What is the "polar coordinate function"? I'm reading Tapp's Differential Geometry of Curves and Surfaces, I have this problem:

$\quad\color{green}{\text{Exercise 1.10.}}$ Prove that the arc length, $L$, of the graph of the polar coordinate function $r(\theta)$, $\theta \in [a, b]$, is

What is the "polar coordinate function"? I tried a few guesses but nothing I thought seemed to make sense nor would give me the arclength function above. Using the definition of arclength, I tried to "reverse engineer" it and I got:
$$\left(\int_{a}^{\theta} r(x)\, dx,\, r(\theta) \right)$$
This seems to be the only function that would make sense here but I have no idea what $r(\theta)$ would be. This is — perhaps — the least worst thing I could think of.
 A: It means that in polar coordinates $(r,\theta)$, your curve is parametrized by $$[a,b]\ni \theta \mapsto (r(\theta),\theta) \in  [0,\infty)\times [0,2\pi).\tag{1}$$Relative to cartesian coordinates $(x,y)$, it means that the parametrization is $$[a,b] \ni \theta \mapsto (r(\theta)\cos\theta, r(\theta)\sin\theta) \in \Bbb R^2.\tag{2}$$Surely you can compute the arclength of $(2)$.
A: Arc length for a curve $C$ in the plane looks like
$$
L = \int_C \mathrm{d}s, 
$$
which you're probably more familiar with when the curve is parametrized, i.e.  the cartesian coordinates $(x, y)$ are expressed as functions of a parameter $t \in [a, b]$:
$$
L = \int_a^b \sqrt{x'(t)^2 + y'(t)^2} \, \mathrm{d}t. 
$$
This formula comes from the chain rule
$$
\mathrm{d}x = x'(t) \, \mathrm{d}t
\quad\text{and}\quad 
\mathrm{d}y = y'(t) \, \mathrm{d}t, 
$$
together with what is essentially the Pythagorean theorem for differentials,
\begin{align}
\mathrm{d}s^2 &= \mathrm{d}x^2 + \mathrm{d}y^2 \\
&= \bigl( x'(t) \, \mathrm{d}t \bigr)^2 
+ \bigl( y'(t) \, \mathrm{d}t \bigr)^2 \\
&= \bigl( x'(t)^2 + y'(t)^2 \bigr) \, \mathrm{d}t^2. 
\end{align}
By the way, in the special case that $y = f(x)$, where the parameter $t$ is extraneous since we can take $t=x$, this reduces to the familiar
$$
\mathrm{d}s^2 = \bigl( 1 + f'(x)^2 \bigr) \, \mathrm{d}x^2.
$$

Now, let's use polar coordinates, related by
$$
\left\{\begin{aligned}
x &= r \cos\theta \\
y &= r \sin\theta
\end{aligned}\right.
$$
First of all,
\begin{align} 
\mathrm{d}x &= \mathrm{d} \bigl( r \cos\theta \bigr) \\
&= \frac{\partial}{\partial r} \bigl( r \cos\theta \bigr) 
\, \mathrm{d}r
+  \frac{\partial}{\partial \theta} \bigl( r \cos\theta \bigr) 
\, \mathrm{d}\theta \\[2pt] 
&= \cos\theta \, \mathrm{d}r - r \sin\theta \, \mathrm{d}\theta
\end{align}
Similarly, we can calculate $\mathrm{d}y$ (try it yourself before revealing the spoiler)

\begin{align} \mathrm{d}y &= \mathrm{d} \bigl( r \sin\theta \bigr) \\ &= \frac{\partial}{\partial r} \bigl( r \sin\theta \bigr) \, \mathrm{d}r +  \frac{\partial}{\partial \theta} \bigl( r \sin\theta \bigr) \, \mathrm{d}\theta \\[2pt] &= \sin\theta \, \mathrm{d}r + r \cos\theta \, \mathrm{d}\theta \end{align}

Putting these together to find $\mathrm{d}s^2$ and simplifying with a bit of algebra (again, try it yourself),

\begin{align} \mathrm{d}s^2 &= \mathrm{d}x^2 + \mathrm{d}y^2 \\ &= \bigl( \cos\theta \, \mathrm{d}r - r \sin\theta \, \mathrm{d}\theta \bigr)^2 + \bigl( \sin\theta \, \mathrm{d}r + r \cos\theta \, \mathrm{d}\theta \bigr)^2 \\ &= \bigl( \cos^2\theta \, \mathrm{d}r^2 + r^2 \sin^2\theta \, \mathrm{d}\theta^2 - 2r\sin\theta\cos\theta \, \mathrm{d}r \, \mathrm{d}\theta \bigr) \\ &\qquad{} + \bigl( \sin^2\theta \, \mathrm{d}r^2 + r^2 \cos^2\theta \, \mathrm{d}\theta^2 + 2r\sin\theta\cos\theta \, \mathrm{d}r \, \mathrm{d}\theta \bigr) \\ &= \mathrm{d}r^2 + r^2 \, \mathrm{d}\theta^2 \end{align}

Now, if the radial coordinate is a function of the argument, i.e. $r = f(\theta)$, then $\mathrm{d}r = f'(\theta) \, \mathrm{d}\theta$ and

 \begin{align} \mathrm{d}s^2 &= \bigl( f'(\theta) \, \mathrm{d}\theta \bigr)^2 + f(\theta)^2  \, \mathrm{d}\theta^2 \\ &= \bigl( f'(\theta)^2 + f(\theta)^2 \bigr) \, \mathrm{d}\theta^2, \end{align}

which gives the arc length formula
$$
L = \int_a^b \sqrt{f'(\theta)^2 + f(\theta)^2} \, \mathrm{d}\theta 
\tag{$^\dagger$}
$$
for $\theta \in [a, b]$.
$^\dagger$This is the same as the formula you're looking for, but with the functional relationship made more explicit by using the symbol $f$. It can be confusing when $r$ is the name of the function and the output values at the same time, where you write things like $r = r(\theta)$. Tomayto. Tomahto. It's up to you how you write it.
