In Uniform Circular Motion, why does the normal accelaration not increase the magnitude of velocity? This very simple question was posed by a high-school student in the class.
Consider a particle going in a uniform circular motion (uniform implies that the speed is constant). We know that there is a centripetal acceleration at every point in the motion, which changes the direction of the velocity. 
Now, consider an infinitesimal period of time dt for this motion at some given instant. Let's say that the initial velocity is $v_1\vec{i}$ in X-direction. Then, the acceleration will add up a component $a.dt\vec{j}$ to this velocity. Adding these 2 quantities should give us our final velocity $\vec{v_2}$. See this diagram:

Now, it is clear from this reasoning that $|\vec{v_2}| > v_1$. So, the magnitude of the velocity should increase!! 
Now, I know that this is WRONG. I just don't know where the fault lies. Clearly, there is some fault in the mathematical modeling done here. 
I am thinking that there must be some fault in the causation modeling. Acceleration is not CAUSING the change in velocity. It is an EFFECT of the change in velocity. That is one probably answer. But I'd like you to shed more light on this conundrum. How should one go about explaining this to a high-school kid? 
 A: After time $dt$, the particle will have moved a distance $v_1 dt$ so the two velocity vectors should not start at the same point. If you do a proper drawing and take the limit $dt\to 0$, as you should, if you are talking about infinitesimal time, you see that $v_2 \to v_1$
In your sketch (Pythagoras):
$$v_2=v_1\sqrt{1+\left(\frac{adt}{v_1} \right)^2}=v_1\left(1+... dt^2+...dt^4+...\right)$$ is quadratic in $dt$. If you had the acceleration in direction of the velocity, you'd get the correction linear in $dt$. That is why in the limit $dt\to 0$,
$$\frac{dv}{dt}=\frac{v_2-v_1}{dt}$$ disappears for your case, but does not disappear for acceleration parallel to the velocity.
A: The problem here is that you take some small but still not zero time and think that acceleration is constant during this time. This is not true. So what you get is an approximation. Because this is an approximation you do not get correct answer "magnitude of velocity remains constant". You get an approximated answer "magnitude of velocity slightly increases".
What you should do is to estimate how much the speed increases according to your approximated calculations. Then you should increase the accuracy of your approximation: let's say split the period of time by 1000, you would have 1000 steps now, and you should account that during each step acceleration is different. You would see, the she shorter are your steps the closer your approximated answer is to the correct one. Difference of speed would be closer and closer to 0.
