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Hyperbolic functions

  1. Sep 21, 2007 #1
    The hyperbolic functions are defined as follows:

    coshz = e[tex]^{z}[/tex] + e[tex]^{-z}[/tex] /2

    sinhz = e[tex]^{z}[/tex] - e[tex]^{-z}[/tex] /2

    a.)Show that coshz = cos (iz). What is the corresponding relationship for sinhz?
    b.)What are the derivatives of coshz and sinhz? What about their integrals?
    c.)Show that cosh^2z - sin^2 =1
    d.)Show that the integral of dx/sqrt[1+x^2 = arcsin x.

    Hint : substitution x = sinhz.



    I'd LOVE starters on showing this, we're told to assume z is real.
    I get the idea that theres an imaginery aspect to hyperbolic functions, since coshz = cos(iz) ?
     
  2. jcsd
  3. Sep 21, 2007 #2

    learningphysics

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    depends on what you're allowed to start with.

    can you use [tex]e^{iz} = cos(z) + isin(z)[/tex]
     
  4. Sep 21, 2007 #3
    I'm not sure.
    Beforehand these values were sketched over a range of real values of z.
    I don't know if that answers whether 'were allowed' ...
     
  5. Sep 21, 2007 #4
    Correction: part a.) isnt necessary. My bad.
     
  6. Sep 21, 2007 #5
    I have really no idea how to derive hyperbolic functions.
    Is d/dx of cosh = -sinh?
     
  7. Sep 21, 2007 #6

    learningphysics

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    No. Get the derivative of [tex]\frac{e^z + e^{-z}}{2}[/tex] with respect to z.
     
  8. Sep 21, 2007 #7
    =2e^z + 2e^-z / 4

    ?
     
  9. Sep 21, 2007 #8

    learningphysics

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    check your work...
     
  10. Sep 21, 2007 #9
    is d/dx of e^z + e^-z the same thing or not?
     
  11. Sep 21, 2007 #10
    nm reading now, i know thats wrong
     
  12. Sep 21, 2007 #11
    is d/dx 0?
     
  13. Sep 21, 2007 #12
    orrr..

    e^z - e^-z / 4
    ?
     
  14. Sep 21, 2007 #13

    learningphysics

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    Show your steps...
     
  15. Sep 21, 2007 #14
    e[tex]^{z}[/tex] + e[tex]^{-z}[/tex] / 2

    d/dx e[tex]^{z}[/tex] = e[tex]^{z}[/tex] * d/dx (z)
    z=real number, d/dx = 1

    d/dx e[tex]^{-z}[/tex] = e[tex]^{-z}[/tex] * d/dx (-z)
    z=real number, d/dx = -1

    Denominator ^2 by quotient rule...

    =e[tex]^{z}[/tex] - e[tex]^{-z}[/tex] / 4

    ?
     
  16. Sep 21, 2007 #15

    learningphysics

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    just do d/dz...

    it's just (1/2)(e^z + e^-z)

    taking d/dz you get [tex]\frac{1}{2}(\frac{d}{dz}e^z + \frac{d}{dz}e^{-z})[/tex]

    so the answer is just (e^z - e^-z)/2

    if you do it using the quotient rule... you need to do derivative of the numerator by the denominator, minus the numerator*deriavative of the denominator divided by the denominator squared so...

    [tex]\frac{(e^z - e^{-z})2 - (e^z + e^{-z})(0)}{2^2}[/tex]

    and you get the same result.
     
  17. Sep 21, 2007 #16
    I dont know the rule of putting a half there...
     
  18. Sep 21, 2007 #17

    learningphysics

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    It's just taking out the constant.

    if z = A*y

    then taking the derivative of both sides with respect to x...

    dz/dx = A*(dy/dx)

    For example... the derivative of 5e^(2x) = 5*d/dx(e^(2x)) = 5*2e^(2x) = 10e^(2x)
     
  19. Sep 21, 2007 #18
    alrighty, i think i get it.
    basically what i get is that d/dx of cosh is sinh and vice versa right?
     
  20. Sep 21, 2007 #19
    I forget integrating quotients.. and e...
     
  21. Sep 21, 2007 #20

    learningphysics

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    Yeah.
     
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