@dmckee Maybe I've missed installing something. I need to start from scratch. Can you suggest me an IDE which is coupled with compiler for direct download? I think some compiler files are missing
@Blue The answer to that question depends on the platform you have available, and I probably don't know if (the answer); I've never set up a development environment on DOS or an old version of windows.
@Blue If you don't mind that it is resource hungry like hell, you could just download microsoft visual studio (Community/Epresss edition is free) and only install the C++ part.
Okay. It seems to run on this online IDE codechef.com/ide. But I think the problem was that most C compilers don't recognize <conhio.h>. Or that the clrscr() and getch() are implicit
@Blue Your problem this time is that the compile look for each of the include files in turn, didn't find them and just kept trying to compiler the program even so.
Don't even look at the errors about function needing prototypes—they are simply consequences of the earlier failure.
Anonymous
BTW it is strange that Coursera and EdX don't have a single C programming MOOC. I'm learning it from an ancient youtube video lecture. Maybe I should just read from a book. :/
Certainly the file conhio.h is a hack used in a number of development environments to enble feature tht the c-standard library expects but the host environment doesn't provide.
Anonymous
@dmckee I get it. Probably most modern IDE's don't need the conhio.h hack?
Those environments that needconio provide it. Those that don't need it simply don't have a file named that way. You can alway provide a black file with that name if you need to use the same code base on those platforms.
@Blue If you want to run K&R exercises on Mac OS you can do it from the terminal$^1$ (because Mac OS X is unix), but if you want to run it from the windowing envrionment then you need exactly that kind of hack.
Similar comments must apply to windows, but I don't know the details because I never program on Windows.
@AccidentalFourierTransform Yes but I happen to have a very strong opinion about this: cutting off a piece of a child's genetalia without their consent seems horrible.
I have battled depression from being circumcised for much of my life.
I had a piece of my body removed from me... from a part of myself that is extremely important, and I never had any say in it.
I will forever and irreversibly be unable to experience some aspects of the most important human activity because of this. That is a deeply depressing thing.
A very simple one. The glans is a mucous membrane. It's supposed to protected. Circumcision leads to the glans becoming dried, cracked, and less sensitive. This is not good. The mechanics of sex are altered as well.
@AccidentalFourierTransform Yes, wisdom teeth cause all kinds of horrible problems. Foreskins do not.
We don't habitually remove wisdom teeth until it's clear that they're not going to come in properly, for example.
We don't remove the appendix or the gall bladder until it's a problem either.
Habitually cutting off a part of a person's sex organs seems completely ridiculous to me.
I'll be back later.
Meeting.
Please don't circumcise your children. Please.
It's not your decision to remove their sex organs.
for acceleration vectors pointing backward the speed decreases.
as we swing the acceleration vector slowly from forward pointing to backward pointing it must pass through a point at which the change in speed passes from being positive to being negative.
At that point the speed isn't changing.
Anonymous
@JakeRose The centripetal acceleration just changes the direction of velocity vector such that the particle moves in a circle. If it was not present the particle would move out in a straight line path.
@JakeRose So the effects of adding a small perpendicular to an existing vector are (a) to change the direction and (b) to lengthen it.
In the limit that the addition is small, effect (a) is proportional to the magnitude $\epsilon$ of the addition, while effect (b) is proportional to the square of $\epsilon$.
But we just said we're doing the limit of small $\epsilon$, so $\epsilon > \epsilon^2$. The first effect is larger than the second. As $\epsilon$ get arbitrarily small we are safe in just ignoring the second effect.
As a workaround while this request is pending, there exist several client-side workarounds that can be used to enable LaTeX rendering in chat, including:
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@JakeRose Well, I personally like the "it has to switch" I gave above argument for the existence of speed-maintaining accelerations, but I don't know of a way to show that perpendicular is that condition except by taking limits by hand until you get nauseated enough to just give up.
@JakeRose Keep in mind that people here are often chatting concurrently with doing other things in the real world, so sometimes they have to step away from the computer without warning.
OK, I think I see where you're going... if I understand correctly, you're confused because according to the Pythagorean theorem, when you add a perpendicular change in velocity ($\Delta v = at$) to the initial velocity $v_i$, it should be the case that $\sqrt{v_i^2 + (at)^2} > v_i$, and you're wondering why that doesn't mean the magnitude of $v$ increases?
It doesn't, really. All I'm doing here is showing why the intuitive way of thinking about this, with the Pythagorean theorem, doesn't work the way you'd think it would.
One thing to check: we're assuming that the force stays perpendicular to the velocity, right? So that if the velocity changes the direction, the force will also change direction?
Cool. That makes a difference because $\vec{v}_f = \vec{v}_i + \vec{a}\Delta t$ only works for constant-acceleration motion. But a centripetal force situation, the moment the velocity changes direction, the force also changes direction, which means the acceleration changes.
So if you want to add $\vec{a}t$ to $\vec{v}_i$ (that's where the Pythagorean theorem comes in), you have to limit yourself to considering a time span where the direction of the velocity is constant.
OK. So that's the point I was making before: when you have a perpendicular force, the direction of the velocity is always changing. (I'm not saying anything about the magnitude just yet.) The maximum amount of time for which the velocity and acceleration are constant is 0.
And that's why, if you're going to add $\vec{a}t$ to $\vec{v}_i$, you can only plug in 0 for $t$. If you try to consider a nonzero amount of time, you're breaking the assumptions that underlie the equation $\vec{v}_f = \vec{v}_i + \vec{a}t$ (and the Pythagorean theorem).
So that's why the Pythagorean theorem argument you were making earlier, to show that the speed should increase, doesn't work.
Well, if the object kept its acceleration constant for a nonzero amount of time, then sure, you could use a nonzero amount of time.
But as we said earlier, an object in circular motion doesn't keep its acceleration constant for any nonzero amount of time. Does that make sense? We can go back to it if need be.
Maybe just forget about the equations for a second and think about an object moving in a circle. How long does that object keep moving in the same direction?
Anyway, the next logical step is that the equation $\vec{v}_f = \vec{v}_i + \vec{a}t$ - which corresponds to that triangle diagram you posted earlier - can only be used if the acceleration is constant for the timespan $t$
I actually prefer to write it $\vec{v}_f = \vec{v}_i + \vec{a}(t_f - t_i)$, and then you have to have constant acceleration between $t_i$ and $t_f$
Cool. So if you understand that the acceleration does not stay constant for any nonzero amount of time, and that you can't use $\vec{v}_f = \vec{v}_i + \vec{a}(t_f - t_i)$ unless the acceleration stays constant between $t_i$ and $t_f$, then it follows that you can't use $\vec{v}_f = \vec{v}_i + \vec{a}(t_f - t_i)$ in this situation, unless $t_i = t_f$ (or in other words, $\Delta t = 0$)
Yeah, I see what you're saying. I only suggested circular motion because it's simple, though. For now, we can just assume the object is moving along some arbitrary curve where the acceleration is perpendicular to the velocity and stays that way.
@JakeRose Well... okay you can have conceptual clarity without calculus, in some cases, but I don't think anything you can do without math will match the rigor that you can achieve with math (in this case, calculus)
As a workaround while this request is pending, there exist several client-side workarounds that can be used to enable LaTeX rendering in chat, including:
ChatJax, a set of bookmarklets by robjohn to enable dynamic MathJax support in chat. Commonly used in the Mathematics chat room.
An altern...
