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# Ejercicio de razon de cambio3

Module by: DIEGO DIAZ LEON. E-mail the author

Summary: Ejercicio de razon de cambio

RAZON DE CAMBIO

2. Un recipiente tiene la forma de un cono circular recto invertido y la longitud de su altura es el doble de la de su diametro. Al recipiente le esta entrando agua a una rapidez constante, por lo que la profundidad del agua va en aumento. cuando la profundidad es de 1m, la superficie sube a razon de 1 cm po minuto.?`a que rapidez la esta entrando agua al recipiente?

v = π 3 r 2 h v = π 3 r 2 h
(1)
r h = R H r h = R 4 R r = h 4 r h = R H r h = R 4 R r = h 4
(2)
v = π 3 ( h 4 ) 2 h π 48 h 3 v = π 3 ( h 4 ) 2 h π 48 h 3
(3)
dv dt = π 48 3 h 2 ( dh dt ) π 16 h 2 ( dh dt ) dv dt = π 48 3 h 2 ( dh dt ) π 16 h 2 ( dh dt )
(4)

en el instante en que h=100cm=10dm y dh/dt=0.1dm/min:

dv dt = π 16 h 2 ( dh dt ) π 16 ( 10 dm ) 2 ( 0 . 1 dm / min ) 10 π 16 dm 3 / min dv dt = π 16 h 2 ( dh dt ) π 16 ( 10 dm ) 2 ( 0 . 1 dm / min ) 10 π 16 dm 3 / min
(5)
dv dt = 5 π 8 l / m 1 . 963 l / min dv dt = 5 π 8 l / m 1 . 963 l / min
(6)

DIEGO DIAZ LEON

CALCULO DIFERENCIAL

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