# I Integrating a velocity vector

1. Mar 5, 2017

### rabbed

Hi

(1/sqrt(4t²+1), 2t/sqrt(4t²+1)) gives a unit tangent to the curve y=x^2 at point (t,t^2).
Viewing the vector as velocity, shouldn't I be able to integrate it and get a parameterization for y=x^2?

2. Mar 5, 2017

### Orodruin

Staff Emeritus
If you try to integrate it with respect to $t$ you will not recover the curve since you have normalised the vector. What you have is
$$\frac{\frac{d\vec x}{dt}}{\left\lvert\frac{d\vec x}{dt}\right\rvert},$$
not $d\vec x/dt$.

3. Mar 5, 2017

### rabbed

Ah, okay.
So if I want to arc-length parameterize, I would need to convert "sqrt(4t^2+1)" from the expression sqrt(4t^2+1)*(cos(atan2(2t,1)), sin(atan2(2t,1))) into some change of the angle atan2(2t,1)..

4. Mar 5, 2017

### Orodruin

Staff Emeritus
If you want to parametrise with the arc length you should define a function s(t) such that $d\vec x/ds= (d\vec x/dt)(dt/ds)$ is a unit vector. This will give you a differential equation to solve.

5. Mar 5, 2017

### rabbed

Okay,
but to have a constant speed of 1, changing the velocity angle is what you would actually do at each point to stay on the curve, right?
And this would involve taking arccos and arcsin of some input r where it's possible that |r| > 1, so that the result can be imaginary?

6. Mar 10, 2017

### rabbed

Can you show me how you would set up and solve the differential equation in my case?