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B Length of the graph of e^x

  1. Nov 5, 2016 #1
    Hi. I derived a formula for the length of the graph of e^x in the first quadrant between x=a and x=b. It is:
    sqrt(1+e^(2b)) - sqrt(1+e^(2a)) + (a-b) + log [(sqrt(1+e^(2b)) - 1) / (sqrt(1+e^(2a)) - 1)]

    I think it works because it gave a value of approx 2.003 units for a=0 and b=1. For a=0 and b=1, we're moving from (0,1) to (1,e), so, there's a straight line distance of 1.987 units. And, considering that we're going along a curve, that would account for the additional length. Is there some way to check if it works or not?
     
    Last edited: Nov 5, 2016
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  3. Nov 5, 2016 #2

    Simon Bridge

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    What do you mean by "perimeter"?
    To my mind, "perimeter" means "the continuous line forming the boundary of a closed geometrical figure" ... or maybe the length of that line.
    How is e^x a closed geometrical figure?

    Without the definition you are using, and your reasoning/working, there is no way to evaluate your result.
    Do you mean the distance along the curve e^x between x=a and x=b?
     
  4. Nov 5, 2016 #3
    I just edited it. I'm sorry. But I think it was clear from the question that I was talking about the length of the curve in the first quadrant between any two values of x.
     
  5. Nov 5, 2016 #4

    Simon Bridge

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    Well, considering that many people asking questions here make mistakes and category errors, it wasn't clear. You have yet to show your reasoning either - without which there is still no way to evaluate your work. Please understand that nobody here knows the context.

    So do you have an expression for the length ds (say) along the curve between x and x+dx?
     
  6. Nov 5, 2016 #5
    No, I don't have any expression like that ( I think). BUT, I do have an expression for the length along the curve between any two values of x. And, I'm talking about only the part of the graph in 1st quadrant. I'm not sure if it will work in other quadrants. I mean, just like we have expression for the area under the curve. The expression that I posted gives you the length. The value of the length between x=0 and x=1 came out to be 2.003 which is close to the straight line distance between the points. So, I'm certain that it works.
    EDIT: And, I have derived a similar expression for the length of log(x). And, the length which it gave between x=1 and x=e was equal to the length that the expression for e^x gave between x=0 and x=1. SO, I'm even more certain that it works.
     
  7. Nov 5, 2016 #6
    Have I found something new?
     
  8. Nov 5, 2016 #7
    I think this may be the length 'ds' that you're asking:
    ds = sqrt(1+e^(2x))
    Because I integrated that thing to get the expression for length.
    I found it by the following reasoning:
    The derivative or the value of slope of the tangent at any point (x,y) on the graph of e^x is : e^x
    I assumed that every point on the graph of e^x is touched by an infinitesimally sized right triangle by its hypotenuse. The triangle has its base equal to dx and perpendicular equal to dy and hypotenuse equal to ds. Now, I have to add all the hypotenui (which I'm assuming is the plural of hypotenuse) to get the length of the curve. In my infinitesimal triangle:
    dy/dx or tan(a) = e^x (I'm calling the angle 'a')
    so, sec(a) = sqrt(1+e^(2x))
    so, the length of the hypotenuse is given by sec(a)*dx
    so, ds= sqrt(1+e^(2x))dx
    And, I integrated this to get the expression for the length. I used the same method to get the length of y=x^2 and y=log(x). For the circle y= sqrt(1-x^2), it gave a length of pi/2 for the length in the first quadrant. So, it should be working for other functions as well.
     
  9. Nov 5, 2016 #8

    PeroK

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    In general, ##ds = \sqrt{1 + f'(x)^2} dx##

    You can find a fuller analysis here:

    http://tutorial.math.lamar.edu/Classes/CalcII/ArcLength.aspx

    By the way, I would recommend Paul's online math notes as a great reference for all things calculus.
     
  10. Nov 5, 2016 #9
    This is the third time that I thought I had invented something new. But everything's been already done in maths.
     
  11. Nov 5, 2016 #10

    PeroK

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    This is only "Calc II", so you're a long way off the pace!
     
  12. Nov 5, 2016 #11
    I'm again getting the feeling that I had when I thought I invented formulas for area under the curve of polynomials. It feels nice that I would have been the inventor of these if I were, maybe, 200 years ago. I know only high school calculus. Is it 'calculus I' or is it even more basic stuff than that?
     
  13. Nov 5, 2016 #12

    Simon Bridge

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    If you are working this stuff out yourself though, you are doing well.
    Best use of your time may be to figure out what seems a reasonable way to go about a problem, then see how others do it, before hetting heavily into the actual process.
    Meantime, there are unsolved problems in maths, some with cash prizes.
    https://en.m.wikipedia.org/wiki/List_of_unsolved_problems_in_mathematics
     
  14. Nov 14, 2016 #13
    You never know when you might solve a famous unsolved math problem.

    But, it's far more likely that you will make progress on a problem under the guidance of a Ph.D. advisor. There are far more highly worthwhile unsolved math problems than just the famous ones on the list in Wikipedia. (All of these have been worked on by many people, often over many years, and so are automatically the least likely problems that someone can just learn about and solve in short order — though it is not impossible.)

    An advisor is likely to understand in great detail the landscape of their particular area of research, and be able to steer you to some relatively solvable yet important areas of research. Getting people started in math research is, of course, the purpose of a Ph.D. program in mathematics. So that is highly recommended.

    Meanwhile, don't hesitate to keep trying to come up with new and worthwhile things. Even if you may come up with something that someone else did before you, it can still be a valuable learning experience to go through that kind of discovery on your own.

    (However, the formula for the arclength of a differentiable curve, mentioned in #8, has been known for hundreds of years, as has the function f(x) = ex, so simply applying a very well known formula to a very well known function is very unlikely to be something nobody has done before.)
     
  15. Nov 17, 2016 #14
    I didn't even know that that the formula of arclength existed. I didn't know about it so I didn't just apply the arc length formula to e^x. My question was about the length of e^x, but that doesn't mean that I had the formula only for e^x. What I did was deriving the arc length formula without knowing it had been done before. I used the reasoning that I've written in #7. So, I just derived the arc length formula for any function f(x).
     
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