Motion in two dimensions, in a plane can be studied by expressing position, velocity and acceleration as vectors in cartesian co-ordinates $A=A_{x} \hat{i}+A_{y} \hat{j},$ where $\hat{i}$ and $\hat{\jmath}$ are unit vector along $x$ and $y$ - directions, respectively and $A_{x}$ and $A_{y}$ are corresponding components of $A$. Motion can also be studied by expressing vectors in circular polar co-ordinates as $\overrightarrow A \, = \,{A_r}\widehat r\,\, + \,{A_\theta }\hat \theta $ where $\hat{r}=\frac{r}{r}=\cos \theta \hat{i}+\sin \theta \hat{\jmath}$ and $\hat{\theta}=-\sin \theta \hat{i}+\cos \theta \hat{j}$ are unit vectors along direction in which $\hat{r}$ and $\hat{\theta}$ are increasing.
$(a)$ Express ${\widehat {i\,}}$ and ${\widehat {j\,}}$ in terms of ${\widehat {r\,}}$ and ${\widehat {\theta }}$ .
$(b)$ Show that both $\widehat r$ and $\widehat \theta $ are unit vectors and are perpendicular to each other.
$(c)$ Show that $\frac{d}{{dr}}(\widehat r)\, = \,\omega \hat \theta \,$, where $\omega \, = \,\frac{{d\theta }}{{dt}}$ and $\frac{d}{{dt}}(\widehat \theta )\, = \, - \theta \widehat r\,$.
$(d)$ For a particle moving along a spiral given by $\overrightarrow r \, = \,a\theta \widehat r$, where $a = 1$ (unit), find dimensions of $a$.
$(e)$ Find velocity and acceleration in polar vector representation for particle moving along spiral described in $(d)$ above.
(a) Given, unit vector
$\hat{r}=\cos \theta \hat{i}+\sin \theta \hat{j}$
$\hat{\theta}=-\sin \theta \hat{i}+\cos \theta \hat{j}$
Multiplying Eq. (i) by $\sin \theta$ and Eq. (ii) with $\cos \theta$ and adding
$\hat{r} \sin \theta+\hat{\theta} \cos \theta=\sin \theta \cdot \cos \theta \hat{i}+\sin ^{2} \theta \hat{j}+\cos ^{2} \theta \hat{j}-\sin \theta \cdot \cos \theta \hat{i} \hat{j}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=\hat{j}$
$\Rightarrow \hat{r} \sin \theta+\hat{\theta} \cos \theta=\hat{j}$
$\text { By Eq. (i) } \times \cos \theta-\text { Eq. (ii) } \times \sin \theta$
$n(\hat{r} \cos \theta-\hat{\theta} \sin \theta)=\hat{i}$
$(b)$ $\hat{r} \theta=(\cos \theta \hat{i}+\sin \theta \hat{j}) \cdot(-\sin \theta \hat{i}+\cos \theta \hat{j})=-\cos \theta \cdot \sin \theta+\sin \theta \cdot \cos \theta=0$
$\Rightarrow \theta=90^{\circ}$ Angle between $\hat{r}$ and $\hat{\theta}$.
$(c)$ Given,
$\hat{r} =\cos \theta \hat{i}+\sin \theta \hat{j}$
$\frac{d \hat{r}}{d t} =\frac{d}{d t}(\cos \theta \hat{i}+\sin \theta \hat{j})=-\sin \theta \cdot \frac{d \theta}{d t} \hat{i}+\cos \theta \cdot \frac{d \theta}{d t} \hat{j}$
$=\omega[-\sin \theta \hat{i}+\cos \theta \hat{j}] \quad\left[\because \theta=\frac{d \theta}{d t}\right]$
$(d)$ Given, $r=a \theta \hat{r}$ here, writing dimensions
$[r]=[a][\theta][\hat{r}]$
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