# Why do we neglect $\Delta t^2(\frac{d\vec{r}}{dt}\frac{d\vec{\hat{r}}}{dt})$ at Taylor Expansion?

I'm just started to Ankara University Physics Department two weeks ago. I have missed my 2 hours of PHY105 course that is the last week Wednesdey. The subject that i missed was Derivatives of Vectors. I'm trying to fill the gap. I get the main idea about position vectors, the change of position over time of a particle, finding the velocity of a particle with derivation. But at the end of the concept, the book is used samething for the other expression of position vector ($$\vec r=r\hat{\vec{r}}$$). Let me show you.

**

At my book,

We can write $$\vec r(t)$$ as $$r(t)$$ magnitude and $$\hat{\vec{r}}(t)$$ unit vector form.

$$\vec r(t)=r(t)\hat{\vec{r}}(t)$$

$$\frac{d\vec r}{dt}=\frac{d}{dt}[r(t)\hat{\vec{r}}(t)]=\lim_{\Delta t\to 0}\frac{r(t+\Delta t)\hat{\vec{r}}(t+\Delta t)-r(t)\hat{\vec{r}}(t)}{\Delta t}$$

If we open the series with Taylor Expansion and take the first two terms, numerator of the fraction is,

$$[r(t)+\frac{dr}{dt}\Delta t][\hat{\vec{r}}(t)+\frac{d\hat{\vec{r}}}{dt}\Delta t]-r(t)\hat{\vec{r}}(t)$$

$$=\Delta t(\frac{dr}{dt}\hat{\vec{r}}+r\frac{d\hat{\vec{r}}}{dt})+{\Delta t}^2(\frac{dr}{dt}\frac{d\hat{\vec{r}}}{dt})$$

At here, if we neglect the second term when $$\Delta t\to 0$$,

$$\vec{V}=\frac{d\vec{r}}{dt}=\frac{d\vec{r}}{dt}\hat{\vec{r}}+r\frac{d\hat{\vec{r}}}{dt}$$

remains.

**

## Questions

1) Why do we use Taylor Expansion, what does it do?

2) When $$\Delta t\to 0$$ why do we only neglect the second term and why not the first term?

That is not a Taylor's series expansion. That is just simple multiplication into a quadratic, i.e. $$(x+a)(x+b)=x^2 + ax + bx + ab$$, but with the "$$ab$$" part subtracted out. Try applying that to your $$\left[ r(t) + \Delta t \frac{dr}{dt}\right]\left[ \hat {\vec r} (t) + \Delta t \frac{d\hat {\vec r}}{dt}\right] - r(t) \hat {\vec r} (t)$$.
2) When $$\Delta t \to 0$$ why do we only neglect the second term and why not the first term?
How fast does $${\Delta t}^2$$ shrink as $$\Delta t \to 0$$ compared to $${\Delta t}^2$$?
• Hey! Thank you for your answer! I know that multiplication but I mean that the part $r(t+\Delta t)\hat{\vec{r}}(t+\Delta t)$ of the numerator of the limit is expanded as $r(t+\Delta t)\hat{\vec{r}}(t+\Delta t)=(r(t)+\Delta t\dot{r}+\frac{1}{2!}(\Delta t)^2\ddot{r}+...)(\hat{\vec{r}}(t)+\Delta t\dot{\hat{\vec{r}}}+\frac{1}{2!}(\Delta t)^2\ddot{\hat{\vec{r}}}+...)$ neglect $(\Delta t)^2)$ as 0 when $\Delta t\to 0$. So, $=(r(t)+\Delta t\dot{\vec{r}})(\hat{\vec{r}}(t)+\Delta t\dot{\hat{\vec{r}}})=r(t)\hat{\vec{r}}(t)+r(t)\dot{\hat{\vec{r}}}\Delta t+\dot{r}\hat{\vec{r}}(t)\Delta t+$ – ICCQBE Sep 28 '19 at 19:41
• $...\dot{r}\dot{\hat{\vec{r}}}(\Delta t)^2=r(t)\hat{\vec{r}}+(r(t)\dot{\hat{\vec{r}}}+\hat{\vec{r}}(t)\dot{r})\Delta t \Rightarrow \vec{V}=\lim_{\Delta t\to 0}\frac{r(t)\hat{\vec{r}}+(r(t)\dot{\hat{\vec{r}}}+\hat{\vec{r}}(t)\dot{r})\Delta t}{\Delta t}\Rightarrow \vec{V}=\dot{\vec{r}}=r(t)\dot{\hat{\vec{r}}}+\dot{r}\hat{\vec{r}}(t)$. That is the Taylor Expansion used on my book. My question was about this. Why we use this and what does it do? Thanks! – ICCQBE Sep 28 '19 at 19:52
• The reason that any terms with ${\Delta t}^n, n > 1$ are ignored is because for any finite $x$ and $n > 1$, $\lim_{\Delta t \to 0} {\Delta t}^n / {\Delta t} = 0$ You should be able to see this pretty much by inspection if you remember that part of your calculus classes. – TimWescott Sep 29 '19 at 1:51
• Thank you. I get the idea about the $\Delta t$ but I could not understand why we use Taylor Expansion on my comments as you can see.. – ICCQBE Sep 29 '19 at 8:39