What is the arc length of #f(x)=secx*tanx # in the interval #[0,pi/4]#?
1 Answer
The arc length formula is basically the distance formula adapted to a function. Whereas the distance formula would be
#\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
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.
