Question about integration with inverse trigonometric functions

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I'm self-studying Calculus and would like to ask some doubts about the following question:

Homework Statement


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.

Homework Equations



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.
 

Answers and Replies

  • #2
ehild
Homework Helper
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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.

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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