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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.

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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.

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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?

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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 ${\Delta t}^2$?

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  • $\begingroup$ 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, $=(r(t)+\Delta t\dot{\vec{r}})(\hat{\vec{r}}(t)+\Delta t\dot{\hat{\vec{r}}})=r(t)\hat{\vec{r}}(t)+r(t)\dot{\hat{\vec{r}}}\Delta t+\dot{r}\hat{\vec{r}}(t)\Delta t+$ $\endgroup$
    – M. Çağlar TUFAN
    Commented Sep 28, 2019 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! $\endgroup$
    – M. Çağlar TUFAN
    Commented Sep 28, 2019 at 19:52
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    $\begingroup$ 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
    Commented Sep 29, 2019 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.. $\endgroup$
    – M. Çağlar TUFAN
    Commented Sep 29, 2019 at 8:39

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