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Calculus with Analytic Geometry by edwards & Penney soln ch10

# Calculus with Analytic Geometry

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Section 10.1 C10S01.001: The given line with equation y = 1 2 x + 5 2 has slope 1 2 , so the parallel line through (1 , 2) has equation y +2= 1 2 ( x 1); that is, x +2 y +3=0. C10S01.002: The equation 4 y =3 x 7 of the given line may be rewritten in the form y = 3 4 x 7 4 . Hence every perpendicular line has slope 4 3 . Thus the perpendicular through the point ( 3 , 2) has equation y 2= 4 3 ( x + 3); that is, 4 x +3 y +6=0. C10S01.003: The radius of the circle terminating at (3 , 4) has slope 4 3 . Hence the line L tangent to the circle at that point (because L is perpendicular to that radius) has equation y +4= 3 4 ( x 3); that is, 3 x 4 y = 25. C10S01.004: If y 2 = x + 3, then 2 y dy dx =1 , so that dy dx = 1 2 y . Therefore the slope of the line L tangent to the given curve at (6 , 3) is 1 6 . Hence an equation of L is y +3= 1 6 ( x 6); that is, x +6 y +12=0. C10S01.005: Given x 2 y 2 = 6, we have by implicit diFerentiation 2 x +4 y dy dx =0 , so that dy dx = x 2 y . Therefore the tangent to the given curve at (2 , 1) has slope 1, so the normal to the curve there has slope 1 and thus equation y +1=( 1)( x 2) = x + 2; that is, x + y =1. C10S01.006: The segment S with endpoints A ( 3 , 2) and B (5 , 4) has slope 4 2 5 ( 3) = 6 8 = 3 4 and its midpoint is (1 , 1). Therefore the perpendicular bisector of S has equation y +1= 4 3 ( x 1); that is, 4 x 3 y =7. C10S01.007: Given x 2 x + y 2 = 4, complete the square in each variable to ±nd that x 2 x +1+ y 2 =5; that is, ( x +1) 2 +( y 0) 2 = 5. Hence the circle has center ( 1 , 0) and radius 5. C10S01.008: Given x 2 + y 2 4 y = 5, complete the square in each variable to ±nd that x 2 + y 2 4 y +4=9; that is, ( x 0) 2 y 2) 2 = 9. Hence the circle has center (0 , 2) and radius 3. C10S01.009: Given x 2 + y 2 4 x y = 3, complete the square in each variable to ±nd that x 2 4 x +4+ y 2 y + 9 = 16; that is, ( x 2) 2 y +3) 2 =16 . Therefore the circle has center (2 , 3) and radius 4. C10S01.010: Given x 2 + y 2 +8 x 6 y = 0, complete the square in each variable to ±nd that x 2 x +16+ y 2 6 y + 9 = 25; that is, ( x +4) 2 y 3) 2 =25 . 1

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Thus the circle has center ( 4 , 3) and radius 5. C10S01.011: Given 4 x 2 +4 y 2 4 x = 3, complete the square in each variable as follows: x 2 + y 2 x = 3 4 ; x 2 x + 1 4 + y 2 =1; µ x 1 2 2 +( y 0) 2 =1 . Consequently the circle has center ( 1 2 , 0 ) and radius 1. C10S01.012: Given 4 x 2 y 2 +12 y = 7, complete the square in each variable as follows: x 2 + y 2 +3 y = 7 4 ; x 2 + y 2 y + 9 4 =4; ( x 0) 2 + µ y + 3 2 2 =4 . Therefore the circle has center ( 0 , 3 2 ) and radius 2. C10S01.013: Given 2 x 2 +2 y 2 2 x +6 y = 13, complete the square in each variable as follows: x 2 + y 2 x y = 13 2 ; x 2 x + 1 4 + y 2 y + 9 4 =9; µ x 1 2 2 + µ y + 3 2 2 =9 . Thus the circle has center ( 1 2 , 3 2 ) and radius 3. C10S01.014: Given 9 x 2 +9 y 2 12 x = 5, complete the square in each variable as follows: x 2 + y 2 4 3 x = 5 9 ; x 2 4 3 x + 4 9 + y 2 µ x 2 3 2 y 0) 2 .
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Calculus with Analytic Geometry by edwards & Penney soln ch10

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