What is the arclength of the polar curve $f (θ) = 2 θ sin (5 θ) - cot θ$ $f(theta) = 2thetasin(5theta)-cottheta$ over $θ \in [\frac{π}{12}, \frac{π}{6}]$ $theta in [pi/12,pi/6]$ ?

Question

What is the arclength of the polar curve $f (θ) = 2 θ sin (5 θ) - cot θ$ $f(theta) = 2thetasin(5theta)-cottheta$ over $θ \in [\frac{π}{12}, \frac{π}{6}]$ $theta in [pi/12,pi/6]$ ?

1 Answer

Impact of this question

1703 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-02T12:45:09

$\approx 1.76588$ $approx 1.76588$

Explanation:

We have given
$r (θ) = 2 θ \cdot sin (5 θ) - cot (5 θ)$ $r(theta)=2theta*sin(5theta)-cot(5theta)$
then
$r'(theta)=2sin(5theta)+10cos(5theta)-1/sin^2(theta)$
and we get

$int_(pi/12)^(pi/6)sqrt((2theta*sin(5theta)-cot(theta))^2+(2sin(5theta)+10cos(5theta)-1/sin(theta)^2)^2)d theta$
I have found only a numerical value of that integral
$approx 1.76588$

Science

Math

Humanities

... and beyond

What is the arclength of the polar curve $f (θ) = 2 θ sin (5 θ) - cot θ$ $f(theta) = 2thetasin(5theta)-cottheta$ over $θ \in [\frac{π}{12}, \frac{π}{6}]$ $theta in [pi/12,pi/6]$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

What is the arclength of the polar curve f(theta) = 2thetasin(5theta)-cottheta f(θ)=2θsin(5θ)−cotθf(theta) = 2thetasin(5theta)-cottheta over theta in [pi/12,pi/6] θ∈[π12,π6]theta in [pi/12,pi/6] ?

1 Answer

Explanation:

Related questions

Impact of this question

What is the arclength of the polar curve $f (θ) = 2 θ sin (5 θ) - cot θ$ $f(theta) = 2thetasin(5theta)-cottheta$ over $θ \in [\frac{π}{12}, \frac{π}{6}]$ $theta in [pi/12,pi/6]$ ?