Johnny5

Do you know any quantum physics 5614? No, let me rephrase that, how much quantum mechanics do you know?

Awesome.

I don't know anything about it.

I didn't understand this right away' date=' I think mainly because you didn't specify where the clock is, which is measuring dt. But it is inferrable from the fact that it has to be the system dm is defined/measured in. you have added dt to t, and you are calling that "total time" Let me think... I think of 'dt' as an infinitessimal [i']amount of time[/i], and t as the time coordinate of the reference frame in which dm is measured in (since I don't use SR i dont usually worry about a time coordinate anyway, I am usually focused on amounts of time as measured by specific clocks, be they accelerating through the coordinates an IRF(as in this case) or otherwise). I didn't understand this definition of total time, you will have to explain it to me more. Moving on... In amount of time dt' date=' the total mass interior to the Gaussian sphere which is ship-centered decreases (because matter was ejected, no matter is entering obviously, because the ship's hull is impenetrable) You switched dm on me i think. dm is positive, you previously made it that way remember? when you wrote -dm, what you effectively did was to make dm strictly positive, so that -dm would be strictly negative. See this is why mathematical symbolism is so important in physics. Here is all you need: [math'] dm = m2-m1 [/math] dm is the change in mass, m1 is the initial mass, and m2 is the mass one moment in time later. From the equation above we have: [math] m1+dm = m2 [/math] We are clear that m2 is less than m1 already. So in the formula above dm is negative. That's what you mean, i know it is. You next say that the rockets momentum is: (m+dm)(v+dv) I have absolutely no clue where you came up with that. It really would help your argument, if you labeled which moments in time the values in your formulas correspond to. So that simultaneous quantities would be identifiable. In the formula above, i think m denotes the mass interior to the Gaussian surface at some moment in time, and v is the speed speed at the same moment in time, where v is defined in the CM rest frame. At the very next moment in time, the total mass interior to the Gaussian sphere has changed, and so has the speed of the center of the sphere in the CM frame. The new mass is: m+dm The new speed is: v+dv The magnitude of the old momentum was: mv The new magnitude is : (new mass)(new speed) = (m+dm)(v+dv) Ok I am with you this far. Ok so, since the CM rest frame is an inertial reference frame, the law of conservation of momentum is true in that frame, as you stated quite nicely momentum before = momentum after In other words the momentum value doesn't change. next you say Here was the initial momentum' date=' as defined by you: [math'] \vec P_1 = m_1 \vec v_1 [/math] Now, the momentum at the very next moment in time is going to be denoted by: [math] \vec P_2 [/math] So that the law of conservation of momentum can be expressed as: [math] \vec P_1 = \vec P_2 [/math] So a blob of mass was emitted in one direction, and the rocket accelerated in the other direction. The final momentum P2, is the sum of the momentum's of the parts, the new momentum of the rocket, and the new momentum of the blob of ejected material. We already know the magnitude of the new momentum of the rocket in the CM frame, that quantity is (m+dm)(v+dv). As for the blob of ejected fuel... We need the momentum of the blob at the moment in time the momentum of the rocket is (m+dm)(v+dv). Let's assume that mass is conserved. Recall: [math] dm = m2-m1 [/math] The LHS is negative, and mass cannot be negative, therefore the LHS cannot be the "mass of the blob" It makes more sense to say that -dm is the mass of the blob, since mass is supposed to be a strictly positive quantity. So now, what we need to know, is the speed of the center of inertia of the blob, in the CM rest frame, at the moment in time that the momentum of the rocket is (m+dm)(v+dv). You have that as being equal to v- vex. It is very important to know what v is in that formula right there. v is the speed/velocity of the center of the gaussian sphere one moment in time before the sphere has momentum (m+dm)(v+dv). Ok I think I have it now... Focus on the speed of a particular water molecule in the CM frame, and forget about the rocket completely. Either the water molecule is being carried along with the rocket, or it just got ejected out the back. Before the engines are turned on, the speed of this water molecule in the CM rest frame is identical to the speed of the rocket in the CM frame. They are moving together. Suppose that initially, the rocket is coasting through interstellar space at a constant speed v. Let the trajectory be a straight line. Therefore, the rocket is initially in an inertial reference frame, that is the CM rest frame is an inertial reference frame, just as we previously stated. So before the engines are turned on the rocket has momentum P1 = m v in this inertial reference frame, the one in which the CM of the system is moving with a constant velocity. So the center of mass of the blob (which is about to be ejected out the back of the rocket) has this same velocity v, as the center of inertia of the rocket, at least initially. Now, the speed of the ship increases in this frame. Therefore, the speed of the center of mass of the blob decreases because it is ejected out the back. Now, we are given that the exhaust velocity vex of the rocket is known. v is the initial speed of the center of mass of the blob in the inertial frame. I left for awhile and now I am back to finish this up. I left off at a place where I had to determine what the speed of the center of inertia of the blob is, in the CM rest inertial frame. And that's the system CM, not the CM frame of either of the parts. Now this part of the problem has to use the exhaust velocity, because one that is easy to measure or know in advance, and two it's quite intuitive. Its the speed of ejected matter in the rest frame of the rocket/Gaussian sphere. Not to mention it's part of the solution here: Rocket science Before the blob is ejected outside of the Gaussian sphere, it's speed is v, which is identical to the speed of the center of inertia of the Gaussian sphere in the CM rest inertial frame. Then the blob is ejected, and the blob is now moving AWAY from the Gaussian sphere/rocket, with speed vex. I think the intelligent thing to do now, is write some formulas for coordinate transformations. I am going to be a bit more mathematical now. Many times throughout this argument, I have referred to the CM rest inertial frame. As I recall, I stipulated that before the engines of the rocket are turned on, the velocity of the center of mass of the rocket is constant, so as to ensure that this frame is inertial. Furthermore, I had to do this because Klaynos has made the rocket have an arbitrary initial velocity v. He did not specify the initial conditions of the acceleration, but I took for granted that initially, the acceleration of the rocket is zero in the frame. CM inertial frame: I am going to refer to this frame as frame S. Set up a three dimensional rectangular coordinate system, with its origin somewhere in space. Stipulate that this reference frame is an inertial reference frame. That means that all three of Newton's laws are true statements in this frame. Most importantly, the law of inertia is true. Law of inertia: An object will either remain at rest or in motion in a straight line at a constant speed, unless acted upon by an external force. So initially, the rocket's engines are off, and the object is in outer space, in the vacuum, so there is no external force upon it. Now, the trajectory of the center of inertia of the rocket/Gaussian sphere through this frame must obey Newton's first law in the frame. Let the object be coasting at a constant speed v, in the i^ direction. Thus, the initial velocity of the rocket/Gaussian surface is: [math] \vec v = v \hat i [/math] The letter v in the statement above, is the speed of the object in frame S. Speed is a strictly nonnegative quantity. Definition: The speed of the center of mass of an object is loosely defined as the distance travelled by the center of mass in some frame, divided by the amount of time of travel as measured by a clock which is stationary (not moving) in the frame. But we can also define speed in terms of the coordinates of the reference frame the object is moving in (or at rest). In the example here, the object is always located on the positive x axis of the frame. So in order for its position in the frame to change, its x coordinate in the frame must change. So the distance it has travelled in the frame, over any two consecutive moments in time, is found by subtracting the lesser X coordinate from the greater. Since the object is moving in the direction of increasing X coordinates, it follows that over any two consecutive moments in time, the lesser X coordinate comes first in temporal order. We can represent a tiny change in X coordinates by: [math] dx [/math] Let the initial X coordinate of the center of mass of the Gaussian sphere/rocket in frame S be denoted by: [math] X_1 [/math] And let the X coordinate of the center of mass of the Gaussian sphere/rocket in frame S be denoted by: [math] X_2 [/math] Thus we can write: [math] dx = X_2 - X_1 [/math] And we obtain the speed of the object in S by dividing both sides by dt. That is: [math] v = \frac{dx}{dt} = \frac{(X_2 - X_1)}{dt} [/math] And we have for the initial velocity (before blob is emitted) of the Gaussian sphere/rocket in frame S... [math] \vec v = v \hat i = \frac{dx}{dt} \hat i= \frac{(X_2 - X_1)}{dt} \hat i [/math] So right now, the blob is happily inside the sphere, and the sphere is coasting down the X axis of inertial reference frame S. Now, there is another frame in which we can view things, and that frame is the rest frame of the rocket. The rest frame of the rocket is attached to the rocket, and the center of inertia of the rocket is permanently located at the origin. So wherever the rocket goes, so does the rest frame of the rocket. I am going to refer to this frame as S` Now, the origin of frame S` is the center of mass of the Gaussian sphere/rocket, and we just found its velocity in frame S. Now I don't want the axes of frame S` spinning wildly in frame S, so let the x,y,z axes of frame S` be parallel to the x,y,z axes of S, and let the basis vectors (i,j,k) have the same direction as S's basis vectors. I am using the Galilean transformations. Now, all I want to do is focus on the speed of the center of mass of the blob of fuel that gets ejected from the rocket. There comes a moment in time when the blob exerts a force on the rocket, and vice versa. The rocket speeds up in frame S, and the blob slows down in frame S. The initial speed of the blob in frame S is v. All I want now is the final speed of the blob in frame S, in terms of the exhaust velocity vex. The exhaust velocity only has meaning in frame S`. The initial speed of the blob in frame S` is zero. At the instant the blob is ejected from the rocket, the speed of the blob in frame S` is vex. Now, let me introduce a third reference frame S`` At the instant in time at which the blob is ejected from the rocket, we can view the motion of the rocket and the blob in S`` where the origin of S`` is permanently located where the center of mass of the blob+rocket is. So in other words, suppose for the sake of clarity, that the blob has a mass which is identical to the mass of the Gaussian sphere/rocket. Let us use the letter M to denote the mass of either object. Now before the blob is ejected from the rocket, the mass of the rocket/blob system is given by: [math] M+M=2M [/math] It will help out a whole lot if I introduce some numbers. Let the initial speed of the (blob+rocket) be 200 meters per second. Therefore, the center of mass of the (blob+rocket) has the following velocity in reference frame S: [math] \vec v = 200 \hat i [/math] Now, let the rocket have a mass of 15,000 kilograms. Since the blob's mass is identical to the rockets mass, the blob also has a mass of 15,000 kilograms. Suppose that at the moment the blob is ejected from the rocket, the center of mass of the (rocket + blob) is located at X coordinate 80 meters, in inertial reference frame S. Since origin and all points of reference frame S` were initially moving in a straight lines at constant speed through S, reference frame S` is also inertial. Now, even after the blob is ejected from the rocket, the origin of S`` will continue to move through S in a straight line at a constant speed. The rocket will have speed V1 along the positive x axis of frame S``, and the blob will have speed V2 along the negative x axis of frame S``. If the masses of the blob and rocket are equal then V2=V1. IN the case where the mass of the blob and the mass of the rocket are different, the speeds of each in reference frame S`` can be found using conservation of momentum in frame S``. Since frame S`` is inertial, we know that conservation of momentum is true in S``. let Mb = mass of blob let Mr = mass of rocket initially both are at rest in the frame so the total momentum of the (rocket+blob) in frame S`` is zero. From conservation of momentum in S``, after the blob is ejected from the rocket, the sum of the individual momenta of the parts must still be zero. The rocket is going to accelerate in the i^ direction of S, which is also the i^ direction of S``. The blob is going to accelerate in the -i^ direction of S, which is also the -i^ direction of S``. let the final speed of the rocket in S`` be Vr let the final speed of the blob in S`` be Vb From conservation of momentum in S`` we have: MbVb = MrVr So the speed of the blob in S`` is: Vb = MrVr/Mb Keep in mind that the speed of the origin of S`` through S is constant, from memory the speed of S`` through S is 200 meters per second. Since I am using the Galilean transformations, it follows that after the blob is emitted from the Gaussian sphere/rocket, the speed of the blob in reference frame S will be given by: 200 - Vb And the speed of the rocket in S will be given by: 200 + Vr And the speed of the blob relative to the rocket is Vb+Vr which is just the exhaust velocity vex. So we can write this: vex = Vb+Vr Vb = MrVr/Mb MbVb/Mr = Vr From which it follows that: vex = Vb+Vr = vb+MbVb/Mr = vb (1+Mb/Mr) From which it follows that: vex = vb (1+Mb/Mr) vex/ (1+Mb/Mr) = vb Now, the speed of the blob in inertial reference frame S we have as: 200 - Vb And now we can write the speed of the blob in frame S right after it is ejected from the rocket as: 200 - [vex/ (1+Mb/Mr) ] Now, 200 was a specific choice for v in the original argument given by Klaynos, so that we have instead: v - [vex/ (1+Mb/Mr) ] = speed of blob in S right after emission from rocket. And that is what I have been looking for. So the instant the blob is ejected from the Gaussian sphere/rocket, the momentum of the blob/fuel in frame S has the following magnitude: Mb v - [ vex Mb/ (1+Mb/Mr) ] Which we can also write as follows: Mb v - [vex MbMr/ (Mr+Mb) ] That mixture of masses inside there looks like the formula for the reduced mass, I think thats what that is. MbMr/ (Mr+Mb) = reduced mass = m The point of all this work was to find P2. Klaynos has the answer as P2 = (m+dm)(v+dv) + (-dm)(v-vex) I wanted to double check this. The initial momentum in frame S, is given by Klaynos as: P1 = Mv And the final momentum is P2, which is the sum of the momentum of the rocket in frame S, and the momentum of the ejected fuel in frame S. We already had the momentum of the rocket in frame S, which is: (M+dm)(V+dv) and Klaynos has the momentum of the fuel as being: -dm(v- vex) Now, since dm is negative, the mass of the ejected fuel is given by (-dm) just as Klaynos has. So all we need to do is multiply by its velocity in the frame, and then we can add it to (M+dm)(V+dv) to have P2. The value I got for speed in the frame after emission was: v - [vex / (1+Mb/Mr) ] Let me multiply this by the inertial mass of the ejected blob of fuel, to obtain the magnitude of the momentum of the blob in frame S after emission. We have: Magnitude of momentum of fuel = (-dm) [ v - [vex / (1+Mb/Mr) ] ] Therefore: Magnitude of momentum of fuel = (-dm) v - (-dm)[vex / (1+Mb/Mr) ] And I also used Mb to denote the mass of the blob therefore: Mb = (-dm) Therefore: Magnitude of momentum of fuel = (-dm) v - Mb[vex / (1+Mb/Mr) ] Therefore: Magnitude of momentum of fuel = (-dm) v - MbMr[vex / (Mr+Mb) ] Which we can write in terms of the reduced mass as: Magnitude of momentum of fuel = (-dm) v - mvex So we finally have P2... P2 = (m+dm)(v+dv)+ (-dm) v - mvex Now this is not the formula you got Klaynos, so did I make an error, or was there an error in the formula you gave me? Here was your result for comparison: The only way they are equivalent is if: (-dm) = m And that isn't the case. So where is the error located? Kind regards

Swansont... If I may... you say here that photons are energy, not matter. Can you expand upon that? Specifically, is energy something that can instantaneously exist? You see, the units of energy puzzle me... [math] \frac{Kg m^2}{s^2} [/math] I sort of understand the unit of inertia... the kilogram, or if you prefer... inertial mass. And of course I understand the unit of meter, that's just a unit of distance. But you also see the unit of second... A second is an interval of time. So that energy is something which is defined in terms of "amount of time" so I am perplexed about what energy is. Thank you

Dave I don't know where to put this post, you can delete it later. I would like to know if there are any plans to implement a drawing program into this forum (you know like MSpaint only your own), so that you could draw circles and straight lines, and have the drawing incorporated neatly into your post, rather than attaching a .jpg or a .bmp. Possibly even a graph program, so that you just enter an equation, and its graph comes up. How hard would something like that be? Just a thought... Regards

Minor error there' date=' that should read... u = -mv/(M-m) PS: If you made the error intentionally, that's a good way to check to see if I followed your work. I still have more reading to do. Kind regards I'm just looking at this formula now. u = -mv/(M-m) M is the system mass, which is composed of three parts, the mass of the trolly, your mass, and the mass of the marble m. So clearly, the system mass M exceeds the marble mass. This means that M-m is greater than zero. Taking that inertial mass is strictly a positive quantity, means that m/(M-m) is a positive quantity. Ok so... initially the system is being observed in a frame in which its center of mass isn't moving. So in this frame, initially, the speed of the trolly is absolute zero. After the marble has been thrown, the trolly will move [i']in this frame [/i] opposite the direction which the marble is thrown, by action/reaction principle. Since it will now be moving in this frame, you could introduce a symbol for the nonzero speed which it will then have, and something to indicate direction, but instead, you used the letter u, to denote the velocity of the trolly in this frame. Following so far... Here is the formula again: u = -mv/(M-m) Since u is a vector quantity, and v is a vector quantity, I am going to boldface them, so that I can follow your argument more easily... u = -m v /(M-m) Sine m/(M-m) is positive, multiplication of any quantity by it does not change algebraic sign. I am going to throw in some specific values. Let M=100 Let m = 1 Therefore m/(M-m) = 1/(100-1) = 1/99 So we now have this... u = -(1/99)v Now, suppose that the LHS is pointing in the i^ direction of the reference frame. In order to have the equation above be a true statement in that reference frame, it absolutely has to be the case that the direction of v is given by [math] - \hat i [/math] You can see why, by writing each vector as its magnitude times its direction. Let us use nonbold to indicate magnitude. THerefore: [math] \mathbf u = u \hat i [/math] [math] \mathbf v = v ? [/math] [math] \mathbf u = -(1/99) \mathbf v [/math] [math] u \hat i = -(1/99) (v?) [/math] The scalar parts of each side of the equation above must be equivalent, and the direction of each side must be equivalent. Now, speed is a scalar quantity. It is a measure of how fast something is moving through a frame. Negative speed is nonsense. Speed is always a positive quantity. So for example, the initial speed of the trolly was zero in the frame, and the final speed of the trolly might be something like 2 meters per second. But the final speed of the trolly could never be -2 meters per second, because as I said, negative speed is meaningless. (this fact comes straight out of the definition of speed, which is this (speed is distance travelled divided by time of travel). There is no such thing as a negative amount of time, else time would run in reverse. And there is no such thing as negative distance travelled, which dovetails with the fact that distance is measured by rulers, and rulers can only measure positive quantities. The point is, that speed u must be positive, and speed v must be positive. So now we can solve for the direction of the marble in the frame, in terms of the direction of the trolly in the frame. [math] u = (1/99) v [/math] [math] \hat i = -? [/math] [math] ? = - \hat i [/math] Where the symbol ? was used to denote the direction of the trolly in the frame. It is now plainly evident that the direction of the trolly in the frame, and the direction of the marble in the frame are antiparallel, that is they are exact opposite directions. And as you said, if you throw one off the back, and one off the front, the trolly will remain stationary in the frame. Ok I followed your first post. Now onto the second.

Thank you Do you think you could help me say what I am trying to say better, using mathematics?

Dapthar, do you know how to derive the formula for the inverse Laplace transform? It certainly doesn't just pop into one's head. Thank you

Well figure out how many different logical connectives there are for me if you can. I think the answer is nine. Regards 1. nor 2. nand 3. not 4. and 5. or 6. xor 7. if-then 8. if and only if 9. nif (not if-then) I think this is all of them

I know all of the ones you listed. Let me ask you something, how many different ones are possible? Do you know, is it 5,6,7,8,9 how many? Oh and by the way the symbol for XOR is... [math] \underline \vee [/math] Also do you use any other symbol besides [math] \neg [/math] to denote the logical connective... NOT ? Also I use the sheffer stroke / for NAND And umm... oh yes this symbol for NOR | A nor B = A|B NAND does not appear in my written work for some reason though. But I do use XOR on occasion, it seems to alleviate some difficulties every now and then. Oh, and how good are you in first order logic?

Can someone explain the rocket equation to me in as much detail as they are capable of? Thank you

There is nothing wrong with changing, as long as it's for the better.

Ok, i have my answer. Would you mind listing the symbols which you use to denote the logical operators, or perhaps you know them as logical connectives? Thank you

You guys are not answering my question here... I want to know if... I have said it 10 different ways so far... I want to know if... force and acceleration can be decoupled??? Yes I think that is what I am asking. For the nth time... If force and acceleration can be decoupled, would that mean that the principle of equivalence is false? I am trying to ask it the best way I know how... If it is possible to change speeds, without having your weight altered, is the principle of equivalence false? There try answering one of them.

I know how to answer my own question then... Dave, what is the definition of [math] \Rightarrow [/math] Now I will find out what I want to know. Thank you

Actually the critical cylinder radius was: [math] R = (2 \pi)^{-1/3} [/math] At least that's what I got, and... for the surface area at that value we have: [math] S = 3 (2 \pi)^{1/3} [/math] I guess I don't know what a turning point is. Do you mean inflection point? Regards

No problem Dave. Kind regards

I meant when using them as logical operators.

Ok Dave, do you differentiate between the meaning of [math]\rightarrow[/math] and the meaning of [math] \Rightarrow [/math] Just a simple yes or no will do. You can add an explanation if you feel it is warranted. Thank you.

Now you say so. Yes, just about everything is trivial after you already know it. Well lets just do it, since it is so trivial. To save myself some writing, i will write S instead of S(R,h), or S instead of S®, because it is inferrable from looking at the RHS what variables S depends upon. With that in mind... We can begin with the formula for the surface area of a closed cylinder... [math]S = 2\pi R^2 + 2\pi R h[/math]. The first term on the RHS is the area of two circles of radius R (which is the surface area of the top of the cylinder plus the surface area of the bottom of the cylinder) and the second term on the RHS is the surface area of the cylinder wall, which we can figure out if we cut the cylinder in a direction parallel to its axis, and view as a rectangular area. OK SO, to answer part b, we have to use the given fact that the volume of the cylinder is to be 1 unit. From memory, the volume of a cylinder is pi times the radius squared times the height, that is: [math] V = \pi R^2 h [/math] Letting V=1, we have as before: [math] 1 = \pi R^2 h [/math] Now, PROVIDED THAT R IS NONZERO, we have: [math] h = \frac{1}{\pi R^2} [/math] And we already have: [math]S = 2\pi R^2 + 2\pi R h[/math] So the rest is trivial, just substitute properly. After substitution we obtain: [math]S = 2\pi R^2 + 2\pi R h = 2\pi R^2 + 2\pi R (\frac{1}{\pi R^2})[/math] Now, after simplifying the formula on the far right, we finally get this: [math]S = 2\pi R^2 + \frac{2}{R} [/math] And we are done. We have expressed the surface area of a closed cylinder of volume 1, in terms of its radius R. This answered part b. It is interesting to note that had we have divided by R without thinking clearly, it would have caught up to us at the end, since we have a 2/R term. Part C: Calculate the turning points for S, and find a minima of S. I am not exactly sure what I am being asked for here, but I know it involves taking a derivative, because that is the only way that you can compute either the maxima, or minima of some function of one variable. Recall: [math]S = 2\pi R^2 + \frac{2}{R} =2\pi R^2 + 2R^{-1} [/math] The derivative of S with respect to R is... [math] \frac{dS}{dR} =4 \pi R -2R^{-2} = 4 \pi R - \frac{2}{R^2} [/math] Let us have used a rectangular coordinate system to graph the function S. Let the horizontal axis be the R axis, and let the vertical axis be the S axis. Since negative radius has no meaning, there is no portion of the S-curve in the left half of the S-R plane. Also, S is not defined if R=0. Now, if a particular radius value minimizes the surface area of the cylinder S, then the tangent line to the curve will be horizontal at that specific radius value. Now, the derivative of S with respect to R gives the slope of the tangent line, for different R values. So if there is a radius which makes dS/dR=0 then the slope of the tangent line is zero, which means the tangent line to the curve is horizontal. Setting dS/dR =0 gives us this: [math] 0 = 4 \pi R - \frac{2}{R^2} [/math] Or equivalently... [math] 4 \pi R = \frac{2}{R^2} [/math] Now, we can solve the equation above for R, to obtain the critical values. Multiplying both sides of the statement above by R^2 we obtain: [math] 4 \pi R^3 = 2 [/math] Dividing both sides of the statement above by 4 pi we obtain: [math] R^3 = \frac{2}{4 \pi} = \frac{1}{2 \pi} [/math] So we finally have this: [math] R^3 = \frac{1}{2 \pi} [/math] Taking the cubed root of both sides of the equation above we obtain: [math] R= (2 \pi)^{-1/3} [/math] Choosing this value for the radius either maximizes or minimizes the value of S. We can now express the surface area S of the cylinder in terms of this particular radius value. The surface area of a cylinder of volume one is related to its radius as follows: [math]S = 2\pi R^2 + 2R^{-1} [/math] Now, at the critical radius the surface of the cylinder is related to R as follows: [math]S = 2\pi ((2 \pi)^{-1/3})^2 + 2((2 \pi)^{-1/3})^{-1} [/math] Where I have done nothing other than substitute the value just found into the original expression for surface area in terms of radius. The previous expression can be simplified a bit... [math]S = 2\pi (2 \pi)^{-2/3} + 2(2 \pi)^{1/3} [/math] [math]S = (2 \pi)^{1/3} + 2(2 \pi)^{1/3} [/math] [math]S = (2 \pi)^{1/3}[1+ 2] [/math] [math]S = 3 (2 \pi)^{1/3} [/math] Now, if we look at the second derivative of S with respect to R, we can determine whether or not the previous S value is a maxima or minima. Recall... [math] \frac{dS}{dR} =4 \pi R -2R^{-2} [/math] Taking the derivative with respect to R of both sides of the equation above we obtain: [math] \frac{d}{dR}(\frac{dS}{dR} ) = \frac{d^2S}{dR^2} = 4 \pi + 4 R^{-3} [/math] Now, if we evaluate the second derivative at the critical R value we just found, we find that the second derivative is positive, which means that the function is concave up, like a U, which means that the critical value we found is a minima. I don't know what a turning point is. Regards