標題:
中5 Amath 兩題 15分 幫幫手
發問:
1.For the baseball diamond shown in the figure below, suppose the player is running from first to second at a speed of 25 feet per second. Find the rate at which the distance from home plate is changing when the player is 35 feet from second base. (Round to two decimal places.)... 顯示更多 1. For the baseball diamond shown in the figure below, suppose the player is running from first to second at a speed of 25 feet per second. Find the rate at which the distance from home plate is changing when the player is 35 feet from second base. (Round to two decimal places.) [/img]http://www.webassign.net/larson/2_06-34.gif[img] 2. A certain water fountain has a conical water cup. The height of the cup is 2.75 inches and a radius of 0.9 inches. The water cup that I get has a hole in the bottom that leaks water at a rate of 0.2 cubic inches. How fast is the water level dropping when the height of the water is at 1 inch.
1. Let P be a point between first and second bases, its distance from second base be x feet and its distance from home be y. y2 = 902 + ( 90 – x )2 Differentiate wrt t : 2y dy/dt = 2( 90 – x ) (–dx/dt ) dy/dt = –( 90 – x ) / y dx/dt When x = 35 dx/dt = –25 y = √[902 + ( 90 – 35 )2] = √11125 = 105.475 dy/dt = –( 90 – 35 ) / 105.475 x (–25) = 13.036 The rate at which the distance from home plate is changing is 13.04 feet per second. 2. The water cup that I get has a hole in the bottom that leaks water at a rate of 0.2 cubic inches per second. Let the volume of the conical water cup be V cubic inches, the height be h inches and the radius be r inches at a certain moment. r / h = 0.9 / 2.75 r = ( 18 / 55 ) h V = (1/3) (pi) r2 h = (1/3) (pi) (18/55)2 h3 = (108/3025) (pi) h3 dV/dt = (108/3025) (pi) (3h2 dh/dt) = (324/3025) (pi) h2 dh/dt dh/dt = 3025/[324 (pi) h2] dV/dt When h = 1 dV/dt = –0.2 dh/dt = 3025/[324 (pi)] x (–0.2) = –0.5944 the water level dropping at a speed of 0.59 inches per second.
其他解答:63D0B758E2D502CC
中5 Amath 兩題 15分 幫幫手
發問:
1.For the baseball diamond shown in the figure below, suppose the player is running from first to second at a speed of 25 feet per second. Find the rate at which the distance from home plate is changing when the player is 35 feet from second base. (Round to two decimal places.)... 顯示更多 1. For the baseball diamond shown in the figure below, suppose the player is running from first to second at a speed of 25 feet per second. Find the rate at which the distance from home plate is changing when the player is 35 feet from second base. (Round to two decimal places.) [/img]http://www.webassign.net/larson/2_06-34.gif[img] 2. A certain water fountain has a conical water cup. The height of the cup is 2.75 inches and a radius of 0.9 inches. The water cup that I get has a hole in the bottom that leaks water at a rate of 0.2 cubic inches. How fast is the water level dropping when the height of the water is at 1 inch.
此文章來自奇摩知識+如有不便請留言告知
最佳解答:1. Let P be a point between first and second bases, its distance from second base be x feet and its distance from home be y. y2 = 902 + ( 90 – x )2 Differentiate wrt t : 2y dy/dt = 2( 90 – x ) (–dx/dt ) dy/dt = –( 90 – x ) / y dx/dt When x = 35 dx/dt = –25 y = √[902 + ( 90 – 35 )2] = √11125 = 105.475 dy/dt = –( 90 – 35 ) / 105.475 x (–25) = 13.036 The rate at which the distance from home plate is changing is 13.04 feet per second. 2. The water cup that I get has a hole in the bottom that leaks water at a rate of 0.2 cubic inches per second. Let the volume of the conical water cup be V cubic inches, the height be h inches and the radius be r inches at a certain moment. r / h = 0.9 / 2.75 r = ( 18 / 55 ) h V = (1/3) (pi) r2 h = (1/3) (pi) (18/55)2 h3 = (108/3025) (pi) h3 dV/dt = (108/3025) (pi) (3h2 dh/dt) = (324/3025) (pi) h2 dh/dt dh/dt = 3025/[324 (pi) h2] dV/dt When h = 1 dV/dt = –0.2 dh/dt = 3025/[324 (pi)] x (–0.2) = –0.5944 the water level dropping at a speed of 0.59 inches per second.
其他解答:63D0B758E2D502CC
文章標籤
全站熱搜
留言列表
