Lang Frazier
11/09/2023 · Elementary School
Aproxima el valor de \( \sqrt{36.12} \) haciendo uso de diferenciales.
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Step-by-step Solution
Para aproximar el valor de \( \sqrt{36.12} \) usando diferenciales, primero identificamos un valor cercano a \( 36.12 \) que sea fácil de calcular. En este caso, \( 36 \) es un buen candidato, ya que \( \sqrt{36} = 6 \).
Definimos la función \( f(x) = \sqrt{x} \). Luego, calculamos la derivada de \( f \):
\[
f'(x) = \frac{1}{2\sqrt{x}}
\]
Evaluamos la derivada en \( x = 36 \):
\[
f'(36) = \frac{1}{2\sqrt{36}} = \frac{1}{2 \cdot 6} = \frac{1}{12}
\]
Ahora, utilizamos la aproximación lineal. La fórmula para la aproximación lineal es:
\[
f(x) \approx f(a) + f'(a)(x - a)
\]
donde \( a = 36 \) y \( x = 36.12 \). Entonces, tenemos:
\[
f(36) = 6
\]
\[
x - a = 36.12 - 36 = 0.12
\]
Sustituyendo en la fórmula de la aproximación:
\[
f(36.12) \approx f(36) + f'(36)(36.12 - 36)
\]
\[
f(36.12) \approx 6 + \frac{1}{12} \cdot 0.12
\]
Calculamos \( \frac{1}{12} \cdot 0.12 \):
\[
\frac{1}{12} \cdot 0.12 = \frac{0.12}{12} = 0.01
\]
Por lo tanto, la aproximación es:
\[
f(36.12) \approx 6 + 0.01 = 6.01
\]
Así que, usando diferenciales, podemos concluir que:
\[
\sqrt{36.12} \approx 6.01
\]
Quick Answer
\[
\sqrt{36.12} \approx 6.01
\]
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