Ap Calculus

If dy/dx= sin^2(πy/4) and y=1 when x = 0, then find the value of x when y = 3.

A) 0
B) 8/π
C) - π/8
D) None of these

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asked by Karen
  1. y' = sin^2(πy/4)
    dy/sin^2(πy/4) = dx
    csc^2(πy/4) dy = dx
    Think you can take it from here?
    You know what ∫(csc^2 u) du is, right?

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    posted by oobleck
  2. In case you get stuck there, the rest is
    4/π (-cot π/4 y) = x+c
    y = -4/π cot^-1(π/4 (x+c))
    Now, y(0) = 1, so
    -4/π cot^-1(π/4 c) = 1
    cot^-1(π/4 c) = -π/4
    since cot^-1(-1) = -π/4,
    π/4 c = -1
    c = -4/π
    y = -4/π cot^-1(π/4 (x-4/π))
    y = -4/π cot^-1(π/4 x - 1)
    So, when y=3,
    -4/π cot^-1(π/4 x - 1) = 3
    cot^-1(π/4 x - 1) = -3π/4
    cot -3π/4 = 1
    so,
    π/4 x - 1 = 1
    π/4 x = 2
    x = 8/π

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    posted by oobleck
  3. Thank you very much.

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    posted by Karen

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