What is the arc length of #f(x)=x^2/(4-x^2) # on #x in [-1,1]#?

Question

Luke Alvarez · Accepted Answer

#L=2+2sum_(n=1)^oosum_(m=0)^oo((1/2),(n))((-4n),(m))(-1)^m/((2n+2m+1)*4^(n+m))# units.

#f(x)=x^2/(4-x^2)=1-4/(4-x^2)# #f'(x)=(8x)/(4-x^2)^2# Arc length is given by: #L=int_-1^1sqrt(1+(64x^2)/(4-x^2)^4)dx# For #x in [-1,1]#, #(64x^2)/(4-x^2)^4<1#. Take the series expansion of the square root: #L=int_-1^1sum_(n=0)^oo((1/2),(n))((64x^2)/(4-x^2)^4)^ndx# Isolate the #n=0# term and simplify: #L=int_-1^1dx+sum_(n=1)^oo((1/2),(n))4^(3n)int_-1^1x^(2n)/(4-x^2)^(4n)dx# Rearrange: #L=2+sum_(n=1)^oo((1/2),(n))1/4^nint_-1^1x^(2n)(1-1/4x^2)^(-4n)dx# For #x in [-1,1]#, #1/4x^2<=1#. Take another series expansion: #L=2+sum_(n=1)^oo((1/2),(n))1/4^nint_-1^1x^(2n){sum_(m=0)^oo((-4n),(m))(-1/4x^2)^m}dx# Simplify: #L=2+sum_(n=1)^oosum_(m=0)^oo((1/2),(n))((-4n),(m))(-1)^m/4^(n+m)int_-1^1x^(2n+2m)dx# Integrate directly: #L=2+sum_(n=1)^oosum_(m=0)^oo((1/2),(n))((-4n),(m))(-1)^m/4^(n+m)[x^(2n+2m+1)]_-1^1/(2n+2m+1)# Insert the limits of integration and simplify: #L=2+2sum_(n=1)^oosum_(m=0)^oo((1/2),(n))((-4n),(m))(-1)^m/((2n+2m+1)*4^(n+m))#

Charles Arocha · Answer

undefined To find the arc length of the function $ f(x) = \frac{x^2}{4-x^2} $ on the interval $ [-1, 1] $, you can use the formula for arc length:

$$ L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx $$

where $ f'(x) $ represents the derivative of the function $ f(x) $ with respect to $ x $, and $ a $ and $ b $ represent the endpoints of the interval.

First, find the derivative of $ f(x) $, then plug it into the formula along with the limits of integration.

After integrating, you should get the arc length.