# Calc-based conceptual problem with centripetal motion

Note: This is more of a math question than a physics question, but I'm posting it here since it's in the context of physics.

I've been thinking about the classic example of a ball attached to a string and moving at constant speed in a circle. The acceleration is v^2/r and always facing in the direction of the string. I've derived this result in the past before by breaking the acceleration into its x and y components and integrating to find v. In these coordinates the result is intuitive to me and makes sense.

However something bothers me about thinking of the problem using vectors in polar coordinates. At any given time, there is a velocity vector tangential to the circle, and the acceleration vector normal to it. At some infinitesimal moment later, there is an infinitesimally small velocity vector pointing the same direction as the acceleration, and it adds with the normal vector to create a new velocity vector. If a = v^/r, this new vector will be tangential to the next bit of the circle; this is why the object moves in a circle. However, the thing that bothers me is, that when you add these two vectors, shouldn't the resultant velocity vector be bigger than the original velocity vector, even if only by an infinitesimal amount? And in that case, wouldn't all the infinitesimal errors eventually add up, making the velocity vector no longer constant?

The only thing I could think of as to why this thinking is wrong is because when you add the vectors, you have to use the Pythagorean Theorem to find the new vector (as the vectors are normal), and that involves the square of dt, which does not integrate into any finite number. Is this reasoning correct?

If it is, I'd like to know to what extent thinking about infinitesimal vectors being added is relevant to the question whether time is continuous or discrete. In the case of centripetal motion, If time was discrete, would the magnitude of velocity vector change in a noticeable way after a long enough time?

Thanks.