Find the acceleration for the given position vector, $\vec { r } ( t ) = \left\langle 2 t , \left( 1 + t ^ { 2 } \right) ^ { 3 } , t \right\rangle$
A) $\vec { a } ( t ) = \left\langle 2,6 t \left( 1 + t ^ { 2 } \right) ^ { 2 } , t \right\rangle$
B) $\vec { a } ( t ) = \left\langle 0,6 \left( 1 + t ^ { 2 } \right) ^ { 2 } + 24 t ^ { 2 } \left( 1 + t ^ { 2 } \right) , 0 \right\rangle$
C) $\vec { a } ( t ) = \left\langle 0,2 t \left( 1 + t ^ { 2 } \right) , 0 \right\rangle$
D) $\vec { a } ( t ) = \left\langle 0,6 t ^ { 2 } ( 1 + 2 t ) ^ { 2 } , 0 \right\rangle$

Question

Quizplus · Accepted Answer

Acceleration, $\vec{a}(t)$, is the second derivative of the position vector, $\vec{r}(t)$, with respect to time. For $\vec{r}(t) = \langle 2t, (1 + t^2)^3, t \rangle$, the first derivative (velocity) is $\vec{v}(t) = \langle 2, 3(1 + t^2)^2(2t), 1 \rangle$, and the second derivative (acceleration) is $\vec{a}(t) = \langle 0, 6(1 + t^2)^2 + 24t^2(1 + t^2), 0 \rangle$.

Find the Acceleration for the Given Position Vector r⃗(t)=⟨2t,(1+t2)3,t⟩\vec { r } ( t ) = \left\langle 2 t , \left( 1 + t ^ { 2 } \right) ^ { 3 } , t \right\rangler(t)=⟨2t,(1+t2)3,t⟩

Find the Acceleration for the Given Position Vector $\vec { r } ( t ) = \left\langle 2 t , \left( 1 + t ^ { 2 } \right) ^ { 3 } , t \right\rangle$