sample-midterm2-with-solutions-v2

# We can use similar triangles x 20 to relate x and y

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Unformatted text preview: eters Let this be length of his shadow, Say y meters W WL represents the distance of 20 meters. L M Let the distance of the man from the spotlight be x meters dx dy when x = 5, that is, = 2.5m / s and we need to find the value of dt dt when the man is 15 meters from the building. We can use similar triangles x 20 to relate x and y as follows: = Þ xy = 40. (When x is 5, y is 8). 2y dy dx Differentiating both sides, we get: x + y =0 dt dt dy dy -(8)(2.5) dy Þ 5 + 8(2.5) = 0 Þ = Þ = -4 dt dt 5 dt Hence the man’s shadow is decreasing at 4 m/s when he is 15 meters from the building. 5 Q11. For what values of the constants a and b is (1,6) a point of inflection of the curve y = x 3 + ax 2 + bx + 1? Solution f '( x) = y ' = 3 x 2 + 2ax + b. f ''( x) = 6 x + 2a. Since (1,6) is a point of inflection, f ''(1) = 0 Þ 6(1) + 2a = 0 Þ 2a = -6 Þ a = -3 Since (1,6) is a point on the graph, it satisfies the equation of the graph. Hence, f (1) = 6 Þ 13 + a ×12 + b ×1 + 1 = 6 Þ a+b = 4 \ b = 4 - a = 4 - (-3) =...
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## This document was uploaded on 02/19/2014 for the course MATH Math 113/1 at Grant MacEwan University.

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