Reparameterizing a curve in path length parameter

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In summary, the conversation discusses the parameterization of an inverted cycloid and the process of parameterizing the curve in its natural parameter. It is mentioned that the modulus of the squared velocity is used in the integral and that the choice of parameter ##t_0## can be arbitrary, but may result in different intervals for the parameter. It is also noted that ##t_0 = 0## can be a reasonable choice.
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Marioweee
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Homework Statement
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Relevant Equations
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Given the parameterization of an inverted cycloid:
$$x(t)=r(t-\sin t)$$
$$y(t)=r(1+\cos t)$$
where $$t \in [0, 2\pi]$$.
I am asked to parameterize the curve in its natural parameter. To do it:
$$s=\int_{t_0}^{t} ||\vec{x}'(t*)||dt*$$
The modulus of the squared velocity is:
$$||\vec{x}'(t*)||=2r\sin(t/2)$$
Therefore, the integral is:
$$s=\int_{t_0}^{t} 2r\sin(t/2)dt*=-4r\cos(t/2)|_{t_0}^{t}=-4r(cos(t/2)-cos(t_0/2))$$
My doubt is, I can take any arbitrary value for the parameter $$t_0$$ for example $$t_0=\pi$$ which would simplify the expression or does it have to be $$t_0=0$$ since the parameter $$t$$ starts on 0.
Thank you very much for all the help.
 
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Both will be parametrizations of the curve in a parameter that is path length along the curve. You will get a different interval for the parameter depending on your choice though as they will be path lengths from different points on the curve.

The ##t_0 = 0## choice does not seem half bad to me either to be honest. Note that ##\cos(0) - \cos(t/2) = 1 - \cos(t/2) = 2 \frac{1-\cos(t/2)}2 = 2 \sin^2(t/4)##.
 
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Related to Reparameterizing a curve in path length parameter

1. What is reparameterization of a curve?

Reparameterization of a curve is the process of changing the parameterization of a curve, which is the way the points on the curve are defined in terms of a parameter. This can involve changing the parameter itself, such as from time to distance, or changing the range of the parameter, such as from 0 to 1 to -1 to 1.

2. Why is reparameterization of a curve important?

Reparameterization of a curve is important because it can simplify the representation of a curve and make it easier to work with. It can also make it easier to compare different curves, as they can be reparameterized to have the same parameter values. Additionally, reparameterization can help with curve fitting and interpolation.

3. How is reparameterization of a curve done?

Reparameterization of a curve can be done in various ways, depending on the specific needs and goals. One method is to use a change of variables, where the original parameter is replaced with a new one. Another method is to use a reparametrization function, which maps the original parameter to a new one. Some software programs also have built-in tools for reparameterizing curves.

4. What are the benefits of reparameterization of a curve?

Reparameterization of a curve can have several benefits, such as making the curve easier to work with, simplifying its representation, and allowing for easier comparison with other curves. It can also help with curve fitting and interpolation, as well as improving the accuracy and precision of calculations involving the curve.

5. Are there any limitations or drawbacks to reparameterization of a curve?

While reparameterization of a curve can have many benefits, there are also some limitations and drawbacks. For example, the reparameterization process can sometimes introduce errors or distortions in the curve. Additionally, not all curves can be easily reparameterized, and some may lose important characteristics or features in the process.

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