What is the arclength of (tant,sect*csct) on t in [pi/8,pi/3]?

Apr 15, 2017

$L = 1.905$

Explanation:

To find arc length, use the formula:

L=int_a^b sqrt((dx/dt)^2+(dy/dt)^2

It helps to recall a few trig derivatives for this problem:

$x \left(t\right) = \tan \left(t\right)$. Take the derivative of that to find $\frac{\mathrm{dx}}{\mathrm{dt}}$

$\frac{\mathrm{dx}}{\mathrm{dt}} = {\sec}^{2} \left(t\right)$

For $y \left(t\right) = \sec \left(t\right) \csc \left(t\right)$, use product rule:

$\frac{\mathrm{dy}}{\mathrm{dt}} = \sec \left(t\right) \tan \left(t\right) \csc \left(t\right) + \sec \left(t\right) \left(- \csc \left(t\right) \cot \left(t\right)\right)$

Plug into our equation:

L=int_(pi/8)^(pi/3) sqrt((sec^2t)^2 + (sec(t)tan(t)csc(t) + sec(t) (-csc(t)cot(t)))^2

This doesn't look like it'll be a clean integral so using a graphing calculator, this becomes:

$L = 1.905$