# HW8sol - M408C Homework 8 Solution Instructor Guoliang Wu 1...

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M408C, Homework 8 — Solution Instructor: Guoliang Wu 1. A fence 8 ft tall runs parallel to a tall building at a distance of 4 ft from the building. What is the length of the shortest ladder that will reach from the ground over the fence to the wall of the building? Solution: Suppose the bottom of the ladder is x ft away from the fence, and its top is y ft from the ground. Then, in terms of x and y , the length of the ladder is x 2 + y 2 . But 8 y = x x + 4 y = 8( x + 4) x . Therefore, the square f ( x ) of the length is f ( x ) = ( x + 4) 2 + ( 8( x + 4) x ) 2 = ( x + 4) 2 + 64 x - 2 ( x + 4) 2 , x > 0 To minimize the length of ladder, it’s the same to minimize its square f ( x ). f 0 ( x ) = 2( x + 4) + 128 x - 2 ( x + 4) - 128 x - 3 ( x + 4) 2 = 2 x - 3 ( x + 4)( x 3 - 256) = 0

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Unformatted text preview: ⇒ x = 4 3 √ 4 . 1 If x < 4 3 √ 4, f ( x ) < 0, and if x > 4 3 √ 4, f ( x ) > 0. Thus, f ( x ) has an absolute minimum at x = 4 3 √ 4. In this case the length of the ladder is √ f ( x ) = ± (4 3 √ 4 + 4) 2 + 64(4 3 √ 4)-2 (4 3 √ 4 + 4) 2 ≈ 14 . 5ft 2. Find f ( θ ) that satisﬁes f 00 ( θ ) = sin θ + cos θ, f (0) = 3 , f (0) = 4 . Solution: f ( θ ) =-cos θ + sin θ + C 1 f ( θ ) =-sin θ-cos θ + C 1 θ + C 2 Since f (0) =-1 + C 2 = 3, C 2 = 4. And f (0) =-1 + C 1 = 4, C 1 = 5. Therefore, f ( θ ) =-sin θ-cos θ + 3 θ + 4 . 2...
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• Fall '06
• Calculus, 8 FT, 4 ft, ladder, ft tall runs, Guoliang Wu

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