exercises5-1 - Critical points and their classification,...

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Unformatted text preview: Critical points and their classification, chain rule (the simplest version) 1. Find all critical points and classify them as local max, local min, or saddle point. (a) f (x, y ) = 2x2 + y 2 + 4x − 4y + 5 (b) f (x, y ) = 6xy 2 − 2x3 − 3y 4 (c) f (x, y ) = x3 + y 3 + 3xy + 3 (d) f (x, y ) = 4xy − 2x4 − y 2 2. Find dz/dt both by using the chain rule and by expressing z explicitly as a function of t before differentiating. √ 2 2 (a) z = e−x −y , x = t, y = t (b) z = xy 2 , x = e−t , y = sin t (c) z = xey , x = 2t, y = 1 − t2 (d) z = ln(x2 + y 2 ), x = 1/t, y = Answers: 1(a) The only critical point is (−1, 2) (local min). 1(b) The critical points are: (0, 0) (for which the test fails), (1, 1) (local max), and (1, −1) (local max). 1(c) The critical points are: (0, 0) (saddle point) and (−1, −1) (local max). 1(d) The critical points are: (0, 0) (saddle point), (1, 2) (local max), (−1, −2) (local max). 2(a) −(2t + 1)e−t 2 √ t −t 2(b) e−t sin t(2 cos t − sin t) 2(c) 2e1−t (1 − 2t2 ) 2(d) t3 − 2 t + t4 2 ...
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