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

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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#