Question

# The position vector r describes the path of an object moving in space. (a) Find the velocity vector, speed, and acceleration vector of the object. Position Vector: r(t) = ⟨ln t, 1/t, t^4⟩ Time: t = 2

Solution

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#### $\color{#4257b2}{a)}$

Differentiating the given equation with respect to $t$, we get:

\begin{align*} \mathbf{v}(t)=\mathbf{r}^{\prime}(t)&=\dfrac{\mathrm d\mathbf{r}}{\mathrm dt}=\dfrac{\mathrm d}{\mathrm dt}\left[\left(\ln (t)\right)\mathbf{i}+\left(\dfrac{1}{t}\right)\mathbf{j}+\left(t^4\right)\mathbf{k}\right]\\ &=\dfrac{\mathbf{i}}{t}-\dfrac{\mathbf{j}}{t^2}+4t^3\mathbf{k}\tag*{\color{#c34632}(1)} \end{align*}

So,

$\begin{equation*} \boxed{\color{#c34632}\mathbf{v}(t)=\dfrac{\mathbf{i}}{t}-\dfrac{\mathbf{j}}{t^2}+4t^3\mathbf{k}} \quad \rightarrow \quad \textbf{Velocity vector} \end{equation*}$

Differentiating the above equation with respect to $t$, we get:

$\begin{equation*} \boxed{\color{#c34632}\mathbf{a}(t)=-\dfrac{\mathbf{i}}{t^2}+\dfrac{2\mathbf{j}}{t^3}+12t^2\mathbf{k}} \quad \rightarrow \quad \textbf{Acceleration vector } \end{equation*}$

Using the equation $\color{#c34632}(1)$ we get:

\begin{align*} \color{#c34632}\left|\left|\mathbf{v}(t)\right|\right|&=\left|\left|\dfrac{\mathbf{i}}{t}-\dfrac{\mathbf{j}}{t^2}+4t^3\mathbf{k}\right|\right|\\ &=\sqrt{\left(\dfrac{1}{t}\right)^2+\left(-\dfrac{1}{t^2}\right)^2+\left(4t^3\right)^2}\\ &=\sqrt{\dfrac{1}{t^2}+\dfrac{1}{t^4}+16t^6}\\ &=\sqrt{\dfrac{t^2+1+16t^{10}}{t^4}}\\ &=\boxed{\color{#c34632}\dfrac{\sqrt{t^2+1+16t^{10}}}{t^2}} \end{align*}

So,

$\begin{equation*} \boxed{\color{#c34632}\left|\left|\mathbf{v}(t)\right|\right|=\dfrac{\sqrt{t^2+1+16t^{10}}}{t^2}}\quad \rightarrow \quad \textbf{Speed } \end{equation*}$

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