What is the exact length of the spiraling polar curve #r=5e^(2theta)# from #0# to #2pi#?

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Answer 1 · 2018-05-19T20:00:01

#=(5sqrt(5))/2 ( e ^(4 pi ) - 1)#

#bb r ( theta) = 5 e^ ( 2 theta) \ bb hat r#

Arc length:

#s = int_C dot s \ dt qquad = int_C sqrt(bb v * bb v ) \ dt qquad triangle#

#bb v ( theta) = d/(dt) (5 e^ ( 2 theta) \ bb hat r)#

#= 10 e^ ( 2 theta) dot theta \ bb hat r + 5 e^ ( 2 theta) d/(dt) (\ bb hat r) qquad square#

#d/(dt) (\ bb hat r)= d/(dt) ((cos theta),(sin theta)) #

#((- sin theta),(cos theta)) dot theta = bb hat theta dot theta#

So #square# is:

#bb v ( theta) = 10 e^ ( 2 theta) dot theta \ bb hat r + 5 e^ ( 2 theta) dot theta \ bb hat theta#

And #triangle# becomes:

#= int_C sqrt((10 e^ ( 2 theta) dot theta)^2 + (5 e^ ( 2 theta) dot theta \ )^2 ) \ dt#

#= int_C \ e^(2theta) sqrt( 10^2 + 5^2 ) qquad \ dot theta \ dt#

#=5sqrt(5) int_0^(2 pi) e ^(2 theta) qquad \ d theta#

#=(5sqrt(5))/2 ( e ^(4 pi ) - 1)#