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anonymous

  • 5 years ago

deriviative of y= -Ae^-t (Sint - cost)

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  1. anonymous
    • 5 years ago
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    A constant?

  2. anonymous
    • 5 years ago
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    What do you mean? A is a constant yes

  3. anonymous
    • 5 years ago
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    y=-A(e^-t)(sint-cost) y=-A(e^-t)sint + A(e^-t)cost y'=-Ad/dt((e^-t)sint)+Ad/dt((e^-t)cost) d/dt ((e^-t)sint) =(e^-t)cost + (-e^-t)sint d/dt ((e^-t)cost)=(e^-t)(-sint) + (-e^-t)cost y'=A(-((e^-t)cost + (-e^-t)sint)+(e^-t)(-sint) + (-e^-t)cost) y'=A(-(e^-t)cost + (e^-t)sint + (e^-t)(-sint) - (e^-t)cost) y'=A(-2(e^-t)cost)=-2A(e^-t)cost

  4. anonymous
    • 5 years ago
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    Ahhhh yes thanks alot :)

  5. anonymous
    • 5 years ago
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    no prob

  6. anonymous
    • 5 years ago
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    btw can you guess what level calculus this is?

  7. anonymous
    • 5 years ago
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    I would say end of 1st year uni or beginning of 2nd year, if that's what you mean by level.

  8. anonymous
    • 5 years ago
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    im doing this in my senoir year of highschool :( my teacher is crazyyyy

  9. anonymous
    • 5 years ago
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    I got expelled :D:D:D hate those highweeded teachers

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spraguer (Moderator)
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