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anonymous

  • 5 years ago

Find the length of the loop for the given curve: x = 12t - 4t^(3) , y = 12t^(2)

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  1. myininaya
    • 5 years ago
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    \[L=\int\limits_{upperintersection}^{lowerintersection}\sqrt{(x')^2+(y')^2} dt\] I believe that is what I remember do you know how to find the intersections?

  2. anonymous
    • 5 years ago
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    no I dont. To be exact my main problem is finding the values of t that I should use for my limits of integration which I guess are just the intersection point.

  3. anonymous
    • 5 years ago
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    points**

  4. asnaseer
    • 5 years ago
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    yes - the loop in question begins and ends at x=0 (see here: http://www.wolframalpha.com/input/?i=x%3D12t-4t%5E3%2C+y%3D12t%5E2) so find the values for t for which x=0 - they will form the lower and upper limits of the integral referred to by @myininaya

  5. anonymous
    • 5 years ago
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    after plugging in all my stuff I get some thing like this: \[\int\limits_{t=0}^{t=\sqrt{3}}\sqrt{144t ^{4}+288t ^{2}+144}dt\]

  6. anonymous
    • 5 years ago
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    I squred the derivatives and collected liked terms inside the radical

  7. asnaseer
    • 5 years ago
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    yup that looks correct - you should be able to spot that 144 can be factored out. that should leave you with a simpler equation that should also be factorable.

  8. asnaseer
    • 5 years ago
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    the square root in the integral will disappear if you spot how to factorize this.

  9. myininaya
    • 5 years ago
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    sorry i had a thing i had my limits backwards lol

  10. myininaya
    • 5 years ago
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    ok so did you guys find out how to find the intersections? i can show you how dedesigns

  11. asnaseer
    • 5 years ago
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    also @DEdesigns57 - your limits are not right. should be from minus square root 3 to plus square root 3.

  12. myininaya
    • 5 years ago
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    well he could double it instead

  13. asnaseer
    • 5 years ago
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    remember:\[x^2=3\implies x=\pm\sqrt{3}\]

  14. anonymous
    • 5 years ago
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    why? it seems the (0,0) is the other intersection point.

  15. asnaseer
    • 5 years ago
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    we are trying to find the start and end limits of 't' - not x.

  16. asnaseer
    • 5 years ago
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    you should have ended up with something like:\[t^2=3\implies t=\pm\sqrt{3}\]

  17. anonymous
    • 5 years ago
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    if t = 0 than (x=0,y=0) though...

  18. asnaseer
    • 5 years ago
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    that only gives you half the curve - so as @myininaya said, you could do 0 to square root 3 and then double the answer.

  19. anonymous
    • 5 years ago
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    and when I look at the graph (0,0) and (0,36) are the intersection points. so I just found the values of t that made that happen. What am I doing wrong?

  20. myininaya
    • 5 years ago
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    ok if we look at the graph we will see two intersection clearly x is 0 when both intersection happen so x=12t-4t^3 => we have 0=12t-4t^3 at the intersections 0=4t(3-t^2) => t=0, pm sqrt(3) and if we plug these into y we would get y=0 and y=36 And our graph confirms these are the right points of intersection: (0,0) and (0, 36) --------------- based on your graphs symmetry I belive you could just double what you have

  21. myininaya
    • 5 years ago
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    and i'm late you guys already said this

  22. asnaseer
    • 5 years ago
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    thanks for your assistance @myininaya - always appreciated :-)

  23. asnaseer
    • 5 years ago
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    ok @DEdesigns57 - are you following the explanations above?

  24. anonymous
    • 5 years ago
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    so its not from t=0 to t=sqrt(3)?

  25. asnaseer
    • 5 years ago
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    let me draw it to try and help

  26. asnaseer
    • 5 years ago
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    |dw:1327172842200:dw|

  27. anonymous
    • 5 years ago
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    ah so your saying the from t=-sqrt(3) to t = sqrt(3) is one revolution around the curve, and that if I were to use from t=0 to t=sqrt(3) I would have to multiply my integral by 2 since it would only be half of it >?

  28. asnaseer
    • 5 years ago
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    so length of loop can be found either by integrating from \(-\sqrt{3}\) to \(\sqrt{3}\), or by doubling the integral from 0 to \(\sqrt{3}\). yes - I think you have it now - well done :-)

  29. anonymous
    • 5 years ago
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    Awesome! thank you so much, wish my teacher took the time to explain things like this.

  30. asnaseer
    • 5 years ago
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    you're more than welcome. I am glad I was able to explain it well enough for you to understand.

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