What is the arclength of the polar curve $f (θ) = 2 θ sin (5 θ) - θ cot 2 θ$ $f(theta) = 2thetasin(5theta)-thetacot2theta$ over $θ \in [\frac{π}{12}, \frac{π}{2}]$ $theta in [pi/12,pi/2]$ ?

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What is the arclength of the polar curve $f (θ) = 2 θ sin (5 θ) - θ cot 2 θ$ $f(theta) = 2thetasin(5theta)-thetacot2theta$ over $θ \in [\frac{π}{12}, \frac{π}{2}]$ $theta in [pi/12,pi/2]$ ?

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Answer 1 · 2018-01-17T03:21:41

$\infty$ $infty$

Explanation:

For this solution, I am assuming that your $f$ $f$ is actually referring to the radial component.

If we travel an infinitely small angle, the distance will be f( $θ$ $theta$ ) d $θ$ $theta$ . Therefore, we can integrate that function across that whole range to get the value, i.e. $\int_{\frac{π}{12}}^{\frac{π}{2}} f (θ) d θ$ $int_(pi/12)^(pi/2) f(theta)d theta$

However, since cotangent goes to infinity very quickly at $\frac{π}{2}$ $pi/2$ , this integral does not converge to a finite value, hence the length is infinite.

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What is the arclength of the polar curve $f (θ) = 2 θ sin (5 θ) - θ cot 2 θ$ $f(theta) = 2thetasin(5theta)-thetacot2theta$ over $θ \in [\frac{π}{12}, \frac{π}{2}]$ $theta in [pi/12,pi/2]$ ?

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What is the arclength of the polar curve f(theta) = 2thetasin(5theta)-thetacot2theta f(θ)=2θsin(5θ)−θcot2θf(theta) = 2thetasin(5theta)-thetacot2theta over theta in [pi/12,pi/2] θ∈[π12,π2]theta in [pi/12,pi/2] ?

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What is the arclength of the polar curve $f (θ) = 2 θ sin (5 θ) - θ cot 2 θ$ $f(theta) = 2thetasin(5theta)-thetacot2theta$ over $θ \in [\frac{π}{12}, \frac{π}{2}]$ $theta in [pi/12,pi/2]$ ?