What is the exact length of the spiraling polar curve $r = 5 e^{2 θ}$ $r=5e^(2theta)$ from $0$ $0$ to $2 π$ $2pi$ ?

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Answer 1 · 2018-05-19T20:00:01

$= \frac{5 \sqrt{5}}{2} (e^{4 π} - 1)$ $=(5sqrt(5))/2 ( e ^(4 pi ) - 1)$

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

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:

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

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

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

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

What is the exact length of the spiraling polar curve r=5e^(2theta)r=5e2θr=5e^(2theta) from 000 to 2pi2π2pi?