# 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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