Let the position function of a particle be $\mathbf { r } ( t ) = \langle 3 \sin 2 t , 2 \cos 2 t , - \sin 4 t \rangle$ . Find the speed of the particle when $t = \frac { \pi } { 4 }$ .
A)1
B)4
C)3
D) $4 \sqrt { 2 }$
E) $\sqrt { 8 }$
F) $\sqrt { 10 }$
G) $\sqrt { 13 }$
H) $\sqrt { 14 }$

Question

Let the position function of a particle be $\mathbf { r } ( t ) = \langle 3 \sin 2 t , 2 \cos 2 t , - \sin 4 t \rangle$ . Find the speed of the particle when $t = \frac { \pi } { 4 }$ .
A)1
B)4
C)3
D) $4 \sqrt { 2 }$
E) $\sqrt { 8 }$ 
F) $\sqrt { 10 }$
G) $\sqrt { 13 }$  
H) $\sqrt { 14 }$

Quizplus · Accepted Answer

The speed of the particle is the magnitude of its velocity vector, which is the derivative of its position vector. So,$$\mathbf { v } ( t ) = \frac { d \mathbf { r } } { d t } = \langle 6 \cos 2 t , - 4 \sin 2 t , - 4 \cos 4 t \rangle$$and$$\Vert \mathbf { v } ( t ) \Vert = \sqrt { 36 \cos ^ { 2 } 2 t + 16 \sin ^ { 2 } 2 t + 16 \cos ^ { 2 } 4 t } = \sqrt { 36 + 16 } = 4$$Therefore, the speed of the particle when $t = \frac { \pi } { 4 }$ is 4.

Let the Position Function of a Particle Be r(t)=⟨3sin⁡2t,2cos⁡2t,−sin⁡4t⟩\mathbf { r } ( t ) = \langle 3 \sin 2 t , 2 \cos 2 t , - \sin 4 t \rangler(t)=⟨3sin2t,2cos2t,−sin4t⟩

Let the Position Function of a Particle Be $\mathbf { r } ( t ) = \langle 3 \sin 2 t , 2 \cos 2 t , - \sin 4 t \rangle$