1

This is more a mathematical question, but comes from physics during the derivation of centripetal acceleration.

We resolve the velocity vector at an arbitrary point up the circle from the initial position into two components, one parallel to the initial velocity and one perpendicular.

\begin{align*} \parallel:\quad v\cos\Delta\theta\\ \perp:\quad v\sin\Delta\theta \end{align*}

Since $a = \frac{dv}{dt}$, we subtract the corresponding components to find the difference in velocity at an arbitrary $\Delta\theta$ and divide by the corresponding arbitrary $\Delta t$, and then consider what happens as $\Delta t \to 0$ in the limit to find the differential.

The components parallel to the initial velocity are straightforward. As $\Delta t \to 0$, so does $\Delta\theta$. Since $\cos0 = 1$, both numerator and denominator in the fraction tend to $0$, so the limit intuitively evaluates to $0$.

$$\lim_{\Delta t \to 0}\left[\frac{v\cos\Delta\theta - v}{\Delta t}\right] = 0$$

However, I cannot get it with the the perpendicular components. The corresponding limit is:

$$\lim_{\Delta t \to 0}\left[\frac{v\sin\Delta\theta - 0}{\Delta t}\right] = \text{?}$$

My reasoning is that since $\sin0 = 0$, the numerator tends to $0$, and the denominator tends to $0$, so the fraction tends to $0$. Can someone please point out where the error is in this reasoning?

The textbook, however, makes a claim that as $\Delta t$ approaches $0$, $\sin\Delta\theta$ would approach $\Delta\theta$ - why would that be?

It then goes on to conclude that the limit evaluates to $v\frac{\Delta\theta}{\Delta t}$, which is a well-known and correct result. But I cannot understand the evaluation above and it is troubling me.

Mihail
  • 245
  • The acceleration written in tangential and normal components can be written in general form as $\vec{a}=\frac{dv}{dt}\vec{e}t+\frac{v^2}{\rho}\vec{e}_n$. If you write it in polar coordinates, the components parallel (radial) and perpendicular (transverse) of the acceleration are: $\vec{a}=(\ddot{r}-r\dot{\theta}^2)\vec{e}_r+(r\ddot{\theta}+2\dot{r}\dot{\theta}\vec{e}\theta)$. That is not what you wrote. Are you sure your results are correct? – Thales Sep 11 '19 at 12:01
  • @Thales I am sure in asmuch as I took it from a textbook, here. – Mihail Sep 11 '19 at 12:24
  • You may want to look at this question and answers https://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1 to understand why $\sin x$ can be approximated as $x$ when $x$ tends to zero. – Vasili Sep 11 '19 at 12:32
  • @Mihael alright,I see what it was done there. For the particular case of an object moving in a cirular path, $\dot{r}=\ddot{r}=0$, which simplifies to $\vec{a}=-r\dot{\theta}^2\vec{e}r+r\ddot{\theta}\vec{e}\theta$. Using the notation in the book, $\dot{\theta}=\omega$, and considering $\omega=cte$, ie, $\ddot{\theta}=0$, $\vec{a}=-r\omega^2\vec{e}r+0\cdot\vec{e}\theta=-r\omega^2\vec{e}_r$. Also, if you write $v=r\omega$, then $a=-\frac{v^2}{r}\vec{e}_r$. Do you want to prove it using limits or other derivations are welcomed? (use differentiation rules to show how to obtain those equations) – Thales Sep 11 '19 at 12:51
  • @Thales I am failing to understand the evaluation of the limit of the change in velocity in the perpendicular direction. Going to look at the link provided above in another comment to try to understand it. – Mihail Sep 11 '19 at 12:57
  • @Vasya I read through that post, but still can't see how that result can be generalized to the limit I have above. – Mihail Sep 11 '19 at 13:04
  • The source is wrong in the last statement, the limit of $\frac{v\delta θ}{\delta t}$ is $v\frac{dθ}{dt}$, the limit of a difference quotient, so it exists, is the differential quotient. In this case the difference is negligible, as the function under consideration, $θ(t)=θ_0+\omega t$, is linear. – Lutz Lehmann Sep 11 '19 at 13:08

1 Answers1

1

For small $Δθ$ trending to zero, you have $$ \sinΔθ=Δθ-\frac{(Δθ)^3}6+... $$ which justifies the small-angle approximation $\sinΔθ\approx Δθ$.

Then $$ \frac{v\sin(Δθ)}{Δt}=v\frac{Δθ}{Δt}\left(1-\frac{(Δθ)^2}6+...\right) $$ and using $Δθ=\omega Δt$ gives the last factor as $1+O(Δt^2)$, which leads to the claim.

Lutz Lehmann
  • 126,666
  • Is approximation the same as the evaluation at an absolute limit? – Mihail Sep 11 '19 at 11:51
  • What exactly is an "absolute limit"? I assume that $Δθ=θ(t+Δt)-θ(t)$. Then you could also employ what Dr. Sonnhard Graubner wanted to say, that $\lim_{Δθ\to 0}\frac{\sinΔθ}{Δθ}=\cos(0)=1$ and apply the product law of the limes. – Lutz Lehmann Sep 11 '19 at 12:02
  • I am sorry I am a relative novice at math. Could you spell the solution out please? – Mihail Sep 11 '19 at 12:16
  • I do not know what you mean with "solution". You use at some point $Δθ=O(Δt)$ to see that the difference to the limit goes to zero. – Lutz Lehmann Sep 11 '19 at 12:25
  • Yes, by my own reasoning the limit goes to zero, but it must be incorrect, because the textbook says the limit goes to $v\frac{\Delta\theta}{\Delta t}$. Where is the error in my reasoning? – Mihail Sep 11 '19 at 12:32
  • As in all difference quotients, both parts of the fraction tend to zero. That the limit exists is the definition of differentiability. The derivative is in general not zero. – Lutz Lehmann Sep 11 '19 at 13:09
  • I think I found where I made a mistake which was confusing me. The instantaneous acceleration is found not by taking $\lim_{\Delta t \to 0}$, but instead to $\lim_{\Delta t \to dx}$, where $dx$ is the theoretically infinitessimal smallest non-zero value. The use of Maclaurin expansions is then justified. – Mihail Sep 11 '19 at 14:56
  • 1
    No, $\frac{dθ}{dt}$ in full is the symbol for the limit $Δt\to 0$ in the difference quotient. Even if it looks like a fraction, as symbolic picture it is not a fraction. – Lutz Lehmann Sep 11 '19 at 15:08