How do you find the arc length of the curve #y=e^(-x)+1/4e^x# from [0,1]?

1 Answer
Jun 19, 2015

The arc length turns out to be identical to simply integrating the original function. It is:
#e/4 - 1/e + 3/4 ~~ 1.06169#

How you do it is written below:

The arc length formula is derived from a "dynamic" distance formula with an independently increasing x value and a y value that varies with a single-valued function:

#D(x) = sqrt((Deltax)^2 + (Deltay)^2)#

But you know that #(Deltay)/(Deltax)# on really small intervals is equivalent to #(dy)/(dx)#. Thus:

#D(x) = sqrt((Deltax)^2 + (Deltay)^2/(Deltax)^2*(Deltax)^2)#

#= sum_xsqrt(1 + ((Deltay)/(Deltax))^2)Deltax#

#= int_a^bsqrt(1 + ((dy)/(dx))^2)dx#

And now we have the formula!

#y = e^(-x) + 1/4e^x#
#(dy)/(dx) = -e^(-x) + 1/4e^x#

#((dy)/(dx))^2 = (-e^(-x) + 1/4e^x)(-e^(-x) + 1/4e^x)#

#= (-e^(-x))^2 - 1/4(e^xe^(-x)) - 1/4(e^xe^(-x)) + 1/16(e^x)^2#

#= e^(-2x) - 1/2(1) + 1/16e^(2x)#

#= 1/16e^(2x) + e^(-2x) - 1/2#

Now let's simplify it in context.

#D(x) = s = int_a^b sqrt(1+1/16e^(2x) + e^(-2x) - 1/2)dx#

#= int_0^1 sqrt(1/16e^(2x) + e^(-2x) + 1/2)dx#

My best guess on where to go next is getting some sort of perfect square. I'm thinking it might be a quartic, though. Let's just try things.

Getting a common denominator:
#= int_0^1 sqrt((e^(2x) + 16e^(-2x) + 8)/16)dx#

Factoring out #sqrt(1/16)#:
#= 1/4int_0^1 sqrt(e^(2x) + 16e^(-2x) + 8)dx#

Factoring out #sqrt(e^(2x))#:
#= 1/4int_0^1 sqrt(e^(2x)(1 + 16e^(-4x) + 8e^(-2x))dx#

#= 1/4int_0^1 e^xsqrt(1 + 16e^(-4x) + 8e^(-2x))dx#

Factoring out #sqrt(16)#:
#= int_0^1 e^xsqrt(e^(-4x) + 1/2e^(-2x) + 1/16)dx#

...This looks promising. What if we just let #u = e^x#?

#=> int_a^b usqrt(u^(-4) + 1/2u^(-2) + 1/16)dx#

Now let's attempt to factor this. ... Nice, it worked!

#= (u^(-2) + 1/4)^2#

Sweet, so now we can put the real letters back in.

#= int_0^1 e^x(e^(-2x) + 1/4)dx#

This looks much better. Now let's just multiply it out and solve it. Notice though that Wolfram Alpha keeps saying that #sqrt(e^(2x)) ne |e^x|#, but we all know that's not right; #e^x# is always, always positive! Anyways...

#= int_0^1 e^(-x) + 1/4e^xdx#

...wait a minute. Are you serious? That's what we started with! Wow.

#= [-e^(-x) + 1/4e^x]|_(0)^(1)#

#= [-e^(-1) + 1/4e^1] - [-e^(-0) + 1/4e^0]#

#= [-1/e + 1/4e] - [-1 + 1/4]#

#= -1/e + 1/4e + 1 - 1/4#

#= e/4 - 1/e + 3/4 ~~ 1.06169#

tl;dr: You can see that here.