- #1

Marioweee

- 18

- 5

- Homework Statement
- See below

- Relevant Equations
- See below

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.

$$x(t)=r(t-\sin t)$$

$$y(t)=r(1+\cos t)$$

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.