$dr$ in polar co-ordinates

Question

I realise $$dr = dx\hat{i} +dy\hat{j}$$

I'm trying to work out $dr$ in polar co-ordinates.

At first thought $$dx = drcos(\theta)$$

This is of course wrong as I realised

$$ dx = drcos(\theta +d\theta)$$

I've tried using the cos addition rule

$$ dx = drcos(\theta)cos(d\theta)- drsin(\theta)sin(d\theta)$$

but unsure how to recover the result $$ dx = drcos(\theta)-rsin(\theta)d\theta$$

Have I made a mistake in a previous part?

Answer 1

In general, to find the differential for a product, use the product rule: $$\frac{d(xy)}{dz}=\frac{dx}{dz} y+\frac{dy}{dz}x$$So $$d(xy)=y\,dx+x\,dy$$So in this case, $$\begin{align}d(r\cos\theta)&=\cos\theta\,dr+r\,d(\cos\theta)\\&=\cos\theta\,dr+r\,\frac{d\cos\theta}{d\theta}\,d\theta\\&=\cos\theta\,dr-r\sin\theta\,d\theta\end{align}$$

Answer 2

Assume $x=r\cos\theta$ and $y=r\sin\theta$. Then, $dx=dr\cos\theta-r\sin\theta d\theta$ and $dy=dr\sin\theta+r\cos\theta d\theta$

Answer 3

I assume that what you want is $dr$ in cartesian coordinates since $dr$ in polar coordinates is just $dr$

You have 2 coordinate systems $(x,y)$ and $(r,θ)$. You can relate this systems by the transition function $Φ(r,θ)=(r\cosθ,r\sinθ)=(x,y)$ and $Φ^{-1}(x,y)=(x^2+y^2,\tan^{-1}(\frac{y}{x}))=(r,θ)$.

This means that $dr=d(x^2+y^2)=\frac{\partial(x^2+y^2)}{\partial x}dx+ \frac{\partial(x^2+y^2)}{\partial y}dy=2xdx+2ydy$. For the sake of understanding am going to write all the relations:

• $dy=d(r\sinθ)=(\sinθ)dr+(r\cosθ)dθ$
• $dx=d(r\cosθ)=(\cosθ)dr+(-r\sinθ)dθ$
• $dθ=d(\tan^{-1}(\frac{y}{x}))=(\frac{-y}{x^2+y^2})dx+(\frac{x}{x^2+y^2})dy$

I hope this is helpful.