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Un objeto se mueve según la trayectoria \( \vec{r}(t)=\left(3 t^{2}\right) i+(4 t) j+\left(t^{3}-t\right) k \), donde \( t \) está en segundos. ¿Cuál es la velocidad instantánea del objeto en \( t=1 \) segundo? \( 6 i+3 j+1 k \) \( 6 i+4 j+2 k \) \( 6 i+4 j+0 k \)

Ask by Stewart Martin. in Mexico
Mar 15,2025

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La velocidad instantánea del objeto en \( t = 1 \) segundo es \( 6 \, \mathbf{i} + 4 \, \mathbf{j} + 2 \, \mathbf{k} \).

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Mind Expander

Para encontrar la velocidad instantánea del objeto, necesitamos calcular la derivada de la posición \( \vec{r}(t) \) con respecto al tiempo \( t \): \[ \vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt}\left(3 t^{2}\right) i + \frac{d}{dt}\left(4 t\right) j + \frac{d}{dt}\left(t^{3} - t\right) k \] Calculando cada componente: - Componente \( i \): \( \frac{d}{dt}(3 t^{2}) = 6t \) - Componente \( j \): \( \frac{d}{dt}(4 t) = 4 \) - Componente \( k \): \( \frac{d}{dt}(t^{3} - t) = 3t^{2} - 1 \) Entonces, la velocidad \( \vec{v}(t) \) es: \[ \vec{v}(t) = (6t) i + (4) j + (3t^{2} - 1) k \] Ahora, evaluamos en \( t = 1 \): \[ \vec{v}(1) = 6(1) i + 4 j + (3(1)^{2} - 1) k = 6i + 4j + (3 - 1)k = 6i + 4j + 2k \] Por lo tanto, la velocidad instantánea del objeto en \( t = 1 \) segundo es \( 6 i + 4 j + 2 k \). ¡La respuesta correcta es \( 6 i + 4 j + 2 k \)!

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