# Polar coordinates: $dr/dx$. Why am I getting two answers for this? [closed]

I am currently working through some past exercises in polar coordinates $$(r, \theta)$$ and want to take the derivative of $$r$$ with respect to $$x$$, but I have hit a bit of a brick wall:

a) $$x = r\cos(\theta)$$

and

b) $$r = \sqrt{x^2 +y^2}$$,

right? How come then if I solve a) for $$r$$ and take $$d/dx$$ I get $$\sec(\theta)$$ but if I take b) $$d/dx$$ I get $$\cos(\theta)$$ ??? Can anybody help me out here, I have probably made some glaringly silly mistake that I am a bit too frustrated to see right now.

Sorry for not formatting the equations, I'm rather new to this forum and maths in general, so I would also really appreciate an answer that doesn't presume much more experience than Calc 1 or 2. Thanks in advance! :)

• what do you mean by "I solve a) for r and take d/dx " ?
– Surb
Oct 6 at 16:09
• I mean to say x = rcos(theta) -> r = x/cos(theta), and then I take the derivative of r with respect to x. My bad if I wasn't clear enough about that. Oct 6 at 16:15
• Can you update your question to show your work? Oct 6 at 16:21
• $\cos\theta = \frac{x}{\sqrt{x^2+y^2}}$ So the way you are finding $\partial_x r$ using $x = r \cos\theta$ is not correct. Oct 6 at 16:32

The issue here, as I understand your question, is that you are forgetting that $$\theta$$ also depends on $$x$$ via $$\theta = \arctan(y/x)$$ for $$x > 0$$ (a similar argument will work for other cases). Here, then, we have $$\frac{d\theta}{dx} = \frac{1}{1+\left(\frac{y}{x}\right)^2}\cdot \frac{-y}{x^2} = -\frac{y}{x^2 + y^2} = -\frac{r\sin(\theta)}{r^2} = -\frac{\sin(\theta)}{r}.$$ So, differentiating $$r = \frac{x}{\cos(\theta)}$$ gives us \begin{align*}\frac{dr}{dx} &= \frac{\cos(\theta) + x\sin(\theta)\frac{d\theta}{dx}}{\cos^{2}(x)}\\ &= \frac{\cos(\theta) - x\sin(\theta)\frac{\sin(\theta)}{r}}{\cos^{2}(x)}\\ &= \frac{\cos(\theta) - x\sin(\theta)\frac{\sin(\theta)}{r}}{\cos^{2}(x)}\\ &= \frac{\cos(\theta) - r\cos(\theta)\sin(\theta)\frac{\sin(\theta)}{r}}{\cos^{2}(x)}\\ &=\frac{\cos(\theta)(1 - \sin^{2}(\theta))}{\cos^{2}(\theta)}\\ &=\frac{\cos(\theta)\cos^{2}(\theta)}{\cos^{2}(\theta)}\\ &=\cos(\theta). \end{align*} That matches with what you got in the second approach.