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Maths problems!!! please help!!! http://www.uwants.com/attachment.php?aid=3857519
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1(i) When x = a , (x – a)/2 = 0 lim (sin x –sin a) / (x – a) x-)a = lim 2cos[(x+a)/2] sin[(x-a)/2] / [2(x-a) / 2] x - )a = cos (a+a)/2 = cos a (ii) lim [x + 開方(3x^4 – 5x^2 + 121)] / [6x^2 + 開方(5x^4 + x^3 +1) –24] x-)oo = lim [1/x + 開方(3 – 5/x^2 + 121/x^4)] / [6 + 開方(5 + 1/x +1/x^4) –24/x^4] x-)oo = [0 + 開方(3-0+0)] / [6 + 開方(5+0+0) – 0] = 開方3 / (6 + 開方5) = (6.開方3 – 開方15) / 31 2(i) b^2 x^2 + a^2 y^2 = a^2 b^2 2b^2 (x) + 2a^2 (y)dy/dx = 0 dy/dx = - (b^2)(x) / (a^2)(y) d^2y/dx^2 = - [a^2 (y)(b^2) – b^2 (x)(a^2)(dy/dx)] / (a^4)(y^2) = b^2 [x(dy/dx) – y] / (a^2)(y^2) = [b^2 (-b^2)( x^2) / (a^2)(y) – y] / (a^2)(y^2) = [- (b^4)(x^2) – (a^2)(y^2)] / (a^4)(y^3) (ii) y= xe^(x^2) dy/dx = e^(x^2) + x[e^(x^2)(2x)] = e^(x^2)[1 + 2x^2] d2y/d2x = 4xe^(x^2) + ( 1 + 2x^2)[e^(x^2)(2x)] = e^(x^2)[4x + 2x(1+ 2x^2)] = 2xe^(x^2)[2 + 1 + 2x^2] = 2x(3 + 2x^2)e^(x^2) 3. Let A be the enclosed area. x + x + y = 500 y = 500 – 2x A = xy = x(500 – 2x) = 500x – 2x^2 dA/dx = 500 – 4x d^2 A / dx^2 = - 4 < 0 When dA/dx = 0, 500 – 4x = 0 x = 125 When x = 125 , y = 500 – 2(125) = 250 Therefore , The enclosed area is at maximum when x = 125 feet and y = 250 feet.
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Maths problems!!!發問:
Maths problems!!! please help!!! http://www.uwants.com/attachment.php?aid=3857519
最佳解答:
1(i) When x = a , (x – a)/2 = 0 lim (sin x –sin a) / (x – a) x-)a = lim 2cos[(x+a)/2] sin[(x-a)/2] / [2(x-a) / 2] x - )a = cos (a+a)/2 = cos a (ii) lim [x + 開方(3x^4 – 5x^2 + 121)] / [6x^2 + 開方(5x^4 + x^3 +1) –24] x-)oo = lim [1/x + 開方(3 – 5/x^2 + 121/x^4)] / [6 + 開方(5 + 1/x +1/x^4) –24/x^4] x-)oo = [0 + 開方(3-0+0)] / [6 + 開方(5+0+0) – 0] = 開方3 / (6 + 開方5) = (6.開方3 – 開方15) / 31 2(i) b^2 x^2 + a^2 y^2 = a^2 b^2 2b^2 (x) + 2a^2 (y)dy/dx = 0 dy/dx = - (b^2)(x) / (a^2)(y) d^2y/dx^2 = - [a^2 (y)(b^2) – b^2 (x)(a^2)(dy/dx)] / (a^4)(y^2) = b^2 [x(dy/dx) – y] / (a^2)(y^2) = [b^2 (-b^2)( x^2) / (a^2)(y) – y] / (a^2)(y^2) = [- (b^4)(x^2) – (a^2)(y^2)] / (a^4)(y^3) (ii) y= xe^(x^2) dy/dx = e^(x^2) + x[e^(x^2)(2x)] = e^(x^2)[1 + 2x^2] d2y/d2x = 4xe^(x^2) + ( 1 + 2x^2)[e^(x^2)(2x)] = e^(x^2)[4x + 2x(1+ 2x^2)] = 2xe^(x^2)[2 + 1 + 2x^2] = 2x(3 + 2x^2)e^(x^2) 3. Let A be the enclosed area. x + x + y = 500 y = 500 – 2x A = xy = x(500 – 2x) = 500x – 2x^2 dA/dx = 500 – 4x d^2 A / dx^2 = - 4 < 0 When dA/dx = 0, 500 – 4x = 0 x = 125 When x = 125 , y = 500 – 2(125) = 250 Therefore , The enclosed area is at maximum when x = 125 feet and y = 250 feet.
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