How do you find the length of the curve for #y= ln(1-x²)# for (0, 1/2)?
1 Answer
The formula for the arc length of the curve
#s=int_a^bsqrt(1+(dy/dx)^2)dx#
Here, where
Thus, the arc length in question is:
#s=int_0^(1//2)sqrt(1+((2x)/(x^2-1))^2)dx#
#s=int_0^(1//2)sqrt(((x^2-1)^2+(2x)^2)/(x^2-1)^2)dx#
#s=int_0^(1//2)sqrt(x^4-2x^2+1+4x^2)/(x^2-1)dx#
#s=int_0^(1//2)sqrt(x^4+2x^2+1)/(x^2-1)dx#
Note this is factorable:
#s=int_0^(1//2)sqrt((x^2+1)^2)/(x^2-1)dx#
#s=int_0^(1//2)(x^2+1)/(x^2-1)dx#
Rewriting:
#s=int_0^(1//2)(x^2-1+2)/(x^2-1)dx#
#s=int_0^(1//2)(1+2/(x^2-1))dx#
Perform partial fractions on the second piece:
#2/(x^2-1)=A/(x-1)+B/(x+1)#
#2=A(x+1)+B(x-1)#
Letting
#2=2A" "=>" "A=1#
And letting
#2=-2B" "=>" "B=-1#
So
#2/(x^2-1)=1/(x-1)-1/(x+1)#
Then
#s=int_0^(1//2)(1+1/(x-1)-1/(x+1))dx#
Which are all easily integrateable:
#s=[x+lnabs(x-1)-lnabs(x+1)]_0^(1//2)#
#s=[x+lnabs((x-1)/(x+1))]_0^(1//2)#
#s=(1/2+lnabs((1/2-1)/(1/2+1)))-(0+lnabs((0-1)/(0+1)))#
#s=1/2+lnabs((-1/2)/(3/2))+lnabs(-1)#
#s=1/2+ln(1/3)#
Or
#s=1/2-ln3#