Patel Rodriquez
03/11/2024 · Senior High School

Un vaso de papel con agua tiene la forma de un cono circular recto truncado de 15 cm de altura con radios de la base y de la orilla libre de 2 cm y 4 cm , respectivamente. El agua se fuga del va- so a razón de \( 100 \mathrm{~cm}^{3} / \mathrm{h} \). ¿A razón de cuántos centimetros por hora disminuye la profundidad del agua cuando es de 10 cm ? (Nota: El volumen \( V \) de un cono circular recto truncado de altura \( h y \) radios \( a \) y \( b \) en los extremos está dado por \( V=\frac{1}{3} \pi h\left(a^{2}+b^{2}+a b\right) \) ).

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La profundidad del agua disminuye aproximadamente a razón de \( 4.38 \, \text{cm/h} \) cuando es de 10 cm.

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