ANSWER The arc length is equal to 2 ∣ k ∣ k ( e k b − 1 ) \frac { \sqrt { 2 } \left| k \right| }{ k } \left( { e }^{ kb }-1 \right) k 2 ∣ k ∣ ( e kb − 1 )
EXPLANATION. The legs LC of the arc of the curve specified parametrically C={ ( x ( t ) , y ( t ) ) : a ≤ t ≤ b } \left\{ \left( x(t),y(t) \right) :\quad a\le t\le b \right\} { ( x ( t ) , y ( t ) ) : a ≤ t ≤ b }
calculated using the formula L C = ∫ a b ( x ′ ( t ) ) 2 + ( y ′ ( t ) ) 2 d t { L }_{ C }=\int _{ a }^{ b }{ \sqrt { { \left( x'(t) \right) }^{ 2 }+{ \left( y'(t) \right) }^{ 2 } } } dt L C = ∫ a b ( x ′ ( t ) ) 2 + ( y ′ ( t ) ) 2 d t
value of the parameter t=0. Hence , a=0.
x ( t ) = e k t cos t , x ′ ( t ) = k e k t cos t − e k t sin t , y ( t ) = e k t sin t , y ′ ( t ) = k e k t sin t + e k t cos t x(t){ =e }^{ kt }\cos { t } ,\quad x'(t)=k{ e }^{ kt }\cos { t } -{ e }^{ kt }\sin { t } ,\quad y(t)={ e }^{ kt }\sin { t } ,\quad y'(t)=k{ e }^{ kt }\sin { t } +{ e }^{ kt }\cos { t } x ( t ) = e k t cos t , x ′ ( t ) = k e k t cos t − e k t sin t , y ( t ) = e k t sin t , y ′ ( t ) = k e k t sin t + e k t cos t
( x ′ ( t ) ) 2 + ( y ′ ( t ) ) 2 = ( k e k t cos t − e k t sin t ) 2 + ( k e k t sin t + e k t cos t ) 2 { \left( x'(t) \right) }^{ 2 }+{ \left( y'(t) \right) }^{ 2 }={ \left( k{ e }^{ kt }\cos { t } -{ e }^{ kt }\sin { t } \right) }^{ 2 }+{ \left( k{ e }^{ kt }\sin { t } +{ e }^{ kt }\cos { t } \right) }^{ 2 } ( x ′ ( t ) ) 2 + ( y ′ ( t ) ) 2 = ( k e k t cos t − e k t sin t ) 2 + ( k e k t sin t + e k t cos t ) 2
= k 2 e 2 k t ( 2 cos 2 t − 2 cos t sin t + 2 cos t sin t + 2 sin 2 t ) = = { k }^{ 2 }{ e }^{ 2kt }\left( 2\cos ^{ 2 }{ t-2\cos { t } \sin { t } +2\cos { t } \sin { t } +2\sin ^{ 2 }{ t } } \right) = = k 2 e 2 k t ( 2 cos 2 t − 2 cos t sin t + 2 cos t sin t + 2 sin 2 t ) = = 2 k 2 e 2 k t =2{ k }^{ 2 }{ e }^{ 2kt } = 2 k 2 e 2 k t
L C = ∫ 0 b 2 k 2 e 2 k t d t = 2 ∣ k ∣ ∫ 0 b e k t d t = 2 ∣ k ∣ k ( e k b − 1 ) { L }_{ C }=\int _{ 0 }^{ b }{ \sqrt { 2{ k }^{ 2 }{ e }^{ 2kt } } } dt=\sqrt { 2 } \left| k \right| \int _{ 0 }^{ b }{ { e }^{ kt } } dt=\frac { \sqrt { 2 } \left| k \right| }{ k } \left( { e }^{ kb }-1 \right) L C = ∫ 0 b 2 k 2 e 2 k t d t = 2 ∣ k ∣ ∫ 0 b e k t d t = k 2 ∣ k ∣ ( e kb − 1 )
#339914 on Dec 2023