Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 2 down vote accepted

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*}

I don't know how to solve this ODE, but you could always solve it numerically.

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.