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Cha456-1 - Chapter 4 Chapter 5 Joy Zhou Use as a reference...

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Chapter 4, Chapter 5 Joy Zhou 10-12-2009 Use as a reference only. Please DO NOT distribute outside class. Chapter 4 4.9, 4.11, 4.12, 4.14, 4.15 4.9 Following the hint of the problem, let us suppose a is the x -coordinate of the point where line of sight #1 is tangent to the silo. Let us also call that tangent point A. Since point A is on the circle x 2 + y 2 = 8 2 , we can calculate its y -coordinate as follows: replace x with a in the equation x 2 + y 2 = 8 2 ,and get a 2 + y 2 = 64. Move constants to the right hand side to solve for y: y 2 = 64 - a 2 . Taking square roots of both sides: y = 64 - a 2 (we are looking at the tangent point in the upper-half plane so we kept the positive square root.) Therefore the coordinates of A are ( a, 64 - a 2 ). The slope of the line of sight #1 is 64 - a 2 - 0 a - 12 = 64 - a 2 a - 12 (because it passes through ( a, 64 - a 2 ) and (12,0)). The slope of the line 1
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OA where O is the origin (0,0) is 64 - a 2 - 0 a - 0 = 64 - a 2 a . Since the line of sight #1 is perpendicular to the line OA, their slopes are negative reciprocals of each other, meaning 64 - a 2 a - 12 = - a 64 - a 2 . From there, multiplying both sides by a - 12 and 64 - a 2 gives you 64 - a 2 · 64 - a 2 = - a · ( a - 12). Therefore 64 - a 2 = 12 a - a 2 64 = 12 a a = 64 12 = 16 3 So the slope of the line of sight #1 is 64 - a 2 a - 12 = 64 - ( 16 3 ) 2 16 3 - 12 = - 2 5 5 . There-
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