For part a and b the answers are fig.(a) and fig(b) respectively. Can anyone explain me the graphs in more details? Like for a why is there lines in graph on the 2nd and 3rd quadrants and why is the vector going outwards from the origin? Similarly for b


1 Answer 1


The trick is to use polar coordinates. In polar coordinates:

\begin{gather*} x = r\cos\theta\\ y = r\sin\theta\\ \vec{i} = \hat{\vec{r}}\cos\theta - \hat{\vec{\theta}}\sin\theta\\ \vec{j} = \hat{\vec{r}}\sin\theta + \hat{\vec{\theta}}\cos\theta \end{gather*}

Where $\hat{\vec{r}}$ and $\hat{\vec{\theta}}$ are unit vectors in the $r$ and $\theta$ directions respectively.

So now you have, for the first part:

\begin{equation*} \begin{aligned} \vec{V}(x, y) &= x\vec{i} + y\vec{j}\\ &= x\left(\hat{\vec{r}}\cos\theta - \hat{\vec{\theta}}\sin\theta\right) + y\left(\vec{j} = \hat{\vec{r}}\sin\theta + \hat{\vec{\theta}}\cos\theta\right)\\ &= \hat{\vec{r}} r\cos^2\theta - \hat{\vec{\theta}} r \cos\theta\sin\theta + \hat{\vec{r}} r \sin^2\theta + \hat{\vec{\theta}} r \cos\theta\sin\theta\\ &= r\hat{\vec{r}}\left(\sin^2\theta + \cos^2\theta\right)\\ &= r\hat{\vec{r}} \end{aligned} \end{equation*}

In other words $\vec{V}$ is a vector field which points in the $\hat{\vec{r}}$ direction and whose magnitude is $r$. It is then easy to draw it.

Similarly for (b).

In general if you have some vector field which looks like $f(x)\vec{i} + f(y)\vec{j}$,then polar coordinates will almost certainly help see what is going on.


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