes do not exist. How is this possible? Please explain it to me because I didn't understand what he said. References: Article in Nature News: Stephen Hawking: 'There are no black holes' (Zeeya Mera...

black-holes quantum-gravity hawking-radiation event-horizon

user228700

Many pointed out that this definition (of composition of functions) ^ is crap so does anybody know a better one, perhaps? Please..?

yes

user228700

8:58 AM

Yes..?

suppose f(x) is a function that maps set A to set B

and suppose h(x) is a function that maps set B to set C

user228700

No, no, I understand it. I just need a better definition.

so we can only fall into an object that is on its way trying to form a black hole, but then what happen when we do fall into one such dense object?

this is a defintiion

I was providing a definition

user228700

Dyou have something like the one in my textbook? I can write one on my own but everything hurts and I'm dying (ignore) so I was looking for something I could just copy onto my notebook directly.

9:00 AM

what's wrong with the one you linked

looks good to me

user228700

If u click on the link, you'll be able to read the conversation that followed if u feel inclined to find out exactly why people said it was crap (It's not crap, it's just that it could be phrased better)

user228700

So, does anybody have a nice definition? One that u trust to be proper and rigorous and whatnot?

Kaumudi

I clicked the link

i read quite a bit down

where did they say it was crap?

user228700

Like I said, they didn't say it was crap, it's just that it could be phrased better.

where did they say that

user228700

9:04 AM

Jan 3 at 2:07, by Balarka Sen

Let's all agree that your book likes to write crap

user228700

Jan 3 at 2:08, by Balarka Sen

crap as in crap notation/phrasing

user228700

Jan 3 at 2:07, by Natecat

@Kaumudi.H Yeah, that definiton is phrased strangely imo

I dunno

I disagree

I like the textbook definition here

user228700

OK, thanks.

np

9:09 AM

The definition in your book is correct($D$ will be understood as the preimage of $Y_2$). It is just a bit wordy in spelling out that technically $h = g \circ f$ is a function different from $g$ and $f$

which people tend to omit for some reasons

probably because in most usages, we tend to choose the same domain and codomain

user228700

Yeah, exactly. And it also fails to properly explain why the two notes (Um, you'll know about the notes if u followed the other conversation) make any sense at all. Those notes were what I asked about the other day.

user228700

Anyway, OK, I'll manage. Thanks.

yeah I"m not sure about the "notes"

but the definition is spot on

the "notes" may be wrong, i don't know what the "notes" are

user228700

No, the notes aren't wrong. It's just that one can get crazy confused reading the notes after reading this definition. They could've done a better job, IMO.

I haven't seen the "notes" so I can't really say but yeah the notes may be confusing you rather than the definition itself

9:14 AM

This definition can in fact be visualise using the domain and codomain mapping diagrams diagrams

user228700

@Kenshin *may have been. Balarka helped me with the notes. I'm just making notes now.

user228700

(As u can see, I don't feel like typing what the notes are :-P Sorry.)

user image

actually, f(x) may be smaller than Y2, it is only important to note that f(D) is a subset of Y2

user228700

That's really cool. Thanks for drawing it :-)

Note Y2 isn't necessarily a subset of Y1

9:33 AM

Hello. Does anyone here know about how to use reduced mass in mechanics ?

yes

it is used when you compute e.g. the rotation of two masses about some common centre of rotation

Good, I need help with this physics.stackexchange.com/questions/303937/…

0

Q: Proof of reduced mass form work energy theorem

Two blocks having mass $4 kg$ and $10 kg$ respectively are connected by a spring. They have initial velocity and accelerations as shown in the above diagram. In my textbook block A is taken as rest and block B is taken to be the reduced mass $\frac{(4)(10)}{4+10}=\frac{40}{14}=\frac{20}{7} kg$...

homework-and-exercises newtonian-mechanics energy work

sorry too boring

Hmm.

I need some reference to learn the technique

9:37 AM

In physics, the reduced mass is the "effective" inertial mass appearing in the two-body problem of Newtonian mechanics. It is a quantity which allows the two-body problem to be solved as if it were a one-body problem. Note, however, that the mass determining the gravitational force is not reduced. In the computation one mass can be replaced with the reduced mass, if this is compensated by replacing the other mass with the sum of both masses. The reduced mass is frequently denoted by mu), although the standard gravitational parameter is also denoted by μ ...

If I take one mass at rest and the other in motion (with reduced mass) then how to calculate its kinetic energy with the body at rest?

I'm not sure how they took it to be $(1/2)\mu v^2$

$(1/2)\mu v^2$ is the KE from the COM frame

@Kenshin Well, I mentioned in my question that that link does not contain the information I am looking for.

@Secret Do you know?

It has been a long time since I do newtonian mechanics problems, but the centre of mass is obtained by a coordinate transformation. In this case it seemed to be the centre of momentum is important. By doing a galliena transformation such that block A is at rest, block B will then have a velocity determined by the centre of momentum. Here the reduced mass of block B pops up and is the result of the galian transformation to the frame of block A

Center-of-momentum frame

In physics, the center-of-momentum frame (zero-momentum frame, or COM frame) of a system is the unique (up to velocity but not origin) inertial frame in which the total momentum of the system vanishes. The center of momentum of a system is not a location (but a collection of relative momenta/velocities). Thus "center of momentum" means "center-of-momentum frame" and is a short form of this phrase. A special case of the center-of-momentum frame is the center-of-mass frame: an inertial frame in which the center of mass (which is a physical point) remains at the origin. In all COM frames, the center...

so I am guessing $v_{initial}'=v_{initial}-\frac{4*8+4*10}{4+10}$

where since block A is at rest in the new frame, $v_{initial}=10$?

@Secret Here, the reference point is the block A. So, wrt A we are looking at B with velocity 2 m/s I think

B is the reduced mass now

However, I cannot understand why KE of B in A's frame should be $(1/2)\mu (2)^2$

@Secret Initial velocity of B in new frame is (10-8)

9:54 AM

@Kenshin it's not tho.

literally nothing rigorous is said about domains/ranges.

10:04 AM

@Kaumudi.H Here's a rigorous piecemeal definition.
Case-1 (easy): Suppose $f : A \to B$ is a function and $g : B \to C$ is another function. Then $f \circ g$ is the function $A \to C$ defined by $(f \circ g)(a) = f(g(a))$.
Case-2: Suppose $f : P \to Q$ is a function and $g : R \to S$ is another function such that $f(P) \subset R$. Then $f \circ g$ is a function $P \to S$ defined by $f \circ g = f \circ g|_{f(P)}$ (which makes sense due to Case-1) where $g|_{f(P)}$ is the restriction of $g$ to $Q \subset R$.

Case-4: Suppose $f : X \to Y$ is a function and $g : Z \to W$ is another function and $Y \cap Z \neq \emptyset$. Then $f \circ g$ is a function $D \to Z$ where $D = \{x \in X : f(x) \in Z\}$ defined by $f|_{D} \circ g|_{f(X) \cap Z}$.

And that's it. These are incarnations of the same definition (Case-1).

The rigorous definitions above makes clear the relationship of the image of f and the domain of g, which is what the text book's lack

this is important because if the domain of g and image of f are incompatible, one cannot define $g \circ f$

(happens very frequently in matrix multiplications)

I am a bit lazy to define what incompatible means in functions of one variables except the daily life meaning though, someone might fill that gap for me

Err, actually I messed up a bit above. In all the cases I really meant $g \circ f$, not $f \circ g$. Please mentally correct that.

In any case the point is to define $g \circ f$, one needs the range of $f$ to be a subset of the domain of $g$.

Once that's understood, all the cases should be clear.

10:22 AM

@BalarkaSen nothing rigerous needed to be said, no relationship between Y1 and Y2 needs to be defined so long as D is not a null set

@Kenshin You're arguing whether or whether not Kaumudi's defn is correct. It's technically so, but extremely unhelpful. Also, it breaks the convention by defining composite to be defined on a subset of the domain.

@anonymous In that wikipedia link I provided to you, your scenario is u1'=0 (since in A's frame, A is at rest) and u2'=2 which is the velociy of B as you expect. Therefore the momentum of B is given by the equation there, and hence the reduced mass of 4*10/(4+10) pops out. After that you integrate this momentum wrt velocity and you get the kinetic eenergy expression as required

Rigorously, it's not even $g \circ f$. Like I said, it's $g|_{f(X) \cap Z} \circ f|_D$.

It is not clear how we can define $g \circ f$ if image of $f$ is a superset of the domain of g. That will mean g is a partial function and that the g that is needed to define the composite function will be a restriction of g to the image of f

10:47 AM

@JohnRennie So this Gibbs free energy is some sort of potential energy of a thermodynamic system that the system can use to do work?

Sounds about right.

The GFE is like a potential in the sense that the system will try to minimise it.

That may or may not involve doing work.

@JohnRennie So it's not like the more you have it the closer it's to a gas for example.

@LuBu I'm not sure what you're asking there ...

@JohnRennie Good.

@JohnRennie I sometimes may ask stupid question.

user228700

11:16 AM

Hallo hallo, again.

user228700

@BalarkaSen Oh my God. I'll read it with full attention in awhile, thanks so much!

Ok, but do note the correction I mentioned. That is to say, switch the places of $f$ and $g$ in each composition $f \circ g$ I mentioned.

user228700

@BalarkaSen Oh :-| Will do.

user228700

For now, I come to ask another one of my classic "quick" questions, which may or may not take longer than 10 minutes to answer.

Sure.

user228700

11:20 AM

Quick sub-question: How to write the integral symbol in LaTeX?

\int

\int_a^b for definite integral

user228700

Followed by {} enclosing the integrand?

{} is not needed

just write \int some expression

and enclose it with \$ \$

user228700

How to end it, then?

e.g. \$ \int_a^b x^2 dx \$ gives $\int_a^b x^2 dx$

user228700

11:23 AM

I dunno if I have asked this before--probs have--but can anybody help me to understand why the average value of a function $f$ is given by $\int_{t_1}^{t_2} f (t) dt/t_2-t_1$ ?

@Kaumudi.H You're "summing" all the values $f(a)$ with $a \in [t_1, t_2]$ and dividing out by "the number of such values", aka $t_2 - t_1$.

user228700

That does make sense but I'm unable to understand where the $dt$ term comes into the picture.

More rigoruosly, given a function $f$ in some interval $[t_1,t_2]$. If it is continuous, then by mean value theorem, there exists a $c \in [t_1,t_2]$ such that $f(c)(t_2-t_1)=\int_{t_1}^{t_2}f(x)dx$. Rearranging gives the required average function formula

@Kaumudi.H $f(t)dt$ is area of a small rectangle of base $dt$ and height $f(t)$. You're summing all such things to get the area of $f$ under $[t_1, t_2]$. Dividing out by $t_2 - t_1$ gives the "average height", which is the height of the rectangle with base $[t_1, t_2]$ of the same area as the curve below $y = f(x)$ over $[t_1, t_2]$ does.

typo: $f(c)$ should be $f'(c)$

11:28 AM

The various "heights" of the functions are averaged out.

user228700

@BalarkaSen Ah, that makes a lot of sense. Thanks very much! :-)

Integration has a nice geometric picture of areas and volumes under functions

No problem

user228700

11:55 AM

@JohnR: What are ur feelings on alternating current? (:-P)

It's shocking! :-)

What do you need to know?

user228700

:-) I could use some help understanding these terms...are u busy?

No, now is good.

user228700

Wokay. Well, just one question, actually.

user228700

So say the current varies with time according to this function:

user228700

11:57 AM

$i=I_m \sin (\omega t + \phi)$

OK

user228700

I understand what $I_m$ is but my textbook says: "The factor $(\omega t+\phi)$ is called "phase".

user228700

What does this term really signify?

It's just where you take your origin i.e. where you choose $t=0$ to be.

user228700

OK..?

11:59 AM

If you draw a graph of a sine wave then it's zero at $x=0$, increases until $x=\pi/2$ then starts decreasing, and so on.

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