Find the arc length of the curve.

Question

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.

Answer 1

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}

Answer 2

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 $\sec^2 x = 1 + \tan^2 x$.

Answer 3

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.

Answer 4

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: $$L(c)(\varphi ) = \int_0^\varphi {\left\| {\dot c(t)} \right\|} {\kern 1pt} dt = \int_0^\varphi {\sqrt {{a^2}{{\tan }^2}\left( {\frac{t}{4}} \right){{\sec }^8}\left( {\frac{t}{4}} \right) + 1} } {\kern 1pt} dt$$

Answer 5

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 $$ \int_{-\pi}^{\pi} ds = 2\int_0^\pi \cos^{-5}(\theta/4) \,d\theta= 8\int_0^{\pi/4}\cos^{-5} t \, dt=12.543617$$

Answer 6

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.