The final answer can be seen here.
The general formula for the arc length is as follows:
#D(x) = sqrt((Deltax)^2 + (Deltay)^2)#
#s = D(x) = sum_a^b sqrt((Deltax)^2 + (Deltay)^2/(Deltax)^2*(Deltax)^2)#
# = sum_a^b sqrt(1 + ((Deltay)/(Deltax))^2)Deltax#
# = int_a^b sqrt(1 + ((dy)/(dx))^2)dx#
Thus, take the derivative and simplify.
#(dy)/(dx) = 4*(1/((x/4)^2 - 1))*(2(x/4)*1/4)#
#= (x/((x/4)^2 - 1))*1/2#
#= x/(2(x/4)^2 - 2)#
#= x/(x^2/8 - 2)#
#= (8x)/(x^2 - 16)#
Now, plug it in and square it:
#s = int_7^8 sqrt(1 + ((8x)/(x^2 - 16))^2)dx#
#= int_7^8 sqrt(1 + (64x^2)/(x^2 - 16)^2)dx#
#= int_7^8 sqrt(((x^2 - 16)^2 + 64x^2)/(x^2 - 16)^2)dx#
If we multiply this out, we should find that something like #-32x^2# adds with the #64x^2# for a nice and sneaky shift into a perfect square.
#(x^2 - 16)^2 = x^4 - 32x^2 + 256#
#(x^2 - 16)^2 + 64x^2 = x^4 + 32x^2 + 256#
#= (x^2 + 16)^2#
Much better. Now we can get rid of that ugly square root.
#= int_7^8 sqrt(((x^2+16)^2)/(x^2 - 16)^2)dx#
#= int_7^8 (x^2+16)/(x^2 - 16)dx#
Then some manipulation to make this evaluation easier...ish.
#= int_7^8 (x^2+16 - 16 + 16)/(x^2 - 16)dx#
#= int_7^8 (x^2 - 16 + 32)/(x^2 - 16)dx#
#= int_7^8 dx + int_7^8 32/((x+4)(x-4))dx#
Looks like we probably have to do Partial Fraction Decomposition on this, unfortunately. Oh well.
#int 32/((x+4)(x-4)) = A/(x+4) + B/(x-4)#
#= (A(x-4) + B(x+4))/((x+4)(x-4))#
#= (Ax-4A + Bx+4B)/((x+4)(x-4))#
#= ((A+B)x + (-4A + 4B))/((x+4)(x-4))#
Thus, equating it back to the original equation:
#A+B = 0#
#A = -B#
#-4A + 4B = 32#
#-A + B = 8#
#2B = 8#
#B = 4 -> A = -4#
Not too bad, actually. Now we have, overall:
#= int_7^8 dx + (-int_7^8 4/(x+4)dx + int4/(x-4)dx)#
#= int_7^8 dx - int_7^8 4/(x+4)dx + int4/(x-4)dx#
#= [x]|_(7)^(8)# #- [4ln|x+4|]|_(7)^(8)# #+ [4ln|x-4|]|_(7)^(8)#
#= (8-7) - (4ln12 - 4ln11) + (4ln4 - 4ln3)#
#~~ 1.8027 "u"#
The exact answer is:
#color(blue)(1 - 4(ln12 - ln11 - ln4 + ln3))#
