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About psychlone

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  • Location
    Australia (Melb)
  • College Major/Degree
    B Eng (Mech)
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    Pwr Gen Engineer
  1. Amr Morsi, I'm just curious, If you were to rewrite your above equation, it would look like be the following, Or not; I'm just pondering your post. [math]\theta_{n}= S_{n} - \frac{f^{(n)}(0)}{n!}S^{n}[/math] where [math]\theta = theta [/math] [math]a = \theta[/math] [math]x = S[/math] Taking the formula notation from the following Wiki page: http://en.wikipedia.org/wiki/Taylor_series.
  2. The answer to your question; how many polygon(s) to theoretically construct a 2D sphere/ sphere cross section it’s limitless. But check out this link, it might be of some interest to you. Cheers. http://en.wikipedia.org/wiki/Method_of_exhaustion
  3. I didn't make an error, I stand by what I wrote. I'm not talking down to you in any way but! a very important word of advice; if you want answers to questions you seek, then you must make the effort to write questions that make sense! (my apologies to the administrators, and to the Science Forums community)
  4. So, when you say the “optimum shape” you’re actually referring to the geometrical shape that will generate the greatest velocity up and down the slope, assuming that the greater the height achieved up the other side of the sphere (“sphere like shape”) is dependent upon maximising velocity? But from memory the equal frictional forces acting on both the bowl’s and marble’s surfaces are dependant not so much on the frictional coefficient of the surface selected, but dependant more on the mass of the object exerting its self on the stationary surface. The small changes in the magnitude of th
  5. Yes, there is an easy mathematical equation. Here are 3 hints! 1. First, though your shape is 3D, the template to create the 3D shape is still only a 2D template. 2. Think more in terms of geometry. 3. Also, try and think about using sine or cosine functions to determine the 2D template’s shape.
  6. Thanks insane_alien, I don't know why I didn't realize before; I'm using Internet Explorer and found the RSS Feed Icon on the command bar like you said. Psychlone.
  7. Hi to all, I was wondering if any of the scienceforum.net sub forums have any URL addresses written specifically to access RSS Feeds? psychlone.
  8. Your equation dcowboys107 from the substitution of equ 1 into equ 2 is correct. [math]\ h = [/math] [math]\frac{370}{cot(25^{\circ})-cot(38^{\circ})}[/math] What you've done is forget to take the reciprocal of tan to equal cot, and then subtracting the two cot’s and divide into 370.
  9. Just like to correct a simple error on the above post. [imath]\frac{d}{dx}y(u(x))=\frac{dy}{du}\frac{du}{dx}[/imath] & rewrite the expression of the last post in a more familiar notation. [imath]\frac{d}{dx}f(g(x))={f'}{(g(x)) ^ . }{g'(x)}[/imath]
  10. To recap on earlier post. [math] f(r,M,N)=\frac{(1+r)^N M r}{(1+r)^N -1} [/math] where [math]f(r,M,N) [/math] is equal to monthly repayments. Take these numbers for example. $100,000.00 Principle. @ 0.00604166% monthly interest (7.25% year). over 300 months (25 years). Yeilds. $722.80 Monthly repayments. ** ONLY AN EXAMPLE** This equation can calculate the entire amount of interest incurred of the full life of the loan (Instantly). What you must do is when plotting the above function; reduce the amount of time remaining on the loan (from the first repayment payment of
  11. **I'm not sure if I' too late for this discussion, but I'd like to add**. I concur with Cap't Refsmmat, the washer and shell isn't a shape of a volume but are names of two methods of determining a volume. Although, I'd like to revise the statement of; " The shell method can easily find the area of a volume rotated around the y axis when it's in terms of x. " Both the washer and shell methods use rotation around the same axis regardless of taking the rotation with respect to x or y axes. Yes, in your case, the two methods, integrating with respect to x along the x-axis with give a
  12. I don’t mean to be rude, and I’m sorry it come across like that & I greatly appreciate your time and effort and knowledge, but as much of a help you are your beginning to frustrating me. My original question is; *But if [math] \phi [/math] & [math] \eta [/math] are variables but are still dimensionless, taking a partial differential with respect to those variables has to be valid. Because you said that taking partial derivative to these variable are similar to taking the derivative of a constant. I meant to back to you sooner with my qustion but I had to go do something. Don
  13. I've looked at my problem again, and the coordinates are in Cartesian form. The reason for this is the problem is; The derivation is uses a Cartesian coordinate system for fluid flowing around the circumference of a cylinder. Where x, is not flow though a pipe but x is flow of flow of fluid parallel to the surface of the cylinder and y is always perpendicular to the surface of the cylinder, which is the thickness of the film. 2 dimensional only, looking only at a cross section of the cylinder. My apologies once again I should have mentioned all this earlier. The answer to the
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