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http://img840.imageshack.us/img840/9651/ch2b.png Four questions are written on the pic. Please help me to solve them by clear steps n show the steps. Thanks : ) 更新: Answers 1) 4cm 2) l^2/32 3) (√3)x - y - 2√3 = 0 4) y = x ±6
最佳解答:
1. Let AP = y AP2 + PB2 = y2 + (8 - y2) = y2 + (64 - 16y + y2) = 2y2 - 16y + 64 = 2(y2 - 8y) + 64 = 2(y2 - 8y + 16) - 32 + 64 = 2(y - 4)2 + 32 For any real number y, (y - 4)2 ≥ 0 Hence, AP2 + PB2 is a minimum when (y - 4)2 = 0 i.e. Distance of P from A = y = 4 2) Let y be the length of one of the pieces of strength. Total area of the two squares = y2 + (l - y)2 = y2 + l 2 - 2 l y + y2 = 2y2 - 2l y + l 2 = 2(y2 - l y) + l 2 = 2(y2 - l y + l /4) - 2l 2/4+ l 2 = 2(y - l /2)2 + l 2/2 For any real number y, (y - l /2)2 ≥ 0 Hence, when y = l /2, the minimum total area of the two squares = l 2/2 3) cosθ = 1/2 sinθ = √(1 - cos2θ) = √[1 - (1/2)2] = (√3)/2 Slope of the line = tanθ = sinθ/cosθ = √3 x-intercept = 2 The line passes through (2, 0) Equation of the line: (y - 0) = (√3)(x - 2) (√3)x - y - 2√3 = 0 4) Let A = (a, 0) and B = (0, b) Slope of the line l = 1 (0 - b)/(a - 0) = 1 a = -b …… (1) Area of the ΔOAB = 18 (1/2)|ab| = 18 ab = -36 …… (2) or ab = 36 (rejected for a = -b) Subst. (1) into (2) (-b)b = -36 b = 6 or b = -6 When b = 6, B(0, 6) Hence, equation of the line l : y - 6 = 1(x - 0) x - y + 6 = 0 When b = -6, B(0, -6) Hence, equation of the line l : y + 6 = 1(x - 0) x - y - 6 = 0 2010-08-23 01:13:18 補充： In Q.1, the unit is "cm". 2010-08-23 01:22:42 補充： Amendment of Q.2 Total area of the two squares = y2 + [(l - 4y)/4]2 = 2y2 - l y/2 + l 2/16 = 2(y2 - l y/4 + l 2/64) - 2l 2/64+ l 2/16 = 2(y - l /8)2 + l 2/32 For any real number y, (y - l /8)2 ≥ 0 Hence, when y = l /8, the minimum total area of the two squares = l 2/32 2010-08-23 16:22:41 補充： Q.4 x - y + 6 = 0 可寫成 y = x + 6 x - y - 6 = 0 可寫成 y = x - 6
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F4 Math - Equations發問:
http://img840.imageshack.us/img840/9651/ch2b.png Four questions are written on the pic. Please help me to solve them by clear steps n show the steps. Thanks : ) 更新: Answers 1) 4cm 2) l^2/32 3) (√3)x - y - 2√3 = 0 4) y = x ±6
最佳解答:
1. Let AP = y AP2 + PB2 = y2 + (8 - y2) = y2 + (64 - 16y + y2) = 2y2 - 16y + 64 = 2(y2 - 8y) + 64 = 2(y2 - 8y + 16) - 32 + 64 = 2(y - 4)2 + 32 For any real number y, (y - 4)2 ≥ 0 Hence, AP2 + PB2 is a minimum when (y - 4)2 = 0 i.e. Distance of P from A = y = 4 2) Let y be the length of one of the pieces of strength. Total area of the two squares = y2 + (l - y)2 = y2 + l 2 - 2 l y + y2 = 2y2 - 2l y + l 2 = 2(y2 - l y) + l 2 = 2(y2 - l y + l /4) - 2l 2/4+ l 2 = 2(y - l /2)2 + l 2/2 For any real number y, (y - l /2)2 ≥ 0 Hence, when y = l /2, the minimum total area of the two squares = l 2/2 3) cosθ = 1/2 sinθ = √(1 - cos2θ) = √[1 - (1/2)2] = (√3)/2 Slope of the line = tanθ = sinθ/cosθ = √3 x-intercept = 2 The line passes through (2, 0) Equation of the line: (y - 0) = (√3)(x - 2) (√3)x - y - 2√3 = 0 4) Let A = (a, 0) and B = (0, b) Slope of the line l = 1 (0 - b)/(a - 0) = 1 a = -b …… (1) Area of the ΔOAB = 18 (1/2)|ab| = 18 ab = -36 …… (2) or ab = 36 (rejected for a = -b) Subst. (1) into (2) (-b)b = -36 b = 6 or b = -6 When b = 6, B(0, 6) Hence, equation of the line l : y - 6 = 1(x - 0) x - y + 6 = 0 When b = -6, B(0, -6) Hence, equation of the line l : y + 6 = 1(x - 0) x - y - 6 = 0 2010-08-23 01:13:18 補充： In Q.1, the unit is "cm". 2010-08-23 01:22:42 補充： Amendment of Q.2 Total area of the two squares = y2 + [(l - 4y)/4]2 = 2y2 - l y/2 + l 2/16 = 2(y2 - l y/4 + l 2/64) - 2l 2/64+ l 2/16 = 2(y - l /8)2 + l 2/32 For any real number y, (y - l /8)2 ≥ 0 Hence, when y = l /8, the minimum total area of the two squares = l 2/32 2010-08-23 16:22:41 補充： Q.4 x - y + 6 = 0 可寫成 y = x + 6 x - y - 6 = 0 可寫成 y = x - 6
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