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Degger [83]
2 years ago
15

A paper cup has the shape of a cone with height 10 cm and radius 3 cm (at the top. if water is poured into the cup at a rate of

2cm3/s, how fast is the water level rising when the water is 5 cm deep?
Mathematics
1 answer:
Bingel [31]2 years ago
4 0
Let 
 h: height of the water
 r: radius of the circular top of the water 
 V: the volume of water in the cup.
 We have:
 r/h = 3/10
 So,
 r = (3/10)*h
 the volume of a cone is: 
 V = (1/3)*π*r^2*h
 Rewriting:
 V (t) = (1/3)*π*((3/10)*h(t))^2*h(t)
 V (t) =(3π/100)*h(t)^3
 Using implicit differentiation:
 V'(t) = (9π/100)*h(t)^2*h'(t)
 Clearing h'(t)
 h'(t)=V'(t)/((9π/100)*h(t)^2)
 the rate of change of volume is V'(t) = 2 cm3/s when h(t) = 5 cm.
 substituting:
 h'(t) = 8/(9π) cm/s
 Answer: 
 the water level is rising at a rate of: 
 h'(t) = 8/(9π) cm/s

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Answer:

(i) Debemos apilar 5 dados para construir el cubo de mayor tamaño.

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Step-by-step explanation:

(i) Sabemos por la Geometría Euclídea del Espacio que un cubo es un sólido regular con 6 caras cuadradas y longitudes iguales. Cada dado tiene un volumen de 1 dado cúbico y 125 dados dan un volumen total de 125 dados cúbicos.

El volumen de un cubo está dado por la siguiente fórmula:

V = L^{3}

Donde:

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V - Volumen del cubo, medido en dados cúbicos.

Ahora, necesitamos despejar la longitud de la arista para calcular la altura máxima posible:

L = \sqrt[3]{V}

Dado que V = 125\,dados^{3}, encontramos que la altura del cubo de mayor tamaño sería:

L =\sqrt[3]{125\,dados^{3}}

L = 5\,dados

Debemos apilar 5 dados para construir el cubo de mayor tamaño.

(ii) El área cuadrada formada por cubos está determinada por la siguiente fórmula:

A = L^{2}

Donde:

L - Longitud de arista, medida en dados.

A - Área, medida en dados cuadrados.

Puesto que la longitud de arista se basa en un conjunto discreto, esto es, el número de dados disponibles, debemos encontrar el valor máximo de L tal que no supere 125 y de un área entera. Es decir:

L \leq 125\,dados

Si cada cubo tiene un área de 1 dado cuadrado, entonces un cuadrado conformado por 125 dados tiene un área total de 125 dados cuadrados. Entonces:

L^{2}< 125\,dados^{2}

Esto nos lleva a decir que:

L < 11.180\,dados

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n = (11\,dados)^{2}

n = 121\,dados

Se necesita 121 dados cuadrados para formar el cuadrado con la mayor cantidad de dados posibles, quedando 4 dados sobrantes.

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Answer:

See diagram attached.

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