1
$\begingroup$

Given a polar curve such as $r=a^\theta$, we could figue out the corresponding rectangular form by using the fact that $r=\sqrt{x^2+y^2}$ and $\theta=\arctan(\frac{y}{x})$. We would then get the following equation: $$r=a^\theta$$ $$\ln{r}=\theta\ln{a}$$ $$\ln{\sqrt{x^2+y^2}}=(\arctan{\frac{y}{x}})(\ln{a})$$ $$\log_a{\sqrt{x^2+y^2}}=\arctan{\frac{y}{x}}$$ $$\tan{(\log_a{\sqrt{x^2+y^2}})}=\frac{y}{x}$$

However, when this is graphed, https://www.desmos.com/calculator/2bvhqe8ruy, the two equations do not look the same as the rectangular form seems to also show the polar curve rotated $\frac{\pi}{2}$ as well. Could someone please explain why this happens or how I can add some sort of restriction to prevent this from happening. Thanks in advance.

$\endgroup$
  • $\begingroup$ From the graph they seem to be equals! $\endgroup$ – gimusi Dec 8 '17 at 5:34
1
$\begingroup$

Part of the problem is that the "fact" that "$\color{red}{\theta=\arctan\left(\frac{y}{x}\right)}$" is false. (This is an extremely common mistake, unfortunately.) And that is because $\tan$ is not a one-to-one function, so $\arctan$ is only its partial (or restricted) inverse, only within the interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. Recall that by definition,

$\arctan a=\theta$ means that $\tan\theta=a$ and $\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.

Quick example: the point $(x,y)=(-1,-1)$ lies in the third quadrant, and it's easy to see that its polar angle is $\theta=\frac{5\pi}{4}$. However, $\arctan\left(\frac{y}{x}\right)=\arctan\left(\frac{-1}{-1}\right)=\arctan(1)=\frac{\pi}{4}\neq\frac{5\pi}{4}$, i.e. in this case $\color{green}{\theta\neq\arctan\left(\frac{y}{x}\right)}$. Since the range of $\arctan$ lies only in the fourth and first quadrants, $\arctan$ can't possibly yield the correct value of $\theta$ for points in the second and third quadrants.

Now to your actual question. Another way to rephrase this issue with $\tan$ not being one-to-one: when you apply it, $\tan$ of different values may become the same. In other words: $\alpha=\beta$ implies $\tan\alpha=\tan\beta$; but $\tan\alpha=\tan\beta$ does not imply $\alpha=\beta$. For example, $\tan\frac{5\pi}{4}=\tan\frac{\pi}{4}$ even though $\frac{5\pi}{4}\neq\frac{\pi}{4}$. That's exactly what happened in your work: the last step turned previously unequal quantities into equal ones, thus creating a lot of new (extraneous) solutions that were not solutions before.

Once again: $$\tan{(\log_a{\sqrt{x^2+y^2}})}=\frac{y}{x} \tag{1}$$ does not imply that $$\log_a{\sqrt{x^2+y^2}}=\arctan{\frac{y}{x}}. \tag{2}$$ Many more points satisfy equation $(1)$ than equation $(2)$, which is why the graphs are not the same. For example both $(1,0)$ and $(a^{\pi},0)$ satisfy $(1)$; but only $(1,0)$ satisfies $(2)$, while $(a^{\pi},0)$ doesn't. On the other hand, since $(2)$ implies $(1)$, you see that the graph of $(2)$ is part of the graph of $(1)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.