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

Question

Evelyn Agard · Accepted Answer

#L=cot(pi/8)-cot(pi/3)+4sum_(n=1)^oo((1/2),(n))1/16^n{((4n-2),(2n-1))(5pi)/24+sum_(k=0)^(2n-2)((4n-2),(k))(-1)^(2n-1-k)/(2n-1-k)(sin((4n-2-2k)pi/3)-sin((4n-2-2k)pi/3))}# units.

#f(t)=tant-sect*csct=-cott# #f'(t)=csc^2t# Arclength is given by: #L=int_(pi/8)^(pi/3)sqrt(1+csc^4t)dt# Rearrange: #L=int_(pi/8)^(pi/3)csc^2tsqrt(1+sin^4t)dt# Since #sin^4t<1#, take the series expansion of the square root: #L=int_(pi/8)^(pi/3)csc^2t{sum_(n=0)^oo((1/2),(n))sin^(4n)t}dt# Isolate the #n=0# term: #L=int_(pi/8)^(pi/3)csc^2tdt+sum_(n=1)^oo((1/2),(n))int_(pi/8)^(pi/3)sin^(4n-2)tdt# Apply the trigonometric power-reduction formula: #L=[-cott]_ (pi/8)^(pi/3)+sum_(n=1)^oo((1/2),(n))int_(pi/8)^(pi/3){1/2^(4n-2)((4n-2),(2n-1))+2/2^(4n-2)sum_(k=0)^(2n-2)((4n-2),(k))(-1)^(2n-1-k)cos((4n-2-2k)t)}dt# Simplify: #L=cot(pi/8)-cot(pi/3)+4sum_(n=1)^oo((1/2),(n))1/16^nint_(pi/8)^(pi/3){((4n-2),(2n-1))+2sum_(k=0)^(2n-2)((4n-2),(k))(-1)^(2n-1-k)cos((4n-2-2k)t)}dt# Integrate directly: #L=cot(pi/8)-cot(pi/3)+4sum_(n=1)^oo((1/2),(n))1/16^n[((4n-2),(2n-1))t+sum_(k=0)^(2n-2)((4n-2),(k))(-1)^(2n-1-k)/(2n-1-k)sin((4n-2-2k)t)]_(pi/8)^(pi/3)# Simplify: #L=cot(pi/8)-cot(pi/3)+4sum_(n=1)^oo((1/2),(n))1/16^n{((4n-2),(2n-1))(5pi)/24+sum_(k=0)^(2n-2)((4n-2),(k))(-1)^(2n-1-k)/(2n-1-k)(sin((4n-2-2k)pi/3)-sin((4n-2-2k)pi/3))}#

Aurora Alvarado · Answer

undefined To find the arc length of the curve $ y = \tan(t) - \sec(t) \cdot \csc(t) $ over the interval $ t \in [\frac{\pi}{8}, \frac{\pi}{3}] $, you would integrate the square root of the sum of the squares of the first derivative of $ y $ with respect to $ t $ over that interval, which is given by:

$$ L = \int_{\frac{\pi}{8}}^{\frac{\pi}{3}} \sqrt{(\frac{dy}{dt})^2 + 1} \, dt $$

First, compute $ \frac{dy}{dt} $ using the chain rule, then plug it into the formula above. After integration, you will get the arc length.