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

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 "\\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="\\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)" .

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