Поэтому находим
[m] ρ `=(\sqrt{2}e^{ φ })`=\sqrt{2}\cdot (e^{ φ })`=\sqrt{2}e^{ φ }[/m]
[m]( ρ `)^2=(\sqrt{2}e^{ φ })^2=2e^{2 φ}[/m]
[m]( ρ )^2=2e^{2 φ }[/m]
[m]( ρ `)^2+( ρ )^2=2e^{2 φ}+2e^{2 φ}=4e^{2 φ}[/m]
[m]\sqrt{( ρ `)^2+( ρ )^2}=\sqrt{4e^{2 φ}}=2e^{ φ }[/m]
[m] L= ∫_{0}^{\frac{π}{3}}2e^{ φ }d φ =2∫_{0}^{\frac{π}{3}}e^{ φ }d φ =2e^{ φ }|_{0}^{\frac{π}{3}} =2e^{ \frac{π }{3}}-2e^{0} =2e^{ \frac{π }{3}}-2 [/m]