  # csmyth3025

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## Everything posted by csmyth3025

1. Thanks Dr R and Tres for your gentle words. I'm hoping to not only learn the equations used in mathematics, but also how to think more analytically. In the last few months I've literally worked over four hundred questions in the first chapter of the textbook I'm using. As you might imagine, these questions started out simple and became more complex with each set of exercises. I think I've learned as much from this wrong answer as I have from the many right answers I gotten (maybe more). Chris
2. I'm embarrassed to say that you're right. I not only misread the question when I originally worked it, I also misread it again when I posted my question. The entire question - word for word from the textbook is as follows: "Two gasoline distributors, A and B, are 228 miles apart on Interstate 80. A charges $0.85 per gallon and B charges$0.80 per gallon. Each charges 0.05 cents (symbol used, rather than the word) per gallon per mile for delivery. Where on Interstate 80 is the cost to the customer the same?" (bold added by me) I apologize for causing the confusion that my sloppy reading habits have spawned. I can only hope that I learn from this how important it is to read these word problems carefully! Chris
3. You're absolutely right about my wrong calculation and also about the cost of delivery being $0.0005 per mile! The question (as written in the book) gives the cost of gasoline as$0.85 and $0.80 and the cost of delivery as 0.05 cents (using that symbol - I don't know how to make it on my keyboard). This is a good lesson for me to read these problems more carefully and also to think through my approach to the solutions more logically. Thanks Chris 4. I'm tediously working my way through the first chapter of "Basic Technical Mathematics with Calculus", Seventh Edition, by Allyn J. Washington. This is strictly self-study on my part so I don't have an instructor to clarify things for me. The word problem given is: "Two gasoline distributors A and B are 228 miles apart on Interstate 80. A charges$0.85 per gallon and B charges $0.80 per gallon. Each charges$0.05 per mile for delivery. Where on Interstate 80 is the cost to the customer the same?" I first designated the cost of delivered gasoline from A = $0.90/gal*mi and the cost of delivered gasoline from B =$0.85/gal*mi. I then designated x = miles from A. I then calculated as follows: $(0.90)(x) = (0.85)(228-x)$ $0.90x = 193.8-0.85x$ $0.90x+0.85x = 193.8$ $1.75x = 193.8$ $x = 110.743$ (miles from A) $228-x = 117.257$ (miles from B) If I use these mileages and multiply by the cost of delivered gas per gallon per mile I get: $(0.90)(110.743) = 99.67$ for the total cost of gasoline from A. Likewise, the total cost for delivered gas from B would be: $(0.85)(117.257) = 99.67$ This all seemed rather straightforward until I checked my answer against the answer for this exercise given in the back of the book. The given answer is 64 miles from A. I've turned this problem around every way I can think of and there doesn't seem to be any way I can come up with 64 miles from A. What am I missing? Chris
5. I also find the WolframAlpha site to be very useful. In this case, though, I think I learned a lot more by taking the "long way around". Chris.
6. Thank you very much for that information imatfaal. As you can tell, I know very little math beyond the high school algebra I learned 45 years ago. Through self-study I'm trying to remedy this - but it's slow going for me. After reading the Wikipedia entry on inverse trigonometric functions here: http://en.wikipedia....etric_functions and taking a closer look at my TI-83 calculator I now have a rudimentary understanding of the difference between a multiplicative inverse (which I was using) and a compositional inverse as exemplified by terms such as $cos^{-1}$. With this new (to me) knowledge in hand I'm now able to apply the formulae presented in the MathWorld article to the formal calculations that eluded me before. The diagram below is, again, taken from that MathWorld article: R = 3 ft r =0.5 ft h=2.5 ft $\theta = 2cos^{-1}\ (\frac{0.5}{3}) =2.8067$ (radians) With $\theta$ I can use an alternate equation to obtain a : $a = 2\ R\ sin(\frac{1}{2}\theta) = (2)(3)sin[(0.5)(2.8067)] = 5.916 ft$. This agrees with my previous calculation that $a = 2 \sqrt{R^2-r^2} = 5.916 ft$. I can now also calculate the arc length s : $s = R\theta = (3)(2.8067) = 8.4201 ft$ With this information I can now calculate the wetted cross-sectional area A by two methods: $A = \frac{1}{2}R^2(\theta-sin\theta) = (0.5)(9)(2.8067-0.3287) = 11.15 ft^2$ -and- $A = \frac{1}{2}(Rs-ar) = (0.5)[(3)(2.84201)-(5.916)(0.5)] = (0.5)(25.2603-2.958) = 11.15 ft^2$ This agrees with my first approximation of ~11.14 ft2. Thanks again for your help, Chris
7. The influent pump building at the wastewater treatment plant where I work receives flow from a 6 ft diameter interceptor sewer. Our pump controls maintain a sewage level of 2.5 ft in this interceptor. The generally accepted flow velocity in a sewer to prevent grit deposition is 2 ft/sec. A discussion among the plant operators here raised the following question: What rate of flow is needed to maintain a 2 ft/sec flow velocity in this interceptor at a water depth of 2.5 ft. As a practical matter we measure flow in terms of millions of gallons per day (MGD). As an approximation I calculated 1/2 the cross-sectional area of this interceptor (3 ft water depth) and then subtracted an area equivalent to a strip 0.5 ft wide by 6 ft long (3 ft2): Circular segment (water) = $\frac{1}{2}\pi(3 ft)^2 - 3 ft^2 = 11.137 ft^2$ I multiplied this area by 2 ft/sec to obtain a flow rate of 22.274 ft3/sec. Applying the conversion factor 1 MGD=1.547 ft3/sec, simple division provides the answer: ~14.4 MGD. This seems reasonable since the flow through this interceptor rarely drops below 16 MGD. So far, so good. Then I wanted to make this same calculation using a formal geometric approach. I quickly found the equations I needed at the mathworld page here: http://mathworld.wol...larSegment.html This is where I became stumped. The following diagram is taken from mathworld: R = 3 ft r = 0.5 ft h = 2.5 ft The calculation of a is straightforward (and the reasoning behind it is understandable) using the mathworld formula $a = 2 \sqrt {R^2 - r^2} = 2 \sqrt{3^2 - 0.5^2} = 5.916 ft$ When I try to calculate $\theta$ (in radians) by the formulae given I get lost, though. The simplest formula using the information available is: $\theta = 2 \cos^{-1} (\frac{r}{R})$. I can understand how $\theta$ could be equal to the inverse of twice the cosine (r/R), which would be $\theta = (2\times \frac{0.5}{3})^{-1} = 3$. This (in degrees) would be equal to about 171.9o, which is the large angle I would expect. But as I read the given equation, it seems to be saying that $\theta$ equals twice the inverse of the cosine (r/R), which would be $\theta = 2\times(\frac{0.5}{3})^{-1} = 12$. This result (in radians) doesn't make any sense to me. Am I misunderstanding how terms such as $\cos^{-1}$ are used in the formulae given in the mathworld page on this subject? Chris
8. Thanks. I'm ponderously working my way through an introduction to calculus textbook (it's hard to teach old dogs new tricks). I know I can graph equations like this on my T-83 calculator. Once I get the hang of it I would like to give Mathematica a try. This thread has been very helpful to me. Chris
9. First, thanks Daedalus for your thorough and very instructive explanation of this problem. I have a very rudimentary idea of how you're able to display formulae using LaTeX code thanks to the LaTeX tutorial thoughtfully pinned at the beginning of the Mathematics section of this forum . For instance: $c^2=a^2+b^2$. What feature or code produces the excellent graph you were able to include in your post? Chris As an afterthought, I thought I'd try using ["math"][/"math"] encoding instead of ["latex"][/"latex"] coding. I get: $c^2=a^2+b^2$ and: $c^2=a^2+b^2$ in both cases. I suppose it's a matter of personal preference which tag one uses in this forum. Chris
10. You might say that the Earth's atmosphere is "almost closed" at the present time - but this has not always been the case and will not always be the case in the future. You might want to read the rather lengthy article that covers this subject here: http://www.scientifi...eir-atmospheres Also, you may want to keep in mind that entropy is a measure of the energy available to do work. Whether the Earth is gaining or losing mass at present has very little to do with the amount of energy available on the surface of the Earth to do work. This energy is largely supplied either directly or indirectly by the radiation we receive constantly (and in copious quantity) from the sun. Chris
11. I don't know with what level of math you're familiar. The equations and reasoning given by Daedalus should be helpful if your question is typical of the math you're doing now. As a general observation, I'm sure that you realize the the first ball will always be ahead of the second ball on the way up. Only after the first ball has reached the apex of its trajectory and is on the way down will both balls meet (while the second ball is, of course, still going up). The formula given should point you in the right direction to find out how many seconds will transpire until that meeting occurs. Chris
12. (Bold lettering added by me) As Swansont points out, you're delving into territory that is not part of mainstream science. This is quite clear by the phrase I put in bold print. (ref. http://en.wikipedia....re#Significance ) To properly discuss interactions that take place at or above Tp you'll need to formulate a new theory of the physical interactions that occur in this extreme temperature domain. Until such a theory can be peer reviewed and becomes widely accepted in the scientific community at large it will be considered a speculation. As I'm sure you're already aware, a phrase like "melting space-time" has no scientific meaning. You're going to have to be more rigorous in defining your terms. Some math describing the interaction of matter, energy and the distortion of space-time in this high temperature regime would be helpful. Chris
13. different flavours - similar but not entirely the same. the accelerated expansion was the sign of something new - this could be the first sign we have ever had that speed of light being max is flawed, and anything with that as axiom needs to be thought of in new light. ftl was forbidden if we wanted causality and SR - so something needs to change if ftl is a reality (although it probably isn't) Not sure I understand this - I don't think there were any mainstream theories that have anything massless or massive (even very small mass) traveling faster than light. If they were looking to measure speed of neutrinos and show they were superluminal, they would have chosen a longer test, better design (per the new experiments that Tom mentioned above), and if they are shown to have misled their funders (who believed they were researching the oscillations between different flavours) As i have said above - it is both a huge thing and also mindnumbingly unimportant. Einstein's theories of relativity work and have been tested exhaustively so far at virtually every level - for what we use them for now, we will continue to use them; they have not stopped providing correct results and accurate predictions. However there is that first sign (iff the results are shown to be correct) that ftl travel is possible and that would show that relativity is applicable in a limited (albeit vast) scenario and that another bigger, more universal theory is in the background. First, I have to apologize for not proof-reading my original post: I intended the second passage to read as corrected above (actually, I was thinking "...expected to see a neutrino speed slightly less than that of light..." when I started typing that sentence and somehow mentally transfigured "speed" into "time-of-travel" in the process). This is just an embarrassing example of my sloppy writing. I generally agree with your reply. I think the inability to resolve quantum field theory with general relativity is already an indicator that a more comprehensive underlying theory is needed. Although there are, indeed, no mainstream theories that propose FTL travel for massive particles, there have been a number of conjectures about the possible existence of (ftl) tachyons which have an imaginary component to their rest mass. I'm open to the possibility that Opera may have uncovered an unexpected real result. If this is actually the case then we can look forward to some new (or renewed) thinking about tachyons. Chris
14. It's been a while since these seemingly anomalous results of the Opera experiment have been announced. It's gratifying - and a natural response - that conducting this experiment in a slightly different way to possibly eliminate systematic errors has been proposed by the original research team, and that other research teams have indicated their intention to try to replicate the observed original anomaly. This is how science works. If the original results are verified, it would be more accurate to describe this "moment" as similar to that resulting from the work of the High-Z Supernova Research Team and the (independent) Supernova Cosmology Project around 1998. It was originally thought that these two teams would be able to quantify the rate at which the expansion of the universe is slowing down over time. Instead, both teams found that the expansion of the universe has been accelerating for about the last six billion years. Since neutrinos are generally considered to have some very tiny mass (due to observed oscillations in type), I suspect that the Opera researchers expected to see a neutrino time-of-travel slightly less than that of light. Their results certainly warrant careful scrutiny and replication by other research groups. It's possible that there is, in fact, an error in their experiment or their analysis. It's also possible that their results are valid. This doesn't destroy Special or General relativity theory. It would, however, open up the possibility of new physics - just as the observation of an accelerating cosmic expansion has done. If FTL neutrinos are found to be real, I wouldn't consider this finding to be a failure of any scientific theory, but rather a triumph of the scientific method. Chris
15. What course are you taking? What textbook or lab manual are you using? There are many courses that cover the analysis of water and wastewater. One such course is the Sacramento State (University of California) "Wastewater Treatment II" course. Another reference that describes in detail the methodologies about which you're asking is "Standard Methods for the Examination of Water and Wastewater". In the old days this reference was available only in book form, but now it can be accessed on-line (for a subscription fee, of course). You can find it here: http://standardmethods.org/ Chris
16. I didn't intend my comment to be patronizing towards technicians - especially since I am, myself, a technician. There are those who deal with chemical equations who don't find units of ppm meaningful for their purposes. As you've pointed out, there are practical applications in regulatory definitions and industrial applications where ppm units are both meaningful and very practical for the calculations to which they're applied. My reference to ppm units being used mainly by tehnicians refers to the fact that they're used mostly for technical applications rather than theoretical chemistry. Chris
17. High school trigonometry is about the limit of my math knowledge, so please bear with me if my explanation seems overly simplified. Like you, to find an answer I have to "logic it out". The fundamental trigonometric relationships that apply to your problem as originally stated are these: Secant= hypotenuse/side adjacent (sec=c/a) -and- Cosecant=hypotenuse/side opposite (csc=c/b) Where a=side adjacent, b=side opposite, and c=hypotenuse. As normally noted in advanced trigonometry a=x axis, b=y axis, c=r (the length of the line from the origin to the x,y co-ordinates), and theta= the central angle. As a practical example, there is a simple type of right triangle with a 3-4-5 relationship of sides, where we can say that a (the side adjacent)=3, b (the side opposite)=4, and c (the hypotenuse)=5. Putting these numbers in the original equation gives you: (c/a)2 + (c/b)2 = (c/b)2 x (c/a)2 = 25/9 + 25/16 = 25/16 x 25/9, cross-multiplying the left side and multiplying through on the right side gives: 625/144 = 625/144 This example certainly shows the equivalence that the question poses for this particular triangle. The trick is to show algebraically that in all cases (c/a)2 + (c/b)2 = (c/b)2 x (c/a)2. Inverting these relationships, as you've done, isn't necessary to the solution. Hint: a2 + b2 = c2 Chris PS The logic expected by your instructor is probably more sophisticated than I've presented, but it's the best I can do. Edited to add PS
18. You need to think carefully about how you define your concept of "succeed" and "intelligent". For instance, if Amadeus Mozart studied science 10 hours a day, would he eventually produce a work such as Newton's Naturalis Principia Mathematica? Conversely, if Isaac Newton studied music 10 hours a day, would he have eventually produced a body of music as overwhelmingly innovative and impressive as Mozart's? Could either if these geniuses have "succeeded" as well as Steve Jobs, written as prolifically as Isaac Asimov, or created deceptively simple and powerfully evocative paintings such as Norman Rockwell's Saturday Evening Post covers? To me, the human mind is a mysterious realm. Certainly, studying 10 hours a day (in any field) will make a person "smarter". I'm sure that those I've mentioned spent many hours a day studying and thinking about their chosen passion. If not, they might have spent their lives as local curiosities with locally recognized talent and "strange" ideas in the eyes of their contemporaries. I don't consider myself unintelligent. I know, however, that I could spend 20 hours a day studying science, music, literature or art and still not hold a candle next to the brilliance of those I've mentioned. For most of us, study trumps smarts. For some, the combination of study and smarts produces a synergy that transcends the sum of its parts. In the strangely philosophical words of Harry Callahan (Clint Eastwood, "Magnum Force"), "A man's got to know his limitations". On that note, I know that I'll never create a new scientific theory or an inspired application of mathematical relationships. I have faith, though, that if I apply myself and devote enough study to science and mathematics, I'll at least be able to understand some of the work that "smart" men (and women) before me have created. Chris
19. As far as I know, the use of ppm concentrations is mainly restricted to calculations and parameters associated with process control and EPA specified limitations as they apply to water/wastewater treatment and, in most cases, levels of gases such as oxygen and hydrogen sulfide. These units are generally used by technicians. They're not intended to be extremely precise because the monitoring equipment, sampling protocols, chemical dosage equipment and water/wastewater characteristics are imprecise and variable. As an example, the solids in sludges generated by treatment processes is usually given in % solids (to the nearest %) rather than ppm because of the factors mentioned above. A 1% solids content (as a process control parameter) is equal to 10,000 ppm. This degree of uncertainty reflects the imprecise nature of the sampling techniques and the variability of the material being sampled. Nonetheless, this level of precision is sufficient to effectively control the processes to which they are applied. Chris
20. I apologize for the error in my post : 1 ppm does not = 1000 mg/l, 1 ppm = 1 mg/L Parts per million as used in water/wastewater treatment is a measurement of the concentration of a substance (in a solute, usually) by weight. Thus, one part per million is the same as 1 mg/1 million mg, 1mg/L (of water), 1 lb/ Mlb, etc. Calculations involving water/wastewater usually asume a specific gravity of one. Calculations involving sludges usually include the specific gravity of the sludge. Chris Edited to correct errors.
21. I agree with Mooyepoo 100% that Shay's post is obviously a case of a student not reading the course material provided. The questions he posted are very basic water/wastewater treatment questions that don't require any sophisticated knowledge or any math beyond simple arithmetic. If Shay had read the course material he would have encountered the conversion factors that would have enabled him to work these problems. The thread is locked (appropriately), so I won't repeat the questions he posted. I would suggest to Shay that he actually read the course material provided. Pay attention to such things as: 1 L=1000 ml, 1 ppm=1000 mg/L, 1 ppb=1000 ppm, ~0.43 psi=1 ft of H2O (conversely, ~2.31 ft of H2O=1 psi), a BOD bottle contains 300 ml, and the dilution factor (for BOD's) is the volume of the bottle divided by the volume of the sample. Many texts put definitions such as these in bold print. There are some calculations used in water/wastewater treatment that might confuse a student and, thus, warrant a pointer or two to clarify an error that a student might be making. The questions that Shay posted do not fall into this category. They only require that the student to know arithmetic and a few definitions such as those given above. Shay presented no work in his post for the simple reason that he didn't read the text and doesn't know the definitions of the terms used in the questions. Chris
22. Thanks SH. It will take a while for me to fully appreciate the calculus aspect of your explanation. Even without a solid grounding in the math, though, I can still get the gist of the logic involved. I hope to revisit your posts again later when I'll be able to appreciate the step-by-step advanced calculations that explain this rather fundamental concept. In between times, I'll rely on keeping 2 or 3 "guard digits" in my intermediate calculations. For my immediate purposes this should suffice for the textbook questions I'll be working. Chris
23. Thanks SH for your reply. The math is still beyond my level of complete understanding, but I understand the general idea and the logic of your presentation. The graphic representations are very helpful. They illustrate that old adage that "...a picture is worth a thousand words..." I'm encouraged by your response. It's my goal to be able to understand calculus and its associated requisite mathematical concepts so that I have a clear understanding of exactly the type of explanation you've provided. Presenting examples in graphical form is brilliant. Chris