Find the arc length of the curve. How can I find the length of the loop in polar coordinates
$$r=\frac{a}{\cos^{4}( \frac{\varphi}{4})} $$ 
I do not even know where to start. Help please.
 A: Parameterising
\begin{align}
\varphi(t) &= t \\
r(t) &= \frac{a}{\cos^{4}( \frac{t}{4})} 
\end{align}
and transforming to Cartesian coordinates
$$
x = r \cos \phi = \frac{a}{\cos^4(\frac{t}{4})} \cos t \quad\quad
y = r \sin \phi = \frac{a}{\cos^4(\frac{t}{4})} \sin t
$$
The image shows a plot for $a = 2$ and $t \in [-5,5]$.

$t=0$ is the point $(a, 0)$ we need to parameterise to the left crossing at $(x(t), 0)$ for $t = \pm \pi$.
The arc lenght is
$$
ds^2 = dx^2 + dy^2 \quad \Rightarrow \quad
ds = 
\sqrt{
\left(\frac{dx}{d\varphi}\right)^2 + 
\left(\frac{dy}{d\varphi}\right)^2} d\varphi
$$
and the derivatives are
$$
\frac{dx}{d\varphi} = \frac{dr}{d\varphi} \cos \varphi - r \sin \varphi
\quad\quad
\frac{dy}{d\varphi} = \frac{dr}{d\varphi} \sin \varphi + r \cos \varphi
$$
so we get the differential arc lenght in polar coordinates
$$
\left(\frac{ds}{d\varphi}\right)^2 =
\left(\frac{dr}{d\varphi}\right)^2 + r^2
$$
and using the parameterisation for $r$ in terms of $\varphi$:
$$
r'(\varphi) 
= -4 \frac{a}{\cos^5(\frac{t}{4})}
\left(-\sin\left(\frac{t}{4}\right) \frac{1}{4}\right)
= a \frac{\sin(\frac{t}{4})}{\cos^5(\frac{t}{4})} 
$$
Returning to the differential arc length element $ds$ we get
\begin{align}
s'(\varphi)^2 &= 
r'(\varphi)^2 + r^2 \\
&=
\left(\frac{a\sin(\frac{\varphi}{4})}{\cos^5(\frac{\varphi}{4})}\right)^2 +
\left(\frac{a}{\cos^{4}( \frac{\varphi}{4})}\right)^2 \\
&=
\frac{a^2}{\cos^8(\frac{\varphi}{4})}
\left(\tan^2\left({\frac{\varphi}{4}}\right)+ 1\right) \\
&=
\frac{a^2}{\cos^8(\frac{\varphi}{4})}
\frac{1}{\cos^2\left({\frac{\varphi}{4}}\right)} \\
&=
\frac{a^2}{\cos^{10}(\frac{\varphi}{4})}
\end{align}
which turned out simpler than feared. So we can now give the arc length:
\begin{align}
s 
&= \int\limits_{-\pi}^\pi \frac{ds}{d\varphi} d\varphi \\
&= 2\int\limits_0^\pi \frac{ds}{d\varphi} d\varphi \\
&= 2a \int\limits_0^\pi \frac{d\varphi}{\cos^{5}(\frac{\varphi}{4})} \\
&= 8a \int\limits_0^{\pi/4} \frac{du}{\cos^{5}(u)} \\
&= 8a (F(\pi/4) - F(0))
\end{align}
This has a longish anti-derivative which Maxima computes to
$$
F(\varphi) =
\frac{3\,\ln\left( \sin\left( u\right) +1\right) }{16}-\frac{3\,\ln\left( \sin\left( u\right) -1\right) }{16}-\frac{3\,{\sin\left( u\right) }^{3}-5\,\sin\left( u\right) }{8\,{\sin\left( u\right) }^{4}-16\,{\sin\left( u\right) }^{2}+8}
+ C
$$
and
\begin{align}
s 
&= 
a \left(
\frac{3\,\ln\left( \frac{1}{\sqrt{2}}+1\right)}{2}-
\frac{3\,\ln\left( 1-\frac{1}{\sqrt{2}}\right)}{2}+
7\,\sqrt{2}
\right) \\
&=
a \left(
\frac{3}{2}
\ln\left( \frac{\sqrt{2}+1}{\sqrt{2}-1} \right) +
7\,\sqrt{2}
\right) \\
&=
a \left(
\frac{3}{2}
\ln\left((\sqrt{2}+1)^2 \right) +
7\,\sqrt{2}
\right) \\
&=
a \left(
3 \ln\left( \sqrt{2}+1 \right) +
7\,\sqrt{2}
\right) \\
&= 12.54361569767029 \, a
\end{align}
A: Hints: First, you must look at a graph of the curve and figure out what values of $\varphi$ (or $\theta$), say $a$ and $b$, form the endpoints of the closed (loop) portion of the curve. Once you've done that, you apply the arc length formula
\begin{equation*}
  \int_a^b \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta.
\end{equation*}
The integral is not too hard if you remember the identity $\sec^2 x = 1 + \tan^2 x$.
A: Your curve form a  loop fot $-\pi\le \theta \le \pi$.
The arc lenght in polar coordinates is: $ds=\sqrt{r^2+\left(\dfrac{dr}{d\theta}\right)^2} d\theta$.
Integrating for $-\pi \le\theta\le \pi$ ghives the leght of the loop.
A: For $$r(\varphi ) = \frac{a}{{{{\cos }^4}\left( {\frac{\varphi }{4}} \right)}}$$
you get a parametrization like so:
$$c(\varphi ) = \left( {\begin{array}{*{20}{c}}
  {r(\varphi )} \\ 
  \varphi  
\end{array}} \right)$$
It's velocity or tangent vector is given by
$$\dot c(\varphi ) = \left( {\begin{array}{*{20}{c}}
  {\dot r(\varphi )} \\ 
  \varphi  
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  {\dot r(\varphi )} \\ 
  1 
\end{array}} \right)$$
Calculate lenght for tangent vector:
$$\begin{gathered}
  {\left\| {\dot c(\varphi )} \right\|^2} = \dot c(\varphi ) \cdot \dot c(\varphi ) = \dot r{(\varphi )^2} + 1 \hfill \\
  \left\| {\dot c(\varphi )} \right\| = \sqrt {\dot r{{(\varphi )}^2} + 1}  \hfill \\ 
\end{gathered} $$
So you are ready to integrate:
$$L(c)(\varphi ) = \int_0^\varphi  {\left\| {\dot c(t)} \right\|} {\kern 1pt} dt = \int_0^\varphi  {\sqrt {{a^2}{{\tan }^2}\left( {\frac{t}{4}} \right){{\sec }^8}\left( {\frac{t}{4}} \right) + 1} } {\kern 1pt} dt$$
A: you can throw out the $a.$ i will work with $$r = \cos^{-4}(\theta/4), \frac{dr}{d\theta} = \cos^{-5}(\theta/4)\sin(\theta/4), ds = \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}=\cos^{-5}(\theta/4)$$ the length of the closed loop is $$ \int_{-\pi}^{\pi} ds = 2\int_0^\pi \cos^{-5}(\theta/4) \,d\theta= 8\int_0^{\pi/4}\cos^{-5} t \, dt=12.543617$$
A: After various responses of this case if you wish to work further along this class of polar curves I like to mention that these belong to McLauren polar curves.
$ r/a = \cos \theta + i \sin \theta. $ Raising to $n$th power we have by Euler theorem
$ (r/a)^n = \cos n\theta + i \sin n \theta $. Taking the real part,the set of McLauren's curves result.
$ r/a = ( \cos n\theta)^{1/n} $. $n$ can take integral, fractional , positive 
and negative values. 
We have here the particular $ n= -\frac14$ case.
(Examples are a straight line, circle through origin , Lemniscate of Bernoulli, rectangular hyperbola, exponentiated roses, some of which fall to origin and some towards radial asymptotes etc.)
Closed form arc length results can be obtained for a single branch/petal in terms of all $n$ values.
