This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ....
View
Full
Document
This note was uploaded on 12/28/2011 for the course MATH 116 taught by Professor Robertandre during the Spring '11 term at Waterloo.
 Spring '11
 RobertAndre
 Calculus, Rate Of Change

Click to edit the document details