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

Question

Quizplus · Accepted Answer

The velocity vector $\vec{v}(t)$ is found by taking the derivative of the position vector $\vec{r}(t)$ with respect to $t$. For $\vec{r}(t) = \langle 2t, (1 + t^2)^3, t \rangle$, the derivatives are $\frac{d}{dt}(2t) = 2$, $\frac{d}{dt}((1 + t^2)^3) = 3(1 + t^2)^2(2t) = 6t(1 + t^2)^2$, and $\frac{d}{dt}(t) = 1$, resulting in $\vec{v}(t) = \langle 2, 6t(1 + t^2)^2, 1 \rangle$.

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