Circular Motion. Intuition on vectors as a physics interpretation I am a maths undergraduate. I  decided to  start studying some basic physics trying to be rigorous with the math and uderstand the interpretations of some of the mathematical constructs and see how they are modeled to explain physics. Vectors and vector spaces are all well defined in math.
So the first interpretation that I stumble upon is seeing vectors as "arrows" with length and direction and that these arrows follow the rules of vectors (that was done also in math calculus textbooks which I think it might be confusing if you are trying to be rigorous).
At  first it was ok. But now on chapter 4 (Halliday Resnick physics) on circular motion, I found that it is an accelerating motion with constant velocity which i find it very counter intuitive since if you are accelerating means you have to be speeding up, now this is a result of the the vector interpretation giving vectors directions and the direction of the velocity is changing so i have acceleration. Still the acceleration is the rate of change of the velocity with time. And if you told me acceleration is constant I would integrate to find the velocity but that doesnt seem to work on circular motion. Why is that? Trying to understand  it so that it doesnt seem counter intuitive to me. It is like if you are turning something makes you loose speed so in order to keep it constant you have to accelerate every time.Is that the case? but if it is, I want a more mathematical way to see it. Sometimes it feels like these "arrows" are not so good to explain things. Sorry if the question is not appropriate for the site.

(Halliday has not mentioned any forces so far and has explained things only using basic definitions as rate of change for velocity and acceleration and has done a mathematical trick to prove the acceleration of circular motion with similar triangles and some limits.)
 A: I find no trouble in thinking of velocity and acceleration vectors as arrows. First some definitions:


*

*Velocity is a vector. Speed is it's magnitude. 

*Acceleration is a vector. It's magnitude has no new name. 
We agree that acceleration is present if there is a change in velocity:
$$\vec a= d \vec v /dt$$
That is, any change. So if magnitude (speed) and/or direction is changed, then acceleration happened. This is better seen from a components point of view, if the vectors in the expression are split into their x and y components:
$$a_x= d v_x/dt$$
$$a_y= d v_y/dt$$
Which envelopes the expression but still only contains magnitudes. No acceleration in the x direction gives no change in x velocity etc. 


*

*Now, if the velocity vector points North, and the object accelerates North as well, then it is natural that the velocity magnitude (the Speed) is increased. 

*Is acceleration south, then the speed will decrease (the vector is shortened). 

*Is acceleration pointing West or East, exactly Perpendicular, then nothing is pulling the velocity vector forwards or backwards, so it isn't increasing or decreasing. It is only turning or rotating. So the object's direction is rotated and in the next instant it has moved to a location slightly to the side. 
If in this new instant, we turn the acceleration vector slightly as well, so that it again is perpendicular, the same happens: no change in the speed (magnitude of velocity), it only turns a bit more.
If this is repeated and done continuously, the path will be perfectly circular and at no point did we have any parallel acceleration component, so speed is not changed anywhere. 
A: 
I want a more mathematical way to see it. 

Acceleration is defined as the rate of change of velocity with time. Velocity, being vector, can change just by changing direction keeping the magnitude constant. In a circular motion if the angular frequency $\omega$ is constant, then the magnitude of the velocity i.e., speed is constant but the velocity changes due to change in direction of the velocity vector.
Similarly the acceleration being vector can change keeping its magnitude constant. Expressing in circular polar coordinates the equation of motion are 
$$\vec{v} = \dot{r}\hat{r} + r\omega \hat{\theta}$$
$$\vec{a} = \left( \ddot{r} - r^2\omega\right)\hat{r} + \left( 2\dot{r}\omega + r\dot{\omega} \right)\hat{\theta}$$
As you can clearly see if the motion is uniform$(\dot{\omega} = 0)$ circular $(\dot{r} = 0)$ Then the expression for velocity and acceleration respectively reduce to 
$$\vec{v} = r\omega \hat{\theta}$$ $$\vec{a} = -r^2\omega \hat{r}$$ both of whose magnitudes are constant (both $r$ and $\omega$ constant) but the vector is changing because $\hat{\theta}$ and $\hat{r}$ are changing. 
It is clear from the expression of velocity and accleration that they are perpendicular because $\hat{\theta}$ and $\hat{r}$ are always perpendicular. So the component of acceleration in the direction of velocity is zero the speed doesn't change but the velocity changes due to radial acceleration. So accleration is not necessarily speeding up or slowing down, it may be changing direction too, which is what happens in circular motion.
