Asked by Sarah
Find the length of the curve:
y=ln(1-x^2), 0<_x<_1/2
y=ln(1-x^2), 0<_x<_1/2
Answers
Answered by
Steve
s = ∫[0,1/2] ds
= ∫[0,1/2] √(1+y'^2) dx
= ∫[0,1/2] √(1+(2x/(x^2-1))^2) dx
= ∫[0,1/2] (x^2+1)/(x^2-1) dx
now use partial fractions to get
= ∫[0,1/2] 1 + 1/(x-1) - 1/(x+1) dx
= (x + ln|x-1| - ln|x+1|) [0,1/2]
= (1/2 + ln(1/2) - ln(3/2)) - (0+0-0)
= 1/2 - ln3
= ∫[0,1/2] √(1+y'^2) dx
= ∫[0,1/2] √(1+(2x/(x^2-1))^2) dx
= ∫[0,1/2] (x^2+1)/(x^2-1) dx
now use partial fractions to get
= ∫[0,1/2] 1 + 1/(x-1) - 1/(x+1) dx
= (x + ln|x-1| - ln|x+1|) [0,1/2]
= (1/2 + ln(1/2) - ln(3/2)) - (0+0-0)
= 1/2 - ln3
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.