Johnny5

IM not sure, let me go back and check. Well we are going to pour water into it, so i guess it should have a bottom, but no top huh? I think it should have a solid bottom with surface area pi R^2, but not a top.

You want to minimize the surface area of a cylinder, so you will need calc 1. First you need a formula for the surface area of the cylinder. Consider the very bottom of the cylinder, its a circle, and a circle doesn't have a surface area. So instead imagine a cylinder of infinitessimal height dh, and radius R. If its a ribbon, you could cut it in half, to produce a rectangle. The height of the rectangle will be dh (i could say width but for this problem its better to call it height), and the length of the rectangle will be given by [math] L = 2 \pi R[/math] Presume that you have already proven that the area of a rectangle is its length, times its height. Therefore, the surface area of a cylinder of infinitessimal height dh, and radius R is given by: [math] S = L(dh) = 2 \pi R dh[/math] Now, since we are going to be pouring water into it, there has to be a solid bottom, but no top. So, the bottom of the cylinder is a circle, and circles have an area which is equal to pi R^2. So let the tiny cylinder have a bottom but no top. Therefore, its total surface area is given by: [math] S = L(dh) = 2 \pi R dh + \pi R^2 [/math] There we go. Thank you yourdadonapogos for pointing out that the cylinder has to have a bottom but no top; that is one side open, and the other closed. I'm not sure how else to say it. Now, we can see that the surface area depends upon the height (dh), and the radius R. So for a teeny tiny radius dR we have: [math] S = L (dh) = 2 \pi (dr)(dh) + \pi (dr)^2 [/math] Now that looks strange to me because I have (dr)2 you don't usually see things like that in calculus 1, so maybe I should just let its radius be denoted by R, so that I can deal with this formula instead: [math] S = L (dh) = 2 \pi R(dh) + \pi R^2 [/math] You know I would love to use partial derivatives really. Ok here is what I am going to do. Here is the formula for the surface area of a cylinder of radius R, and height h, which is half-open. [math] S(R,h) = 2 \pi Rh + \pi R^2 [/math] Now here is the partial derivative of S with respect to R: [math] \frac{\partial S}{\partial R} = 2 \pi h + 2 \pi R [/math] All I did, was take the ordinary derivative of S(R,h) with respect to R, and held the height h constant. Since the ordinary derivative is from Calculus 1, i see no objection to this. Now, determine the partial derivative of S with respect to h. So again take an ordinary derivative, this time hold R constant instead of h. You get this: [math] \frac{\partial S}{\partial h} = 2 \pi R [/math] Alright, I did that just as an excercise. In the process, I have answered part a, which was to express the surface area of a half-open cylinder in terms of pi,h,R. That was this formula here: [math] S(R,h) = 2 \pi Rh + \pi R^2 [/math] This agrees with what yourdadonapogos got. Now to part b of the question... Dave said, "using the fact that the volume is 1, express the surface area [which is currently being denoted by S(R,h) ] in terms of R." The first question is, "what is the volume of a cylinder?" Now, we have to think of our cylinder as a three dimensional solid. The answer from memory, is pi times the radius squared, times the height. Let V denote the volume of a cylinder. Therefore: [math] V(R,h) = \pi R^2 h [/math] Now, Dave wants us to express the surface area as a pure function of R. Because we were told that the volume of this cylinder is 1, this is achievable. We have: [math] 1 = \pi R^2 h [/math] And we also have: [math] S(R,h) = 2 \pi Rh + \pi R^2 [/math] PROVIDED THAT THE RADIUS R IS NONZERO, we can use the first of the two equations immediately above, to express the height h, in terms of the radius. Dividing both sides of that statement by pi times the radius squared we have: [math] h = \frac{1}{\pi R^2} [/math] Now, we are just a substitution (as Dave suggested) away from answering part b. [math] S(R,\frac{1}{\pi R^2}) = 2 \pi R(\frac{1}{\pi R^2}) + \pi R^2 [/math] Which simplifies a bit, to this: [math] S® = \frac{2}{R} + \pi R^2 [/math]

Ah this...

I looked at his signature, and saw something about squirrels. No website.

Is this the problem you are working on? Find the derivative with respect to x of: iv) [math]\frac{1}{x}\cdot \frac{x^2}{a^2} + a^4[/math], where a is constant.

Yes well sometimes there is a single line arrow used for ifthen When an ifthen is known to be true, then the double arrow is used. Perhaps I didn't say that right... Here is a truth functional definition of --> XY X-->Y 00 1 01 1 10 0 11 1 Here is a truth functional definition of <--> XY X<-->Y 00 1 01 0 10 0 11 1 Now, in the case where A <--> B is a tautology, the thicker arrow is used In the case where A-->B is a tautology, the thicker arrow is used. Do you have any idea what I'm talking about? It's just a notation-type question thats all I'm asking.

Dave, What is the symbol in latex for 'if then'? Also, what is mathbb? I saw it give the symbol for the natural numbers.

Where is Sayo's site? I am interested in real physics you know, but I will have a look.

yes... that was wrong indeed. You mean in there dont you?

Dave, what is the definition of continuity please. Thank you.

Oh ok, now it makes sense. Regards

Ok, I saw it. You introduce the definition of derivative in Calculus I-lesson 2, post number one. The latex was garbled (just as you said), which was why I missed it the first time. Now, in CalcI,lesson2,post1, you give the following as the definition of derivative: [math]\frac{dy}{dx}=\lim_{h\to0}\frac{f(x+h) - f(x)}{h}[/math] Now look at my post #15, where I quote yourdadonapogos, and you see that he has you saying this: I just copied that now. Now his post is this Apparently you guys are having some problems with the computer over there. I noticed problems before, but I didn't say anything.

A moment ago, lesson 2 read lesson 1, i think there is a problem with the server. Whatever... Yes I looked through lesson 2, i didn't see it there, which post?

I looked through the thread entitled, "Calculus I- Lesson 1" and I couldn't find it. Which post?

Can you please direct me to the other thread, where you introduce the definition of derivative. Thank you

Here is perhaps the simplest way to explain my question... What IF Your weight inside a spaceship is not necessarily directly proportional to it's acceleration as defined in some inertial reference frame? Would that mean that the principle of equivalence is false? Or to phrase my question still another way, what if it were possible to go from rest to 4000 miles per hour in just a few seconds, but never know (from inside the Elevator), that you changed speeds. Would that mean that the propulsion system used, demonstrates that the principle of equivalence is false?

Dave' date=' I have a question, isn't the definition of derivative supposed to be: [math']\frac{dy}{dx}=\lim_{h\to0}\frac{f(x+h) - f(x)}{h}[/math] Why does yourdadonapogos have you using a different definition of derivative, than the standard one? Something appears to be going on with the direction from which h is approaching zero. Thank you

Is this the exact definition of limit that you see in calculus books?

Here is a link that looks promising: NASA article on aerodynamics In the article you see two things come up, Newton's laws of motion, and air resistance. Explaining them in your powerpoint essay is a good idea, additionally, you should explain the concept of lift, and possibly include some information about propulsion systems. Choose parts of aerodynamics which seem most fascinating to someone who knows very little about aerodynamics. And remember to site the sources of your information. Regards

Right there you say , and I quote, "A and B will disagree on when events happened." Are you using the term 'event' in a manner consistent with its usage in physics, or its usage in temporal logic? The reason I ask, is because perhaps this is one source of confusion, which can easily be eliminated.

As swansont correctly pointed out, I do not share the mainstream view of special relativity. It is clear that I don't, no disclaimer is necessary. Now, I do not know GR, so I am not the best person in the world to answer this. However, I do have execellent spatial reasoning ability, and I think I understand your question, but first I have to make sure. I want to try and answer it, because the subject matter interests me. Ok so... Let it be stipulated to be true, that space expands in the sense associated with the General theory of relativity. First I want to understand your question, I don't want to answer a different question, I want to answer yours. Let us call any reference frame at rest on the surface of the earth, an earth frame. So observatories are earth frames. We could also reference observations of distant galaxies to the location of the hubble telescope, and call such frames hubble telescope frames. Or we could reference our observations to a frame in which the center of our solar system is at rest, and call them solar reference frames. The mathematical description of the motion is going to have varying complexity, depending on which frame we choose. When actual measurments are made in one of the frames, we have no recourse but to use that frame. Now I think that the idea that space is expanding, has recently been bolstered by the discovery that distant galaxies are accelerating away from us. I need to be corrected if that's wrong. And I also think this, that General relativists explain that acceleration by saying that "the rate of expansion of space is accelerating" So now, to your question... You ask, if space is expanding, how does it "grip the material" and pull the center of mass of the material with it. Did I understand your question right?