I have a small doubt regarding the resolution of forces and vectors.
Suppose, we have our standard cartesian coordinate system, with unit vectors $\hat{i}$ and $\hat{j}$. Now we have defined polar coordinates with the unit vectors $\hat{r}$ and $\hat{\theta}$.
Suppose, we have a vector $\vec{r} = r\hat{r}$. We can write it as $r cos\theta \hat{i} + r sin\theta \hat{j} $. Thus we have managed to find the components of the vector along the $x$ and $y$ directions. Thus, if we had a force, we could find its components along $x$ and $y$ in the same fashion.
However, what if the force or the vector was initially along the $x$ direction, say $\vec{x}=x\hat{i}$. What would be the component of this vector along $\hat{r}$ ? Suppose, this vector represented Force. What would be the component of this force along $\hat{r}$
The first case is like, the wind is blowing towards the north east, finds its component along north and east. However, the second case is like, what if the wind was blowing east in the first place ? How will I find its component along the north east. Do I do the exact same thing and multiply by $cos\theta$ or should I multiply by $sec\theta$ for this second case ?
What I know is, if the original force was acting along $\hat{r}$, and we were given it's x-component, we would have multiplied with $sec\theta$ to get back the original force. However, what if the original force was blowing along the x direction, then what ? Since $r=xsec\theta$, should we multiply $sec\theta$ ?
Any help would be highly appreciated.
Best Answer
What you have identified is a very common situation. In physics we rarely deal with just vectors. More common is to deal with vector fields, i.e. a vector, possibly different one, in each point in space.
A good way to resolve this is to note that $\mathbf{\hat{x}}=\boldsymbol{\nabla}x$ (check it).
So if your force is $\mathbf{F}=x\mathbf{\hat{x}}$, then simply do:
$$ \mathbf{F}=x\mathbf{\hat{x}}=x\,\boldsymbol{\nabla}x=\left(r\cos\theta\right)\boldsymbol{\nabla}\left(r\cos\theta\right)=\left(r\cos\theta\right)\cdot\left(\mathbf{\hat{r}}\partial_r + \frac{1}{r}\boldsymbol{\hat{\theta}}\partial_\theta \right)\left(r\cos\theta\right)=\left(r\cos\theta\right)\cdot\left(\cos\theta\,\mathbf{\hat{r}}-\sin\theta \,\boldsymbol{\hat{\theta}} \right) $$
Where I have evaluated 2d gradient in cylindrical coordinates
So the component of force along $\mathbf{\hat{r}}$ direction in case of $\mathbf{F}=x\mathbf{\hat{x}}$ will be:
$$ \mathbf{\hat{r}}.\mathbf{F}=r\cos^2\theta $$
ADDENDUM
Of course if you "know" that $\mathbf{\hat{x}}=\cos\theta \,\mathbf{\hat{r}}-\sin\theta \, \boldsymbol{\hat{\theta}}$ you can bypass the manipulations with the gradient and deriviatives and simply substitute $x\to r\cos\theta$ and $\mathbf{\hat{x}}\to \dots$