The tangential component $a _ { \mathrm { T } } ( t )$ of the acceleration of a particle moving along the curve $\mathbf { r } ( t ) = t ^ { 2 } \mathbf { i } + \left( 2 t ^ { 2 } - 1 \right) \mathbf { j } - 3 t ^ { 2 } \mathbf { k }$ is

A) $2 \sqrt { 14 }$
B) $\sqrt { 58 }$
C) $\sqrt { 65 }$
D) $\sqrt { 73 }$
E) $\sqrt { 78 }$

Question

The tangential component

a _ { \mathrm { T } } ( t )

of the acceleration of a particle moving along the curve

\mathbf { r } ( t ) = t ^ { 2 } \mathbf { i } + \left( 2 t ^ { 2 } - 1 \right) \mathbf { j } - 3 t ^ { 2 } \mathbf { k }

is

A) $2 \sqrt { 14 }$
B) $\sqrt { 58 }$
C) $\sqrt { 65 }$
D) $\sqrt { 73 }$
E) $\sqrt { 78 }$

Quizplus · Accepted Answer

The tangential component of acceleration is given by $$a_T(t) = \frac{d}{dt} \left[ \frac{ds}{dt} \right] = \frac{d}{dt} \left[ \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2 } \right]$$$$= \frac{1}{ds/dt} \left[ \frac{dx}{dt} \frac{d^2x}{dt^2} + \frac{dy}{dt} \frac{d^2y}{dt^2} + \frac{dz}{dt} \frac{d^2z}{dt^2} \right]$$$$= \frac{1}{\sqrt{ (2t)^2 + (4t)^2 + (-6t)^2 }} \left[ 2t(2) + 4t(4) - 6t(-6) \right]$$$$= \frac{1}{\sqrt{ 4t^2 + 16t^2 + 36t^2 }} \left[ 4t + 16t + 36t \right]$$$$= \frac{1}{\sqrt{ 56t^2 }} \left[ 56t \right]$$$$= \frac{56t}{\sqrt{ 56t^2 }} = \frac{56t}{56t} = 1$$Therefore, the tangential component of acceleration is $a_T(t) = 1$.

The Tangential Component aT(t)a _ { \mathrm { T } } ( t )aT​(t)

The Tangential Component $a _ { \mathrm { T } } ( t )$