lab5_solns_f11

lab5_solns_f11 - Math 116 - Lab 5 - Solutions - Fall 2011....

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Unformatted text preview: Math 116 - Lab 5 - Solutions - Fall 2011. 1. Since dK dt = 2 v dv dt , the rate of change of the speed is dv dt =- 3 8 m/s . 2. The volume is given in terms of the radius by V = 4 3 πr 3 . Hence when V = 64 in 3 the radius is r ≈ 2 . 48 inches. The rate of change of the radius is determined from dV dt = 4 πr 2 dr dt to be dr dt = 0 . 0647 in/sec when r = 2 . 48. Then the rate of change of the area is dA dt = 8 πr dr dt = 4 . 033 in 2 / sec. 3. Using the right triangle with vertices the top of the building, the foot of the building and the end of the shadow away from the building, we can write x = 80cot( θ ) , (1) where x denotes the length of the shadow (i.e., the distance between the foot of the building and the end of the shadow away from the building) and θ is the angle of inclination between the sun and the ground. Differentiating this relation, we have dx dt =- 80csc 2 ( θ ) dθ dt . (2) To compute dx dt , we also need to know csc 2 ( θ ) when x = 60. From (1) we find: cot( θ ) = 60 80 = 3 4 ⇒ csc 2 ( θ ) = 1 + cot 2 ( θ ) = 25 16 . (3) Therefore dx dt =- (80) 25 16 (0 . 25) =- 31 . 25 ft / min (4) 4. a) f ( x ) = 2 x 2 + x- 12 = 0 ⇒ x =- 1 ± √ 97 4 . (5) f ( x ) is decreasing when f ( x ) < 0. One can easily check that f ( x ) = > 0 when x <- 1- √ 97 4 < 0 when- 1- √ 97 4 < x <- 1+ √ 97 4 > 0 when x >- 1+ √ 97 4 ....
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This note was uploaded on 12/28/2011 for the course MATH 116 taught by Professor Robertandre during the Spring '11 term at Waterloo.

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lab5_solns_f11 - Math 116 - Lab 5 - Solutions - Fall 2011....

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