Math 103 Selected Homework Solutions, Section 3.9

Transcription

1 Math 103 Selected Homework Solutions, Section Let s be the distance from the base of the light pole to the top of the man s shadow, and the distance from the light pole to the man. 15 s 6 s We know: = 5 ft/s We want to find: ds when = 40 ft Connecting equation: Similar triangles give us s 15 = s. We can rewrite this as 6 6s = 15s 15, or 5 = 3s. Differentiate: 5 = 3 ds Substitute and solve: It turns out that we can solve for ds (the information that = 40 ft is irrelevant). Then: 5 = 3ds 5(5) = 3 ds ds = 25 3 by simply knowing So the tip of the man s shadow is moving at 25 3 feet per second.

2 11. Let P be the starting point of the cars. Let be the distance from P to the car that goes south, and y be the distance from P to the car that goes west. Let z be the distance between the two cars. y P z We know: = 60 mi/h and = 25 mi/h. We want to find: when t = 2 hours. Connecting equation: 2 + y 2 = z 2 (Pythagorean theorem) Differentiate: 2 + 2y = 2z, or equivalently + y = z. Substitute and solve: Two hours after the cars started moving, = 120 miles, y = 50 miles, and z = 130 miles. Thus, we have: + y = z (120)(60) + (50)(25) = (130) = 65 mi/h So the distance between the cars is increasing at 65 miles per hour.

3 13. Suppose the man starts walking north from P at t = 0 min, and the woman starts walking south from Q, 500 feet to the east of P, at t = 5 min. Let be the distance from point P to the man, and let y be the distance from Q to the woman. Let z be the distance from the man to the woman, as in the diagram: P z 500 ft Q y We know: = 4 ft/s and = 5 ft/s. We want to find: when t = 20 min Connecting equation: By the Pythagorean theorem, ( + y) = z 2, or equivalently 2 + 2y + y = z 2. Differentiate: ( y + ) 2y = 2z, which we can simplify to: + ( y + ) + y = z or simplify further to: ( + y) ( + ) = z Substitute and solve: At t = 20 min, = 4800 ft, y = 4500 ft, and z = Thus, we have: ( ( + y) + ) = z ( )(4 + 5) = = 8674 So the distance is increasing at , or approimately 8.99, feet per second.

4 14. (a) Let be the distance from the runner to first base, and let y be the distance from the runner to second base. 2 nd base 90 3 rd base y 1 st base home plate We know: = 24 ft/s (distance from runner to first base is decreasing) We want to find: when = 45 feet Connecting equation: = y 2 (Pythagorean theorem) Differentiate: = 2y, or equivalently = y. Substitute and solve: When = 45, y = Therefore, = y (45)( 24) = (45 5) = So the distance from the runner to second base is decreasing at approimately 10.7 feet per second. Part (b) is very similar, ecept it uses a different triangle in the diagram. However, the answer is evident by symmetry. When the runner is halfway between home plate and first base, his distance from second base is increasing at eactly the rate that his distance from third base is decreasing. Thus, when the runner is halfway to first base, his distance from third base is increasing at feet per second.

5 16. Let r be the length of the rope between the boat and the pulley, and be the distance from the boat to the dock, as in the diagram. pulley r 1 boat We know: dr = 1 m/s We want to find: when = 8 m Connecting equation: = r 2 (Pythagorean theorem) Differentiate: 2 dr + 0 = 2r, or equivalently = r dr. Substitute and solve: When = 8, r = 65, = r dr (8) = ( 65)( 1) = m/s 8 So the boat is approaching the dock at approimately meters per second.

6 19. Let V be the volume of the water in the tank (cm 3 ), and h be the depth of the water in the tank (cm). Let R be the rate at which water is being pumped into the tank (cm 3 /min). We know: dv = 10, R and dh We want to find: R = 20 when h = 200. Connecting equation: The volume of a cone is V = 1(area of base)(height) = 1π ( ) h h, so V = π 27 h3. = π dh h2 9 Differentiate: dv Substitute and solve: dv = π dh h2 9 10, R = π 9 (200)2 (20) R = 10, , 000 π 9 R 289, 000 cm 3 /min

7 24. Let s be the length of the kite string, the distance from the person holding the string to the spot on the ground directly below the kite, and θ the angle between the string and the horizontal. kite s 100 ft θ ground We know: = 8 ft/s We want to find: dθ when s = 200 ft; that is, when = Connecting equation: tan θ = 100 Differentiate: sec 2 (θ) dθ = Substitute and solve: When = 100 3, θ = π, and we have: 6 sec 2 (θ) dθ = ( π ) dθ sec 2 6 = 100(100 3) 2 (8) 4 dθ 3 = 2 75 dθ = 1 50 = 0.02 So the angle is decreasing at 0.02 radians per second.