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 $\Delta t\to 0$,





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?


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

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 ${\Delta t}^2$?

  • $\begingroup$ 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, $=(r(t)+\Delta t\dot{\vec{r}})(\hat{\vec{r}}(t)+\Delta t\dot{\hat{\vec{r}}})=r(t)\hat{\vec{r}}(t)+r(t)\dot{\hat{\vec{r}}}\Delta t+\dot{r}\hat{\vec{r}}(t)\Delta t+$ $\endgroup$ – ICCQBE Sep 28 '19 at 19:41
  • $\begingroup$ $...\dot{r}\dot{\hat{\vec{r}}}(\Delta t)^2=r(t)\hat{\vec{r}}+(r(t)\dot{\hat{\vec{r}}}+\hat{\vec{r}}(t)\dot{r})\Delta t \Rightarrow \vec{V}=\lim_{\Delta t\to 0}\frac{r(t)\hat{\vec{r}}+(r(t)\dot{\hat{\vec{r}}}+\hat{\vec{r}}(t)\dot{r})\Delta t}{\Delta t}\Rightarrow \vec{V}=\dot{\vec{r}}=r(t)\dot{\hat{\vec{r}}}+\dot{r}\hat{\vec{r}}(t)$. That is the Taylor Expansion used on my book. My question was about this. Why we use this and what does it do? Thanks! $\endgroup$ – ICCQBE Sep 28 '19 at 19:52
  • 1
    $\begingroup$ 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. $\endgroup$ – TimWescott Sep 29 '19 at 1:51
  • $\begingroup$ 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.. $\endgroup$ – ICCQBE Sep 29 '19 at 8:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.