Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

katzndogs

  • 3 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

  • This Question is Closed
  1. katzndogs
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    *x was given as a function of y

  3. beginnersmind
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  5. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy