Question #97482
let r(t)= (e^kt cos t, e^kt sint) find the arc length of r starting at the point (1,0)
1
Expert's answer
2019-10-31T10:18:02-0400

ANSWER The arc length is equal to 2kk(ekb1)\frac { \sqrt { 2 } \left| k \right| }{ k } \left( { e }^{ kb }-1 \right)


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

calculated using the formula LC=ab(x(t))2+(y(t))2dt{ 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)=ektcost,x(t)=kektcostektsint,y(t)=ektsint,y(t)=kektsint+ektcostx(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))2+(y(t))2=(kektcostektsint)2+(kektsint+ektcost)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 } =

=k2e2kt(2cos2t2costsint+2costsint+2sin2t)== { k }^{ 2 }{ e }^{ 2kt }\left( 2\cos ^{ 2 }{ t-2\cos { t } \sin { t } +2\cos { t } \sin { t } +2\sin ^{ 2 }{ t } } \right) = =2k2e2kt=2{ k }^{ 2 }{ e }^{ 2kt } .

LC=0b2k2e2ktdt=2k0bektdt=2kk(ekb1){ 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) .

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