## Find the arc length of the curve (Polar)

1. The problem statement, all variables and given/known data
I was wondering if I did this problem correctly as I don't have the solution, also wanted to make sure that my limits of integration were correct as they tend to be tricky in finding arc length in polar coordinates.

x(t)=arcsint
y(t)=ln(sqrt(1-t^2))

2. Relevant equations
S= integral from a-b of
sqrt((dx/dt)^2+(dy/dt)^2)dt

3. The attempt at a solution
(dx/dt)^2=1/(1-t^2)
(dy/dt)^2=t^2/(1-t^2)^2
I get 1/(1-t^2)^2
Put all of this into the square root as said by the formula
I simplified it to the integral from 0 to 1/2 of dt/(1-t^2)
Factoring the bottom I get dt/((1-t)(1+t))
by Partial Fractions I get
2 separate integrals
(1/2)∫dt/(1-t)+(1/2)∫dt/(1+t)
Finally integrating this I get
(1/2)(ln(1-t)+ln(1+t))
Plugging in my limits of integration I get
(1/2)(ln(1/2)+ln(3/2))
Using the log rule
I get ln(3/4)^(1/2)

Thank you so much to anyone who read through this long problem!

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Recognitions:
Homework Help
 Quote by mathnoobie 1. The problem statement, all variables and given/known data I was wondering if I did this problem correctly as I don't have the solution, also wanted to make sure that my limits of integration were correct as they tend to be tricky in finding arc length in polar coordinates. x(t)=arcsint y(t)=ln(sqrt(1-t^2)) 2. Relevant equations S= integral from a-b of sqrt((dx/dt)^2+(dy/dt)^2)dt 3. The attempt at a solution (dx/dt)^2=1/(1-t^2) (dy/dt)^2=t^2/(1-t^2)^2 adding (dx/dt)^2+(dy/dt)^2 I get 1/(1-t^2)^2 Put all of this into the square root as said by the formula I simplified it to the integral from 0 to 1/2 of dt/(1-t^2) Factoring the bottom I get dt/((1-t)(1+t)) by Partial Fractions I get 2 separate integrals (1/2)∫dt/(1-t)+(1/2)∫dt/(1+t) Finally integrating this I get (1/2)(ln(1-t)+ln(1+t)) Plugging in my limits of integration I get (1/2)(ln(1/2)+ln(3/2)) Using the log rule I get ln(3/4)^(1/2) Thank you so much to anyone who read through this long problem!
What are the limits a, b?

ehild

 Quote by ehild You missed a minus sign in the first integral. What are the limits a, b? ehild
Ah yes you're right. I always make those negative sign mistakes. As for the limits, I'm not 100% sure what they are but it says the curve is defined by the interval 0≤t≤1/2
so I assumed they're 0 to 1/2.
Another mistake I just saw, I never plugged in 0 into my log functions.

## Find the arc length of the curve (Polar)

Alright going back and fixing my mistakes.
using the fundamental theorum of calculus, I have.
(1/2)(-ln(1/2)+ln(3/2)+ln(1)-ln(1))
=(1/2)(ln(1/2)^(-1)+ln(3/2)

On a side note, (1/2)^(-1)=2 right?
If so I end up with
ln(3)^(1/2)

 Recognitions: Homework Help Why didn't you simplify -ln(1-t)+ln(1+t)? ln(1/2)^(-1) means the reciprocal of ln(1/2) and not the logarithm of the reciprocal of 1/2. Write ln[(1/2)^(-1)]. And yes, (1/2)^(-1)=2. And do not forget the closing parenthesis from (1/2)(ln(1/2)^(-1)+ln(3/2)) If you meant $\sqrt{\ln(3)}$ then your solution is correct now. ehild

 Quote by ehild Why didn't you simplify -ln(1-t)+ln(1+t)? ln(1/2)^(-1) means the reciprocal of ln(1/2) and not the logarithm of the reciprocal of 1/2. Write ln[(1/2)^(-1)]. And yes, (1/2)^(-1)=2. And do not forget the closing parenthesis from (1/2)(ln(1/2)^(-1)+ln(3/2)) If you meant $\sqrt{\ln(3)}$ then your solution is correct now. ehild
I didn't simplify because I was unsure of the effect of the - in front of ln, so I didn't continue with something I was unsure of to avoid mistakes early on.
Thank you so much though for taking the time to help me!

 Recognitions: Homework Help Remember, ln(y)-ln(x)=ln(y/x). It appears quite often in case a definite integral. ehild