# Constant velocity of a sine function

I am defining the location of an object based on the sine function. The position of the object at s seconds along the x-axis is defined as x=s and its position along the y-axis is defined as y=sin(x). For example, when one second has passed x=1 and y=sin(1). I think that this works fine except for the fact that I want to have the object moving at a constant velocity. How would I adjust x=s so that the object will move at a constant velocity?

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If I understand correctly, the position of your object at time $t$ is $p(t) = (x(t), \sin(x(t))$. The velocity of this object at time $t$ is $p'(t) = (x'(t),\cos(x(t))x'(t))$. What you want is \begin{align*} \| p'(t)\|^2 &= x'(t)^2 + (\cos(x(t)) x'(t))^2 \\ &= x'(t)^2 (1 + \cos^2 (x(t))) \\ &= 1. \end{align*}
That is exactly what I meant. It has been a while since I have dealt with parametric equations and I didn't actually realize that was what I was trying to do. I should be able to find what $x(t)$ should be equal to from what you just said. I will give my result in the next comment. – russjohnson09 Nov 3 '12 at 21:28
So, Wolfram Alpha gives $x(t) = InverseFunction[EllipticE, 1, 2]((c_1-t)/sqrt(2), 1/2)$ as the solution where $c_1$ is an arbitrary constant. I am thinking there is not a simpler solution unfortunately. I honestly am not sure what this solution means. – russjohnson09 Nov 3 '12 at 22:07
I am somewhat confused by your question. If you want to find a function $g(x)$ so that if $v(x)=f(g(x))$, then $v'(x)=c$ where $c\in \mathbb{R}$, then apply the chain rule so that $v'(x)=f'(g(x))g'(x)$. So, $f'(g(x))g'(x)=\sin (g(x)) g'(x)$, then you can try to solve $c=\sin(g(x)) g'(x)$ where $g'(x)=\frac {c} {\sin(g(x))}$. But I am not sure this is what you are looking for.