I tried to make sure I understand uniform circular motion by deriving the formula for centripetal acceleration, $a_c = v^2/r$, rigorously from first principles. I think I did it correctly, but it seems quite complicated - can you show me better ways?
Given a particle moving uniformly in a circule of radius $r$ around $O$ with period $T$, I take it as given that it has tangential velocity at each point with the same constant magnitude. So let's say its velocity at the topmost point, broken down into Cartesian components, is $v_0=(v,0)$. Now some small amount of time $\Delta t$ passes, and the particle rotates by the angle $\alpha = \frac{2\pi}{T} \Delta t$. The magnitude of velocity remains $v$, and it's now pointed in the new tangential direction, so it's easy to see its components as $v_1=(v\cos\alpha, v\sin\alpha)$.
Now acceleration is the limit of $\frac{v_1-v_0}{\Delta t}$ when $\Delta t \to 0$. I just want the magnitude of acceleration which will be
$$\lim_{\Delta t \to 0}\frac{\sqrt{(v\cos\alpha - v)^2 + (v\sin\alpha)^2}}{\Delta t}$$
I can simplify this, by passing to $\alpha$ as the dependent variable ($\Delta t=\frac{T}{2\pi}\alpha$), to
$$v \frac{2\pi}{T} \lim_{\alpha \to 0}\frac{\sqrt{2-2\cos\alpha}}{\alpha} $$
To evaluate the limit, I bring the denominator under the square sign and break down $\cos\alpha$ by its Taylor series:
$$\lim_{\alpha \to 0}\sqrt{\frac{2-2\cos\alpha}{\alpha^2}} = \lim_{\alpha \to 0}\sqrt{\frac{2-2(1-\frac{\alpha^2}{2} + \frac{\alpha^4}{24} - \cdots}{\alpha^2}} = \lim_{\alpha \to 0}\sqrt{1 - \frac{\alpha^2}{12} + \cdots} = 1$$
So in the end the magnitude of acceleration is $v\frac{2\pi}{T}$. But since the particle traces the circumference of the circle, of length $2\pi r$, under constant speed $v$ in time $T$, we have $vT = 2\pi r$, so $\frac{2\pi}{T} = \frac{v}{r}$, and finally the acceleration is the familiar formula $$a_c=\frac{v^2}{r}$$
Did I do this right? And even if I did, was it necessary to do the Taylor series stuff to obtain the trigonometric limit? What are some ways to make this derivation easier, but still rigorous under the formal definition of acceleration in Cartesian coordinates?
You are making this much more difficult than it has to be. Hints:
$1).\ $Write $\vec r(t)=r\cos \theta(t)\vec i+r\sin\theta(t)\vec j$, note that $r$ is constant, differentiate once to obtain $\vec v(t)$, again to get $\vec a(t).$ You will use the chain rule to do this.
$2).\ $Now, with $\frac{d\theta}{dt}:=\omega, $ the angular velocity, you will get $\vec a(t)=-\omega^2\vec r(t).$
$3).\ $To finish, take the magnitude of both sides, and the formula you want will drop out as soon as you observe that since the speed is constant (why?), one has $\omega=\|\vec v(t)\|/r=v/r.$