What is the arclength of the polar curve $f (θ) = sin (3 θ) - 4 cot 6 θ$ $f(theta) = sin(3theta)-4cot6theta$ over $θ \in [0, \frac{π}{4}]$ $theta in [0,pi/4]$ ?

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What is the arclength of the polar curve $f (θ) = sin (3 θ) - 4 cot 6 θ$ $f(theta) = sin(3theta)-4cot6theta$ over $θ \in [0, \frac{π}{4}]$ $theta in [0,pi/4]$ ?

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Answer 1 · 2016-11-21T18:35:09

The arclength is infinite.

Explanation:

From the reference Arc Length with Polar Coordinates

$L = \int_{α}^{β} \sqrt{{(r (θ))}^{2} + {(\frac{d r (θ)}{d θ})}^{2}} d θ$ $L = int_alpha^beta sqrt{(r(theta))^2 + ((dr(theta))/(d theta))^2}d theta$

Given: $r (θ) = sin (3 θ) - 4 cot (6 θ), α = 0 and β = \frac{π}{4}$ $r(theta) = sin(3theta) - 4cot(6theta), alpha = 0 and beta = pi/4$

$\frac{d r (θ)}{d θ} = 3 cos (3 θ) + 24 {csc}^{2} (6 θ)$ $(dr(theta))/(d"theta) = 3cos(3theta) + 24csc^2(6theta)$

Substituting into the integral:

$L = \int_{0}^{\frac{π}{4}} \sqrt{{(sin (3 θ) - 4 cot (6 θ))}^{2} + {(3 cos (3 θ) + 24 {csc}^{2} (6 θ))}^{2}} d θ$ $L = int_0^(pi/4) sqrt{(sin(3theta) - 4cot(6theta))^2 + (3cos(3theta) + 24csc^2(6theta))^2}d theta$

This integral does not converge.

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What is the arclength of the polar curve $f (θ) = sin (3 θ) - 4 cot 6 θ$ $f(theta) = sin(3theta)-4cot6theta$ over $θ \in [0, \frac{π}{4}]$ $theta in [0,pi/4]$ ?

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Explanation:

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What is the arclength of the polar curve f(θ)=sin(3θ)−4cot6θf(theta) = sin(3theta)-4cot6theta over θ∈[0,π4]theta in [0,pi/4] ?

1 Answer

Explanation:

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What is the arclength of the polar curve $f (θ) = sin (3 θ) - 4 cot 6 θ$ $f(theta) = sin(3theta)-4cot6theta$ over $θ \in [0, \frac{π}{4}]$ $theta in [0,pi/4]$ ?