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svetlana [45]
2 years ago
7

Se desea construir un vaso de papel en forma de cono circular recto que tenga un volumen de 25πcm3

Mathematics
1 answer:
Genrish500 [490]2 years ago
5 0

Answer:

Para un vaso de V = 25\pi\,cm^{3}, las dimensiones del vaso son r \approx 2.321\,cm y h \approx 4.642\,cm.

Para un vaso de V = 1000\,cm^{3}, las dimensiones del vaso son r \approx 5.419\,cm y h \approx 10.839\,cm.

Step-by-step explanation:

El vaso se puede modelar como un cilindro recto. El enunciado pregunta por las dimensiones del vaso tal que su área superficial (A_{s}), en centímetros cuadrados, sea mínima para el volumen dado (V), en centímetros cúbicos. Las ecuaciones de volumen y área superficial son, respectivamente:

V = \pi\cdot r^{2}\cdot h (1)

A_{s} = 2\pi\cdot r^{2} + 2\pi\cdot r\cdot h (2)

De (1):

h = \frac{V}{\pi\cdot r^{2}}

En (2):

A_{s} = 2\pi\cdot r^{2} + 2\pi\cdot \left(\frac{V}{\pi\cdot r} \right)

A_{s} = 2\cdot \left(\pi\cdot r^{2}+V\cdot r^{-1} \right)

Asumamos que V es constante, la primera y segunda derivadas de la función son, respectivamente:

A'_{s} = 2\cdot (2\pi\cdot r -V\cdot r^{-2})

A'_{s} = 4\pi\cdot r - 2\cdot V\cdot r^{-2} (3)

A''_{s} = 4\pi + 4\cdot V \cdot r^{-3} (4)

Si igualamos A'_{s} a cero, entonces hallamos los siguientes puntos críticos:

4\pi\cdot r - 2\cdot V\cdot r^{-2} = 0

4\pi\cdot r = 2\cdot V\cdot r^{-2}

4\pi\cdot r^{3} = 2\cdot V

r^{3} = \frac{V}{2\pi}

r = \sqrt[3]{\frac{V}{2\pi} } (5)

Ahora, si aplicamos este valor a (4), tenemos que:

A_{s}'' = 4\pi + \frac{4\cdot V}{\frac{V}{2\pi} }

A''_{s} = 4\pi + 8\pi

A_{s}'' = 12\pi (6)

De acuerdo con este resultado, el valor crítico está asociado al área superficial mínima. Ahora, la altura se calcula a partir de (5) y (1):

h = \frac{V}{\pi\cdot \left(\frac{V}{2\pi} \right)^{2/3} }

h = \frac{2^{2/3}\cdot \pi^{2/3}\cdot V}{\pi\cdot V^{2/3}}

h = \frac{2^{2/3}\cdot V^{1/3}}{\pi^{1/3}}

Si V = 25\pi\,cm^{3}, entonces las dimensiones del vaso son:

r = \sqrt[3]{\frac{25\pi\,cm^{3}}{2\pi} }

r \approx 2.321\,cm

h = \frac{2^{2/3}\cdot (25\pi\,cm^{3})^{1/3}}{\pi^{1/3}}

h \approx 4.642\,cm

Un litro equivale a 1000 centímetros cúbicos, las dimensiones del vaso son:

r = \sqrt[3]{\frac{1000\,cm^{3}}{2\pi} }

r \approx 5.419\,cm

h = \frac{2^{2/3}\cdot (1000\,cm^{3})^{1/3}}{\pi^{1/3}}

h \approx 10.839\,cm

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<h3>What exactly would the standard deviation indicate?</h3>

The term "standard deviation" (or "") refers to the degree of dispersion of the data from the mean. Data are grouped around the mean when the standard deviation is low, and are more dispersed when the standard deviation is high.

<h3>According to given information:</h3>

The mean is the product of the dataset's total and the sample size. Mathematically.

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The standard deviation:

\sigma=\sqrt{\frac{\sum(x-\bar{x})^2}{N}}

\sum(x-\bar{x})^2 = (1.48-1.292)^2+(1.37-1.292)^2+(0.98-1.292)^2+(1.31-1.292)^2+(1.59-1.292)^2+(1.55-1.292)^2.

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Putting into the formula we get:

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σ = 0.2098

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To know more about standard deviation visit:

brainly.com/question/18521100

#SPJ4

I understand that the question you are looking for is:

You measure the period of a mass oscillating on a vertical spring ten times as follows:

Period (s): 1.06, 1.31, 1.28, 0.99, 1.48, 1.37, 0.98, 1.31, 1.59, 1.55

Required:

What are the mean and (sample) standard deviation?

a. Mean: 1.228, Standard Deviation: 0.2135

b. Mean: 1.325, Standard Deviation: 0.1674

c. Mean: 1.292. Standard Deviation: 0.2211

d. Mean: 1.228, Standard Deviation: 0.2098

e. Mean: 1.292, Standard Deviation: 0.2135

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