Calculus with Analytic Geometry by edwards & Penney soln ch4

# Calculus with Analytic Geometry

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Section 4.1 C04S01.001: 2 x 2 y dy dx = 0, so dy dx = x y . Also, y = ± x 2 1, so dy dx = ± x x 2 1 = x ± x 2 1 = x y . C04S01.002: x dy dx + y = 0, so dy dx = y x . By substituting y = x 1 here, we get dy dx = x 1 x = x 2 , which is the result obtained by explicit differentiation. C04S01.003: 32 x + 50 y dy dx = 0; dy dx = 16 x 25 y . Substituting y = ± 1 5 400 16 x 2 into the derivative, we get dy dx = 16 x 5 400 16 x 2 , which is the result obtained by explicit differentiation. C04S01.004: 3 x 2 +3 y 2 dy dx = 0, so dy dx = x 2 y 2 . y = 3 1 x 3 , so substitution results in dy dx = x 2 (1 x 3 ) 2 / 3 . Explicit differentiation yields the same answer. C04S01.005: 1 2 x 1 / 2 + 1 2 y 1 / 2 dy dx = 0: dy dx = y x . C04S01.006: 4 x 3 +2 x 2 y dy dx +2 xy 2 +4 y 3 dy dx = 0: (2 x 2 y +4 y 3 ) dy dx = (4 x 3 +2 xy 2 ); dy dx = 4 x 3 + 2 xy 2 2 x 2 y + 4 y 3 . C04S01.007: 2 3 x 1 / 3 + 2 3 y 1 / 3 dy dx = 0: dy dx = x y 1 / 3 = y x 1 / 3 . C04S01.008: y 2 + 2( x 1) y dy dx = 1, so dy dx = 1 y 2 2 y ( x 1) . C04S01.009: Given: x 3 x 2 y = xy 2 + y 3 : 3 x 2 x 2 dy dx 2 xy = y 2 + 2 xy dy dx + 3 y 2 dy dx ; 3 x 2 2 xy y 2 = (2 xy + 3 y 2 + x 2 ) dy dx ; dy dx = 3 x 2 2 xy y 2 3 y 2 + 2 xy + x 2 . C04S01.010: Given: x 5 + y 5 = 5 x 2 y 2 : 5 x 4 + 5 y 4 dy dx = 10 x 2 y dy dx + 10 xy 2 ; dy dx = 10 xy 2 5 x 4 5 y 4 10 x 2 y . C04S01.011: Given: x sin y + y sin x = 1: x cos y dy dx + sin y + y cos x + sin x dy dx = 0; dy dx = sin y + y cos x x cos y + sin x . 1

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C04S01.012: Given: cos( x + y ) = sin x sin y : sin( x + y )(1 + dy dx ) = sin x cos y dy dx + sin y cos x ; dy dx = sin y cos x + sin( x + y ) sin( x + y ) + sin x cos y . C04S01.013: Given: cos 3 x + cos 3 y = sin( x + y ): 3 ( cos 2 x ) ( sin x ) + 3 ( cos 2 y ) ( sin y ) dy dx = cos( x + y ) 1 + dy dx ; dy dx = cos( x + y ) + 3cos 2 x sin x cos( x + y ) + 3cos 2 y sin y . C04S01.014: Given: xy = tan xy : x dy dx + y = ( sec 2 xy ) x dy dx + y ; dy dx = y sec 2 xy y x x sec 2 xy = y x . Note: The original equation is of the form tan u = u , which is true only for some isolated constant values of u ; under the assumption that xy = k for some constant k , we also obtain dy dx = y x . C04S01.015: 2 x + 2 y dy dx = 0: dy dx = x y . At (3 , 4) the tangent has slope 3 4 and thus equation y + 4 = 3 4 ( x 3). C04S01.016: x dy dx + y = 0: dy dx = y x . At (4 , 2) the tangent has slope 1 2 and thus equation y + 2 = 1 2 ( x 4). C04S01.017: x 2 dy dx + 2 xy = 1, so dy dx = 1 2 xy x 2 . At (2 , 1) the tangent has slope 3 4 and thus equation 3 x + 4 y = 10. C04S01.018: 1 4 x 3 / 4 + 1 4 y 3 / 4 dy dx = 0: dy dx = ( y/x ) 3 / 4 . At (16 , 16) the tangent has slope 1 and thus equation x + y = 32. C04S01.019: y 2 + 2 xy dy dx + 2 xy + x 2 dy dx = 0: dy dx = 2 xy + y 2 2 xy + x 2 . At (1 , 2) the slope is zero, so an equation of the tangent there is y = 2. C04S01.020: 1 ( x + 1) 2 1 ( y + 1) 2 · dy dx = 0, so dy dx = ( y + 1) 2 ( x + 1) 2 . At (1 , 1) the tangent line has slope 1 and thus equation y 1 = ( x 1). C04S01.021: 24 x + 24 y dy dx = 25 y + 25 x dy dx : dy dx = 25 y 24 x 24 y 25 x . At (3 , 4) the tangent line has slope 4 3 and thus equation 4 x = 3 y .
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