# Explain how $\Delta v_{\perp}=v\Delta\theta$

In The Feynman lectures, under the chapter entitled Vectors, Feynman writes:

My two intimately related questions are: 1) What does he mean by the magnitude of velocity? is he talking about the magnitude of $v_1$, $v_2$ or $\Delta v$?

2)I have struggled a lot on my own, trying to figure out where this equation came from, but I didn't succeed. Can someone explain to me where it did come from?

• I will say it like $\dfrac{\Delta v_\perp}{v_1} = \sin\Delta \theta \approx \Delta\theta$. So, $v_1 = v.$ Feynman here divided the component of the change in velocity in two parts so as to find the centripetal acceleration & tangential acceleration separately; it's quite simple! – user36790 Jul 11 '15 at 20:13

## 1 Answer

$v$ isn't referring to either $v_1$ or $v_2$, necessarily; $v_1$ is representing the vector before it moves, and $v_2$ is the vector after this movement. If we are working in polar coordinates (the reason he is using $v_\perp$ and $v_\parallel$), then let's suppose this small movement isn't changing the magnitude of the vector, it is just changing the direction (hence why we just care about $v_\perp$).

In very small rotations, we can approximate the distance moved in the rotation simply by the magnitude of the vector multiplied by the angle moved. Hence we get

$$\Delta v_\perp\approx v\Delta\theta$$

In other words, look at figure 11-8. Imagine that the vector $v_\perp$ is very small. Then you will notice that $v_1$ and $v_2$ will converge to the same length, we call this $v$. Then try to see if you can convince yourself that what I said before is true.

• I know that $\lim_{\theta\to 0}sin(\theta)=\theta$, and since the triangle bounded by $v_1$ , $v_2$ and $\Delta v_\perp$ is a right triangle, therefore $\frac{\Delta v_\perp}{v_1}=$ $\lim_{\theta\to 0}sin(\theta)=\theta$. is this what you mean? – Omar Nagib Jul 11 '15 at 20:21
• Yes that's a great way to think about it! – Alex Mathers Jul 11 '15 at 20:24