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Given dy/dx=1/square root(36-x^2). What is y ?
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To determine the primitive function y, we'll have to compute the indefinite integral of the function.
If dy/dx = 1/sqrt(36-x^2) => dy = dx/sqrt[(6)^2 - x^2]
We'll integrate both sides:
Int dy = Int dx/sqrt[(6)^2 - x^2]
We'll recognize the formula:
Int dx/sqrt(a^2 - x^2) = arcsin (x/a) + C
Let a = 6 => Int dx/sqrt[(6)^2 - x^2] = arcsin (x/6) + C
The required function y, when dy/dx = 1/sqrt(36-x^2), is: y = arcsin (x/6) + C.
Posted by giorgiana1976 on May 19, 2011 at 4:23 PM (Answer #1)
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