Question about integration with inverse trigonometric functions

In summary, the conversation discusses a differential equation for harmonic simple motion and solving for the velocity and position functions in terms of time. The solution is obtained using integration and justifying the choice of a²k² as the arbitrary integration constant. The ± is eliminated by taking the cosine of both sides of the equation.
  • #1
pc2-brazil
205
3
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.
 
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  • #2
pc2-brazil said:
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
 

1. What is integration with inverse trigonometric functions?

Integration with inverse trigonometric functions is a method used in calculus to find the antiderivatives of functions involving inverse trigonometric functions, such as arcsine, arccosine, and arctangent.

2. Why is integration with inverse trigonometric functions important?

Integration with inverse trigonometric functions is important because it allows us to solve a wide range of problems in physics, engineering, and other fields where inverse trigonometric functions are commonly used. It also helps us to better understand the relationships between different functions and their derivatives.

3. What are some common integration rules for inverse trigonometric functions?

Some common integration rules for inverse trigonometric functions include:

  • The integral of arcsine is x times the arcsine of x plus the square root of 1 minus x squared, plus a constant.
  • The integral of arccosine is x times the arccosine of x minus the square root of 1 minus x squared, plus a constant.
  • The integral of arctangent is x times the arctangent of x minus the natural log of the absolute value of 1 plus x squared, plus a constant.

4. How do you use integration with inverse trigonometric functions to solve problems?

To use integration with inverse trigonometric functions to solve problems, you first need to identify the function that needs to be integrated. Then, you can use the appropriate integration rules to find the antiderivative of the function. Finally, you can evaluate the integral using the limits of integration to find the exact solution.

5. Are there any limitations to using integration with inverse trigonometric functions?

Yes, there are some limitations to using integration with inverse trigonometric functions. For example, there are certain functions that cannot be integrated using this method and require more advanced techniques. Also, when dealing with complex functions, integrating with inverse trigonometric functions may not always provide an exact solution and may require numerical methods instead.

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