Question

For the unit vector $\hat{\theta }$ , geometrically show that $\hat{\theta }=-\sin \theta \hat{i}+\cos \theta \hat{j}$. Essentially, converting from Cartesian to polar, how would I determine the unit vector for $\vec{\theta }$ in terms of $\theta $, $\hat{i}$, and $\hat{j}$?

Answer 1

Hint: With respect to the position of a point in the Cartesian plane, two vectors are defined one is the position vector $\vec{r}$ and the other is $\vec{\theta }$. The position vector is defined as $\vec{r}=r\left( \cos \theta \hat{i}+\sin \theta \hat{j} \right)$ and the vector $\vec{\theta }$ is defined as $\vec{\theta }=\dfrac{d\vec{r}}{d\theta }$. Therefore, the vector $\vec{\theta }$ can be obtained by differentiating the position vector with respect to $\theta $. And for determining the unit vector for $\vec{\theta }$, we need to divide the obtained vector $\vec{\theta }$ by its magnitude, which is equal to r.

Complete step by step solution:
We know that the position vector of a point in the Cartesian plane is defined as
$\Rightarrow \vec{r}=r\left( \cos \theta \hat{i}+\sin \theta \hat{j} \right)........\left( i \right)$
Now, we also know that the vector $\vec{\theta }$ is defined for the same point as
$\Rightarrow \vec{\theta }=\dfrac{d\vec{r}}{d\theta }$
On substituting the equation (i) in the above vector equation, we get
$\begin{align}
  & \Rightarrow \vec{\theta }=\dfrac{d\left[ r\left( \cos \theta \hat{i}+\sin \theta \hat{j} \right) \right]}{d\theta } \\
 & \Rightarrow \vec{\theta }=r\dfrac{d\left[ \left( \cos \theta \hat{i}+\sin \theta \hat{j} \right) \right]}{d\theta } \\
\end{align}$
Separating the $\hat{i}$ and $\hat{j}$ terms, we get
\[\begin{align}
  & \Rightarrow \vec{\theta }=r\left\{ \dfrac{d\left[ \left( \cos \theta \hat{i} \right) \right]}{d\theta }+\dfrac{d\left[ \left( \sin \theta \hat{j} \right) \right]}{d\theta } \right\} \\
 & \Rightarrow \vec{\theta }=r\left( \dfrac{d\left( \cos \theta \right)}{d\theta }\hat{i}+\dfrac{d\left( \sin \theta \right)}{d\theta }\hat{j} \right) \\
 & \Rightarrow \vec{\theta }=r\left( -\sin \theta \hat{i}+\cos \theta \hat{j} \right) \\
\end{align}\]
Clearly, the magnitude of the vector $\vec{\theta }$ is equal to $r$. Therefore, for getting the unit vector $\hat{\theta }$, we divide the above equation by $r$.
\[\Rightarrow \hat{\theta }=-\sin \theta \hat{i}+\cos \theta \hat{j}\]

Hence, we have determined the unit vector for $\vec{\theta }$ in terms of $\theta $, $\hat{i}$, and $\hat{j}$ as \[\hat{\theta }=-\sin \theta \hat{i}+\cos \theta \hat{j}\]

Note: For solving these types of questions, we must remember the important relation between the position vector $\vec{r}$ and the vector $\vec{\theta }$ which is given by $\vec{\theta }=\dfrac{d\vec{r}}{d\theta }$. Also, we must be careful regarding the signs of the derivatives of the trigonometric functions.