Sketch the domain of f(x,y)=arcsin(x-y)*sqrt(9-x^2-y^2) and find it\'s range. Ca

Question

Sketch the domain of f(x,y)=arcsin(x-y)*sqrt(9-x^2-y^2) and find it's range.

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Explanation / Answer

Domain: arcsin(x - y) is defined for all (x, y) in R^2. v(9 - x^2 - y^2) is only defined when 9 - x^2 - y^2 = 0 ==> x^2 + y^2 = 9. So, the domain of f is given by {(x, y) in R^2 : x^2 + y^2 = 9}. This is a closed disk (boundary included) centered at (0, 0) with radius 3. ------------------------- Range: v(9 - x^2 - y^2) yields values in [0, 3]. arcsin(x - y) yields values in [-p/2, p/2]. So, the range of f is contained in at most [-3p/2, 3p/2]. --- To show that this is indeed the range: Let arcsin(x - y) = t; note that t is in [-p/2, p/2]. x - y = sin t x = y + sin t. So, f(x, y) = t v(9 - (y + sin t)^2 - y^2) ...............= t v[9 - 2y^2 - 2y sin t - sin^2(t)] Let z be in [-3p/2, 3p/2]. We need to show that we can find y so that z = t v[9 - 2y^2 - 2y sin t - sin^2(t)] If z = 0, just set t = 0. Otherwise, z/t = v[9 - 2y^2 - 2y sin t - sin^2(t)] (for no extraneous solutions, we choose t to have the same sign as z so that z/t > 0) ==> (z/t)^2 = 9 - 2y^2 - 2y sin t - sin^2(t) ==> 2y^2 + (2 sin t) y + (sin^2(t) - (z/t)^2 - 9) = 0 ==> y = {-(2 sin t) ± v[4 sin^2(t) - 4(sin^2(t) - (z/t)^2 - 9)]} / 4 ==> y = (1/2) {-sin t ± v[(z/t)^2 + 9]}, which does exist for the choices of z, t given. Hence, the range for f is indeed [-3p/2, 3p/2].

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