## katzndogs 2 years ago I'm on exam 4, question 5 b) and asked to set-up an integral for arclength along a curve. I don't understand how the answer parametrizes the curve though without any initial conditions, here's the link: http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/unit-4-techniques-of-integration/exam-4/session-86-materials-for-exam-4/MIT18_01SCF10_exam4sol.pdf

1. katzndogs

I instead found arclength in with respect to y since the x was given as a function of x but I m sure looking through my notes there whenever asked to parametrize initial conditions for t were provided. So any one out there with any idea could you be so kind to help. Thankyou!

2. katzndogs

*x was given as a function of y

3. beginnersmind

The initial condition is implicit in the choice t=y. We could have made the choice s=y-1, which would have given x=s+1+(s+1)^3. We would need to change the limits of integration s=0 and s=3 (the values corresponding to y=1 and y=4). So we would integrate a slightly different function over different limits, leading to the same arc length. It's a matter of choice and if you try anything else you'll see how t=y leads to the simplest result.

4. katzndogs

I understand now! Thanks so much man for that great explanation! :)