A particle moves with position function $\mathbf { r } ( t ) = \left( 21 t - 7 t ^ { 3 } - 5 \right) \mathbf { i } + 21 t ^ { 2 } \mathbf { j }$. Find the tangential component of the acceleration vector.
A) $a _ { T } = 5 t$
B) $a _ { T } = 542 t$
C) $a _ { T } = 42 t + 5$
D) $a _ { T } = - 55 t$
E) $a _ { T } = 42 t$

Question

Quizplus · Accepted Answer

The velocity vector is given by the derivative of the position function, i.e. $\mathbf{v}(t) = (21-21t^2)\mathbf{i} + 42t\mathbf{j}$. The acceleration vector is given by the derivative of the velocity function, i.e.  $\mathbf{a}(t) = -42t\mathbf{i} + 42\mathbf{j}$.

The tangential component of the acceleration vector is given by projecting $\mathbf{a}(t)$ onto the unit tangent vector, which is $\frac{\mathbf{v}(t)}{\|\mathbf{v}(t)\|}$. Thus, we have

$$a_T = \mathbf{a}(t) \cdot \frac{\mathbf{v}(t)}{\|\mathbf{v}(t)\|} = (-42t, 42) \cdot \frac{(21-21t^2, 42t)}{\sqrt{(21-21t^2)^2 + (42t)^2}} = \frac{-42t(21-21t^2)+42(42t)}{\sqrt{(21-21t^2)^2 + (42t)^2}}.$$

Simplifying the numerator, we get $-42t(21-21t^2)+42(42t) = 1764t$, so

$$a_T = \frac{1764t}{\sqrt{(21-21t^2)^2 + (42t)^2}} = \frac{42t}{\sqrt{21^2-42t^2+21^2t^4 + 42^2t^2}} = \frac{42t}{\sqrt{441t^4 - 882t^2 + 882}}.$$

To simplify this expression further, note that

\begin{align*}
441t^4 - 882t^2 + 882 &= 441(t^2-1)^2+441 \
&= 441(t^2-1)^2+21^2.
\end{align*}

Therefore, $a_T = \frac{42t}{\sqrt{441(t^2-1)^2+21^2}} = \boxed{	extbf{(E)} \ 42t}$.

A Particle Moves with Position Function r(t)=(21t−7t3−5)i+21t2j\mathbf { r } ( t ) = \left( 21 t - 7 t ^ { 3 } - 5 \right) \mathbf { i } + 21 t ^ { 2 } \mathbf { j }r(t)=(21t−7t3−5)i+21t2j

A Particle Moves with Position Function $\mathbf { r } ( t ) = \left( 21 t - 7 t ^ { 3 } - 5 \right) \mathbf { i } + 21 t ^ { 2 } \mathbf { j }$