What is the arc length of #f(x)=secx*tanx # in the interval #[0,pi/4]#?

1 Answer
Aug 4, 2016

The arc length formula is basically the distance formula adapted to a function. Whereas the distance formula would be #D = sqrt((Deltax)^2 + (Deltay)^2)#, the arc length formula is:

#\mathbf(s = int_a^b sqrt(1 + ((dy)/(dx))^2)dx)#

So, we need to take the derivative and then square it.

#d/(dx)[secxtanx]#

#= secx*sec^2x + secxtan^2x#

#= secx(sec^2x + tan^2x)#

#= secx(1 + 2tan^2x)#

And squaring it...

#((dy)/(dx))^2#

#= color(green)(sec^2x(2tan^2x + 1)^2)#

And now putting it in the integral.

#s = color(blue)(int_0^(pi"/"4) sqrt(1 + sec^2x(2tan^2x + 1)^2) dx)#

At that point, you should evaluate the integral numerically on your calculator. A TI-89 should be able to do it, but working it out would be far too complicated.

I get #~~ color(blue)(1.6458)# using Wolfram Alpha, and this integral returns the same answer.

As you can see, the setup is usually pretty easy. What's usually hard (and generally nearly impossible on medium-complexity functions) is the actual simplification work.