The speed of a particle moving along the curve $\mathbf { r } ( t ) = 2 t \mathbf { i } - \cos ( 3 t ) \mathbf { j } + \sin ( 3 t ) \mathbf { k }$ is

A) $\sqrt { 8 t ^ { 2 } - 1 }$
B) $\sqrt { 11 }$
C) $\sqrt { 8 t ^ { 2 } + 1 }$
D) $\sqrt { 5 }$
E) $\sqrt { 13 }$

Question

The speed of a particle moving along the curve $\mathbf { r } ( t ) = 2 t \mathbf { i } - \cos ( 3 t ) \mathbf { j } + \sin ( 3 t ) \mathbf { k }$ is

A) $\sqrt { 8 t ^ { 2 } - 1 }$
B) $\sqrt { 11 }$
C) $\sqrt { 8 t ^ { 2 } + 1 }$
D) $\sqrt { 5 }$
E) $\sqrt { 13 }$

Quizplus · Accepted Answer

The speed of a particle is the magnitude of its velocity vector. The velocity vector is the derivative of the position vector $\mathbf{r}(t)$. Differentiating $\mathbf{r}(t)$ gives $\mathbf{v}(t) = 2 \mathbf{i} + 3 \sin(3t) \mathbf{j} + 3 \cos(3t) \mathbf{k}$. The magnitude of $\mathbf{v}(t)$ is $\sqrt{2^2 + (3 \sin(3t))^2 + (3 \cos(3t))^2} = \sqrt{4 + 9(\sin^2(3t) + \cos^2(3t))} = \sqrt{4 + 9} = \sqrt{13}$.

The Speed of a Particle Moving Along the Curve r(t)=2ti−cos⁡(3t)j+sin⁡(3t)k\mathbf { r } ( t ) = 2 t \mathbf { i } - \cos ( 3 t ) \mathbf { j } + \sin ( 3 t ) \mathbf { k }r(t)=2ti−cos(3t)j+sin(3t)k

The Speed of a Particle Moving Along the Curve $\mathbf { r } ( t ) = 2 t \mathbf { i } - \cos ( 3 t ) \mathbf { j } + \sin ( 3 t ) \mathbf { k }$