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 $\overrightarrow{\theta }$ 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 $\overrightarrow{r}$ and the other is $\overrightarrow{\theta }$. The position vector is defined as $\overrightarrow{r}=r(\cos \theta \hat{i}+\sin \theta \hat{j})$ and the vector $\overrightarrow{\theta }$ is defined as $\overrightarrow{\theta }={\displaystyle \frac{d\overrightarrow{r}}{d\theta }}$. Therefore, the vector $\overrightarrow{\theta }$ can be obtained by differentiating the position vector with respect to $\theta$. And for determining the unit vector for $\overrightarrow{\theta }$, we need to divide the obtained vector $\overrightarrow{\theta }$ 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
$\Rightarrow \overrightarrow{r}=r(\cos \theta \hat{i}+\sin \theta \hat{j})........\left(i\right)$
Now, we also know that the vector $\overrightarrow{\theta }$ is defined for the same point as
$\Rightarrow \overrightarrow{\theta }={\displaystyle \frac{d\overrightarrow{r}}{d\theta }}$
On substituting the equation (i) in the above vector equation, we get
$\begin{array}{rl}& \Rightarrow \overrightarrow{\theta }={\displaystyle \frac{d\left[r(\cos \theta \hat{i}+\sin \theta \hat{j})\right]}{d\theta }}\\ & \Rightarrow \overrightarrow{\theta }=r{\displaystyle \frac{d\left[(\cos \theta \hat{i}+\sin \theta \hat{j})\right]}{d\theta }}\end{array}$
Separating the $\hat{i}$ and $\hat{j}$ terms, we get
$$\begin{array}{rl}& \Rightarrow \overrightarrow{\theta }=r\{{\displaystyle \frac{d\left[(\cos \theta \hat{i})\right]}{d\theta }}+{\displaystyle \frac{d\left[(\sin \theta \hat{j})\right]}{d\theta }}\}\\ & \Rightarrow \overrightarrow{\theta }=r({\displaystyle \frac{d(\cos \theta )}{d\theta }}\hat{i}+{\displaystyle \frac{d(\sin \theta )}{d\theta }}\hat{j})\\ & \Rightarrow \overrightarrow{\theta }=r(-\sin \theta \hat{i}+\cos \theta \hat{j})\end{array}$$
Clearly, the magnitude of the vector $\overrightarrow{\theta }$ 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 $\overrightarrow{\theta }$ 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 $\overrightarrow{r}$ and the vector $\overrightarrow{\theta }$ which is given by $\overrightarrow{\theta }={\displaystyle \frac{d\overrightarrow{r}}{d\theta }}$. Also, we must be careful regarding the signs of the derivatives of the trigonometric functions.