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liliy

  • 2 years ago

find the length: x=e^tcos t y=e^tsin t. t is betw 0 and 2

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  1. liliy
    • 2 years ago
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    hahha. if i were the real bieber do u think i would calll myself liliy and would i actually be smart enough to find this site and do math qs?

  2. liliy
    • 2 years ago
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    to find the curve you do sqaure root of(dx/dy)^2+ (dy/dx)^2)

  3. liliy
    • 2 years ago
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    ya?

  4. Australopithecus
    • 2 years ago
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    are you looking for distance between y values?

  5. liliy
    • 2 years ago
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    holy crap. DO nOT!!! come into MY QUESTION and write a whole long question like this. please open ur open question and i will be glad to help u out with these:)

  6. ...qwqwqwqw
    • 2 years ago
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    @liliy :o i did :P and no one answered it and theres only 4 questions but there not bath there career education .-.

  7. liliy
    • 2 years ago
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    fine. im a nice person so ill help: but its really not fair what u did. A,A,A,D

  8. ...qwqwqwqw
    • 2 years ago
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    ;p I'm sorry ill delete mine

  9. liliy
    • 2 years ago
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    relax, its fine

  10. liliy
    • 2 years ago
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    im just saying having it in ur own question is better.

  11. liliy
    • 2 years ago
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    (also you can legit just google the questions)

  12. ...qwqwqwqw
    • 2 years ago
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    i did lmfao and I couldn't find any thing thats why I'm here

  13. liliy
    • 2 years ago
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    alright:). well i hope i helped

  14. ...qwqwqwqw
    • 2 years ago
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    with what o.O?

  15. liliy
    • 2 years ago
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    i just gave u the answers : 1.A 2.A 3.A 4.D

  16. ...qwqwqwqw
    • 2 years ago
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    :O

  17. ...qwqwqwqw
    • 2 years ago
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    @liliy thanks :O

  18. dumbcow
    • 2 years ago
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    to find length of the parametric curve you need to integrate over the arc length of the curve \[\int\limits_{}^{}\sqrt{dx^{2} +dy^{2}}\] to factor in variable t, multiply by dt/dt \[\int\limits_{0}^{2}\sqrt{\frac{dx^{2} +dy^{2}}{dt^{2}}} dt\] which can be written as \[\int\limits_{0}^{2}\sqrt{(\frac{dx}{dt})^{2} +(\frac{dy}{dt})^{2}} dt\] \[\frac{dx}{dt} = e^{t}(\cos t -\sin t)\] \[\frac{dy}{dt} = e^{t} (\sin t + \cos t)\] \[\rightarrow \sqrt{2}\int\limits_{0}^{2}e^{t} dt\]

  19. liliy
    • 2 years ago
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    wait, i dont undestand ur last step. wat happend to the sin and cos?

  20. dumbcow
    • 2 years ago
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    plug in the expressions for dx/dt and dy/dt , square them and combine terms when you do that you notice that sin^2 + cos^2 = 1 , and the rest of the sin and cos terms cancel so it simplifies to that nice easy integral i have above

  21. liliy
    • 2 years ago
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    dx/dt is subtracting betw the cos and sin, so it doesnt cancel out!

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