A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

I need help figuring out the wording to this Calc 2 problem

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Consider the parametric equation \[x = 16(\cos \theta + \theta \sin \theta)\] \[y = 16( \sin \theta - \theta \cos \theta)\] What is the length of the curve for \[0 \le \theta \le \frac{ 11 }{ 8 }\pi\]

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I understand what the arc length means, but the equations have two separate arcs over the interval

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

    @mathstudent55 you seem like the most qualified person answering right now, I just need to understand what it's asking for, I can do the math myself

  4. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[\int_{a}^{b} ds\] given that ds is \[ds=\sqrt{(x')^2+(y')^2}dt\]

  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Where does this equation come from? Just so I have a base of understanding. I don't remember seeing anything like that in lecture

  6. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    its just the result of takeing the segment of an arc ... and comparing it to a right triangle

  7. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1439771344418:dw|

  8. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    as we work our way to infinitesimals we get: |dw:1439771398544:dw| and the pythag takes care of the rest

  9. thomas5267
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Wikipedia is your friend. https://en.wikipedia.org/wiki/Arc_length#Finding_arc_lengths_by_integrating

  10. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    x=16(cos(t) +t sin(t)) x' = 16(cos(t)+sin(t) +t cos(t)) dt y=16(sin(t) -t cos(t)) y'=16(cos(t) -cos(t) +t sin(t)) dt

  11. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    if we can eliminate the parameter, and make y a function of x, then x'=1 ... but thats not always a simple case is it

  12. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So that equation accounts for the two arcs?

  13. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    yes, whether it makes life simpler or not ... thats to be seen :)

  14. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[\int~ds\implies\int_{a}^{b}\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}~dt\]

  15. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    http://tutorial.math.lamar.edu/Classes/CalcII/ParaArcLength.aspx

  16. thomas5267
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[ \begin{align*} x^\phantom{'}&=16\left(\cos(\theta)+\theta\sin(\theta)\right)\\ x'&=16\left(-\sin(\theta)+\sin(\theta)+\theta\cos(\theta)\right)\\ y^\phantom{'}&=16\left(\sin(\theta)-\theta\cos(\theta)\right)\\ y'&=16\left(\cos(\theta)-\cos(\theta)+\theta\sin(\theta)\right)\\ \end{align*} \]

  17. thomas5267
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The integral is much easier to evaluate once I got rid of my error lol.

  18. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    heheh .. my x' got befuddled as well

  19. thomas5267
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I got \(x'=\sin(\theta)+\sin(\theta)+\theta\cos(\theta)\) and got a mess when evaluating the integral.

  20. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    16t dt from a to b ...

  21. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    maybe ... trying to do it in my head bites at my age

  22. thomas5267
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Just use the answer box as scrap paper lol.

  23. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thomas, dumb question but could you walk me through the derivation of x real quick? I know the 16 can just get pulled out first

  24. thomas5267
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[ \frac{d}{dx}\cos(x)=-\sin(x) \] And product rule.

  25. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Okay, thank you so much guys, got the answer right finally!

  26. 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

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.