Cummings Johnson
02/25/2023 · Elementary School
Calcule la longitud de las siguientes curvas: \( x=3 t-t^{3}, \quad y=3 t^{2}, \quad 0 \leq t \leq 2 \)
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Step-by-step Solution
Para calcular la longitud de la curva dada por las ecuaciones paramétricas \( x=3t-t^{3} \) y \( y=3t^{2} \) en el intervalo \( 0 \leq t \leq 2 \), utilizaremos la fórmula de la longitud de una curva paramétrica:
\[ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^{2} + \left(\frac{dy}{dt}\right)^{2}} dt \]
Donde:
- \( x = 3t - t^{3} \)
- \( y = 3t^{2} \)
- \( a = 0 \)
- \( b = 2 \)
Primero, calculamos las derivadas de \( x \) y \( y \) con respecto a \( t \):
\[ \frac{dx}{dt} = 3 - 3t^{2} \]
\[ \frac{dy}{dt} = 6t \]
Sustituimos estas derivadas en la fórmula de la longitud de la curva y evaluamos la integral en el intervalo dado. Vamos a calcular esto.
Evaluate the integral by following steps:
- step0: Evaluate using formulas and rules:
\(\int_{0}^{2} \sqrt{\left(3-3t^{2}\right)^{2}+\left(6t\right)^{2}} dt\)
- step1: Add the terms:
\(\int_{0}^{2} \sqrt{9+18t^{2}+9t^{4}} dt\)
- step2: Evaluate the power:
\(\int_{0}^{2} 3\left(1+t^{2}\right) dt\)
- step3: Simplify:
\(\int_{0}^{2} 3+3t^{2} dt\)
- step4: Evaluate the integral:
\(\int 3+3t^{2} dt\)
- step5: Use properties of integrals:
\(\int 3 dt+\int 3t^{2} dt\)
- step6: Evaluate the integral:
\(3t+t^{3}\)
- step7: Return the limits:
\(\left(3t+t^{3}\right)\bigg |_{0}^{2}\)
- step8: Calculate the value:
\(14\)
La longitud de la curva dada por las ecuaciones paramétricas \( x=3t-t^{3} \) y \( y=3t^{2} \) en el intervalo \( 0 \leq t \leq 2 \) es de 14 unidades.
Quick Answer
La longitud de la curva es de 14 unidades.
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