TITLE: Why does Spivak claim there's a definable tangent line through $(0,0)$ for the graph of the polar coordinates described by $f(\theta)=|\cos(2\theta)|$
QUESTION [0 upvotes]: There is a problem in the Chapter 12 Appendix of Spivak's 4th Ed. Calculus (Problem 6b) that asks the reader to consider the tangent lines of different graphs of polar coordinates. For the polar coordinates described by $r=|\cos(2\theta)|$, Spivak claims (in the solution manual) that the graph's tangent line of the polar coordinate $(0,\frac{\pi}{4})$ is equal to a a line at a $45 \deg$ angle from the horizontal axis that passes through $(0,0)$. However, I think that this must be a typo.
In $6a$, we demonstrated that:

...for the graph of $f$ in polar coordinates, the slope of the tangent line at the point with polar coordinates $(f(\theta),\theta)$ is \begin{align}\frac{f(\theta)\cos(\theta)+f'(\theta)\sin(\theta)}{-f(\theta)\sin(\theta)+f'(\theta)\cos(\theta)} \end{align}

Note that in order for this to be valid, the denominator cannot equal $0$. From the equation $f(\theta)=r=|\cos(2\theta)|$, if $\theta=\frac{\pi}{4}$, we see that $f(\theta)=0$, which means that our fraction can be rewritten as:
$$\frac{f'(\frac{\pi}{4})\cos(\frac{\pi}{4})}{-f'(\frac{\pi}{4})\sin(\frac{\pi}{4})}$$
At first glance, we would claim that this equals $\tan(\frac{\pi}{4})$...i.e. a slope of $1$ with origin $(0,0)$ corresponding to a "line at a $45 \deg$ angle from the horizontal axis". However, I am pretty certain that $f'(\frac{\pi}{4})$ does not exist.
If we consider the left and right limits of (and apply L'hopital's rule), we get:

$\displaystyle \lim_{\theta\to \frac{\pi}{4}^-}\frac{|\cos(2\theta)|-|\cos(\frac{\pi}{4})|}{\theta-\frac{\pi}{4}}=-2$

$\displaystyle \lim_{\theta\to \frac{\pi}{4}^+}\frac{|\cos(2\theta)|-|\cos(\frac{\pi}{4})|}{\theta-\frac{\pi}{4}}=2$


So $f'(\frac{\pi}{4}) \text{ DNE}$.
Interestingly, if we look at $g(\theta)=\cos(2\theta)$ (i.e. with no absolute value bars), even though the graphs of the polar coordinates specified by $g$ is the same as the graph generated by $f$, in this case, $f'(\frac{\pi}{4})=-2$.
Am I thinking about this correctly?

Edit: He does this again when handling the graph of the polar coordinates given by the expression $f(\theta)=r=|\cos(3\theta)|$ (where $\theta=\frac{\pi}{6}$)
...and again when handling the graph of the polar coordinates given by the expression: $r^2=2a^2\cos(2\theta)$ (in particular, when $\theta=\frac{\pi}{4}$). Given that Spivak interprets polar coordinates as being of the form $(f(\theta),\theta)$, if one graphs $r^2=2a^2\cos(2\theta)$ (treating it as a standard expression unrelated to polar coordinates), and then implicitly differentiates the expression treating $r$ as $f(\theta)$, we will see that no such derivative exists at $\theta=\frac{\pi}{4}$ (the slope would be $\infty$).

REPLY [1 votes]: It works in the limiting case. When $f(\theta)=0$, then the slope becomes (slightly abusing the notation) $$\frac{0+f'(\theta)\sin\theta}{0+f'(\theta)\cos\theta}=\tan\theta$$
As I've mentioned, this is a slight abuse, and assumes that we can simplify numerator and denominator by $f'(\theta)$ (which does not exists). But the limit of the slope exists both on the left and right of $\pi/4$.
{"set_name": "stack_exchange", "score": 0, "question_id": 4437677}