Duljina luka ravninske krivulje

Krivulje $\displaystyle y^{3}=x^{2}$ i $y=\sqrt{2-x}$ se sijeku u točkama $A\left( 1,1\right)$ i $B\left( -1,1\right)$ .

**Slika 2.5:** Duljina luka a)
$\begin{figure}\begin{center} \epsfig{file=duljinaluka2.eps, width=9.6cm}\end{center}\end{figure}$

Ukupnu duljinu luka računat ćemo kao (vidi sliku 2.5)

$\displaystyle l=2\left( l_{1}+l_{2}\right) ,$

koristeći formulu za duljinu luka krivulje

[M2, poglavlje 2.6.2.1], pa je

$\displaystyle l_{1}$	$\displaystyle =\int\limits_{0}^{1}\sqrt{1+\frac{9}{4}y}dy=\frac{1}{2} \int\limi... ...t\limits_{0}^{1}\left( 4+9y\right) ^{\frac{1}{2}}\frac{1}{9}d\left( 4+9y\right)$
	$\displaystyle =\frac{1}{18}\cdot \frac{2}{3}\left( 4+9y\right) ^{\frac{3}{2}}\bigg\vert _{0}^{1}=\frac{1}{27}\left( 13\sqrt{13}-8\right)$

$\displaystyle l_{2}$	$\displaystyle =\int\limits_{0}^{1}\sqrt{1+\frac{x^{2}}{2-x^{2}}} dx=\sqrt{2} \int\limits_{0}^{1}\frac{ dx}{\sqrt{2-x^{2}}}=$
	$\displaystyle =\sqrt{2}\arcsin \frac{x}{\sqrt{2}}\bigg\vert_{0}^{1}=\frac{\pi \sqrt{2}}{4 },$

iz čega slijedi

$\displaystyle l=2\left[ \frac{1}{27}\left( 13\sqrt{13}-8\right) +\frac{\pi \sqrt{2}}{4} \right] \approx 5.1.$

$\begin{displaymath} \left\{ \begin{array}{c} x=\frac{1}{3}t^{3}-t \\ y=t^{2}+2 \end{array}\right. \end{displaymath}$

je $\displaystyle\overset{\cdot }{x}\left( t\right) =t^{2}-1$ i $\overset{\cdot }{y}\left( t\right) =2t$ pa iz formule za duljinu luka krivulje zadane u polarnim koordinatama

[M2, poglavlje 2.6.2.2] slijedi

$\displaystyle l$	$\displaystyle =\int\limits_{0}^{3}\sqrt{\left( t^{2}-1\right) ^{2}+4t^{2}} dt... ...-2t^{2}+1+4t^{2}} dt=\int\limits_{0}^{3} \sqrt{\left( t^{2}+1\right) ^{2}} dt$
	$\displaystyle =\int\limits_{0}^{3}\left( t^{2}+1\right) dt=\left( \frac{t^{3}}{3} +t\right) \bigg\vert_{0}^{3}=12.$