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

Question

1780 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-15T18:47:17

$L=1.905$

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:

$tutorial.math.lamar.edu$

$x(t)=tan(t)$ . Take the derivative of that to find $dx/dt$

$dx/dt=sec^2(t)$

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

$dy/dt=sec(t)tan(t)csc(t) + sec(t) (-csc(t)cot(t))$

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:

$L=1.905$

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