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Question about integration with inverse trigonometric functions

  1. Dec 13, 2011 #1
    I'm self-studying Calculus and would like to ask some doubts about the following question:

    1. The problem statement, all variables and given/known data
    If, in t seconds, s is the oriented distance of the particle from the origin and v is the velocity of the particle, then a differential equation for harmonic simple motion is:
    [tex]\frac{ds}{dt}=-k^2s[/tex]
    where k2 is a proportionality constant.
    Since [itex]\frac{dv}{dt}=\frac{dv}{ds}\frac{ds}{dt}=v\frac{dv}{ds}[/itex],
    [tex]v\frac{dv}{ds}=-k^2s[/tex]
    a) Obtain that [itex]v = \pm\sqrt{a^2-s^2}[/itex]. Note: Take a²k² as the arbitrary integration constant and justify this choice.
    b) Taking t = 0 in the instant that v = 0 (s = a), obtain that s = a cos kt.

    2. Relevant equations

    3. The attempt at a solution
    a)
    [tex]\int vdv=\int -k^2sds[/tex]
    [tex]\frac{v^2}{2}=-\frac{k^2s^2}{2}+\text{constant}[/tex]
    [tex]v^2=-k^2s^2+C[/tex]
    [tex]v=\pm\sqrt{C-k^2s^2}[/tex]
    [tex]v=\pm\sqrt{a^2k^2-k^2s^2}[/tex]
    Using C = a²k²:
    [tex]v=\pm k\sqrt{a^2-s^2}[/tex]
    I'm not sure how to justify the choice of C as a²k². Is it because the velocity will be 0 for s = a (the amplitude of the motion)?
    b)
    [tex]\frac{ds}{dt}=\pm k\sqrt{a^2-s^2}[/tex]
    [tex]\frac{ds}{\sqrt{a^2-s^2}}=\pm k dt[/tex]
    [itex]\arcsin\frac{s}{a} = \pm kt + \bar{c}[/itex] where [itex]a > 0[/itex]
    [tex]\frac{\pi}{2} - \arccos\frac{s}{a} = \pm kt + \bar{k}[/tex]
    [tex]-\arccos\frac{s}{a} = \pm kt + \bar{k} - \frac{\pi}{2}[/tex]
    Replacing [itex]\bar{k} - \frac{\pi}{2}[/itex] by [itex]\bar{c}[/itex]:
    [tex]-\arccos\frac{s}{a} = \pm kt + \bar{c}[/tex]
    [tex]\arccos\frac{s}{a} = \mp kt - \bar{c}[/tex]
    When t = 0, s = a, so:
    [tex]\arccos\frac{a}{a} = -\bar{c}[/tex]
    [tex]\arccos 1 = -\bar{c}[/tex]
    [tex]\bar{c}=0[/tex]
    Then:
    [tex]\arccos\frac{s}{a} = \pm kt[/tex]
    Now the problem is that I'm not sure how to get rid of the ±.
    I think that ± is eliminated here because, in the way that the arc-cosine function is defined, its image is from 0 to π (therefore, its image must be positive). Then:
    [tex]\frac{s}{a} = \cos kt[/tex]
    [tex]s = a\cos kt[/tex]

    Thank you in advance.
     
  2. jcsd
  3. Dec 13, 2011 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Cosine is an even function, cos(x)=cos(-x). When you take the cosine of both sides of [itex]\arccos\frac{s}{a} = \pm kt[/itex] it gives the single result [itex]\frac{s}{a} = \cos kt[/itex].


    ehild
     
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