Answer to Question #97482 in Differential Geometry | Topology for Warsi

Question

let r(t)= (e^kt cos t, e^kt sint) find the arc length of r starting at the point (1,0)

Answer 1

ANSWER The arc length is equal to $\frac { \sqrt { 2 } \left| k \right| }{ k } \left( { e }^{ kb }-1 \right)$

EXPLANATION. The legs L_C of the arc of the curve specified parametrically C=$\left\{ \left( x(t),y(t) \right) :\quad a\le t\le b \right\}$

calculated using the formula ${ L }_{ C }=\int _{ a }^{ b }{ \sqrt { { \left( x'(t) \right) }^{ 2 }+{ \left( y'(t) \right) }^{ 2 } } } dt$ . In the task , the point (1,0) corresponds to the

value of the parameter t=0. Hence , a=0.

$x(t){ =e }^{ kt }\cos { t } ,\quad x'(t)=k{ e }^{ kt }\cos { t } -{ e }^{ kt }\sin { t } ,\quad y(t)={ e }^{ kt }\sin { t } ,\quad y'(t)=k{ e }^{ kt }\sin { t } +{ e }^{ kt }\cos { t }$

${ \left( x'(t) \right) }^{ 2 }+{ \left( y'(t) \right) }^{ 2 }={ \left( k{ e }^{ kt }\cos { t } -{ e }^{ kt }\sin { t } \right) }^{ 2 }+{ \left( k{ e }^{ kt }\sin { t } +{ e }^{ kt }\cos { t } \right) }^{ 2 }$ =

$= { k }^{ 2 }{ e }^{ 2kt }\left( 2\cos ^{ 2 }{ t-2\cos { t } \sin { t } +2\cos { t } \sin { t } +2\sin ^{ 2 }{ t } } \right) =$ $=2{ k }^{ 2 }{ e }^{ 2kt }$ .

${ L }_{ C }=\int _{ 0 }^{ b }{ \sqrt { 2{ k }^{ 2 }{ e }^{ 2kt } } } dt=\sqrt { 2 } \left| k \right| \int _{ 0 }^{ b }{ { e }^{ kt } } dt=\frac { \sqrt { 2 } \left| k \right| }{ k } \left( { e }^{ kb }-1 \right)$ .