How do you find the arc length of the curve #y = 4 ln((x/4)^(2) - 1)# from [7,8]?

Question

Aurora Alvarado · Accepted Answer

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))#

Cameron Alexander · Answer

undefined To find the arc length of the curve $ y = 4 \ln\left(\left(\frac{x}{4}\right)^2 - 1\right) $ from $ x = 7 $ to $ x = 8 $, you use the formula for arc length:

$$ L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx $$

where $ a = 7 $ and $ b = 8 $.

First, find $ \frac{dy}{dx} $ by differentiating $ y $ with respect to $ x $. Then, substitute $ \frac{dy}{dx} $ into the formula and integrate from $ x = 7 $ to $ x = 8 $ to find the arc length $ L $.