Calculus III - Velocity and Acceleration

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Section 12.11 : Velocity and Acceleration

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1. An objects acceleration is given by $\vec a = 3t\,\vec i - 4{{\bf{e}}^{ - t}}\,\vec j + 12{t^2}\vec k$ . The objects initial velocity is $\vec v\left( 0 \right) = \vec j - 3\vec k$ and the objects initial position is $\vec r\left( 0 \right) = - 5\vec i + 2\vec j - 3\vec k$ . Determine the objects velocity and position functions.

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To determine the velocity function all we need to do is integrate the acceleration function.

\vec{v} (t) = \int 3 t \vec{i} - 4 e^{- t} \vec{j} + 12 t^{2} \vec{k} d t = \frac{3}{2} t^{2} \vec{i} + 4 e^{- t} \vec{j} + 4 t^{3} \vec{k} + \vec{c}

$\vec v\left( t \right) = \int{{3t\,\vec i - 4{{\bf{e}}^{ - t}}\,\vec j + 12{t^2}\vec k\,dt}} = \frac{3}{2}{t^2}\,\vec i + 4{{\bf{e}}^{ - t}}\,\vec j + 4{t^3}\vec k + \vec c$

Don’t forget the “constant” of integration, which in this case is actually the vector $\vec c = {c_1}\vec i + {c_2}\vec j + {c_3}\vec k$ . To determine the constant of integration all we need is to use the value $\vec v\left( 0 \right)$ that we were given in the problem statement.

\vec{j} - 3 \vec{k} = \vec{v} (0) = 4 \vec{j} + c_{1} \vec{i} + c_{2} \vec{j} + c_{3} \vec{k}

$\vec j - 3\vec k = \vec v\left( 0 \right) = 4\,\vec j + {c_1}\vec i + {c_2}\vec j + {c_3}\vec k\,$

To determine the values of ${c_1}$ , ${c_2}$ , and ${c_3}$ all we need to do is set the various components equal.

\begin{aligned} \vec{i} : 0 = c_{1} \\ \vec{j} : 1 = 4 + c_{2} \\ \vec{k} : - 3 = c_{3} \end{aligned} \Rightarrow c_{1} = 0, c_{2} = - 3, c_{3} = - 3

$\begin{align*}&{\vec i:0 = {c_1}}\\&{\vec j:1 = 4 + {c_2}}\\&{\vec k: - 3 = {c_3}}\end{align*}\hspace{0.5in} \Rightarrow \hspace{0.5in}{c_1} = 0,\,\,\,\,{c_2} = - 3,\,\,\,\,{c_3} = - 3$

The velocity is then,

\vec{v} (t) = \frac{3}{2} t^{2} \vec{i} + (4 e^{- t} - 3) \vec{j} + (4 t^{3} - 3) \vec{k}

$\require{bbox} \bbox[2pt,border:1px solid black]{{\vec v\left( t \right) = \frac{3}{2}{t^2}\,\vec i + \left( {4{{\bf{e}}^{ - t}} - 3} \right)\,\vec j + \left( {4{t^3} - 3} \right)\vec k}}$ Show Step 2

The position function is simply the integral of the velocity function we found in the previous step.

\vec{r} (t) = \int \frac{3}{2} t^{2} \vec{i} + (4 e^{- t} - 3) \vec{j} + (4 t^{3} - 3) \vec{k} d t = \frac{1}{2} t^{3} \vec{i} + (- 4 e^{- t} - 3 t) \vec{j} + (t^{4} - 3 t) \vec{k} + \vec{c}

$\vec r\left( t \right) = \int{{\frac{3}{2}{t^2}\,\vec i + \left( {4{{\bf{e}}^{ - t}} - 3} \right)\,\vec j + \left( {4{t^3} - 3} \right)\vec k\,dt}} = \frac{1}{2}{t^3}\,\vec i + \left( { - 4{{\bf{e}}^{ - t}} - 3t} \right)\,\vec j + \left( {{t^4} - 3t} \right)\vec k + \vec c$ We’ll use the value of

\vec{r} (0)

$\vec r\left( 0 \right)$ from the problem statement to determine the value of the constant of integration.

- 5 \vec{i} + 2 \vec{j} - 3 \vec{k} = \vec{r} (0) = - 4 \vec{j} + c_{1} \vec{i} + c_{2} \vec{j} + c_{3} \vec{k}

$- 5\vec i + 2\vec j - 3\vec k = \vec r\left( 0 \right) = - 4\,\vec j + {c_1}\vec i + {c_2}\vec j + {c_3}\vec k$

\begin{aligned} \vec{i} : - 5 = c_{1} \\ \vec{j} : 2 = - 4 + c_{2} \\ \vec{k} : - 3 = c_{3} \end{aligned} \Rightarrow c_{1} = - 5, c_{2} = 6, c_{3} = - 3

$\begin{align*} &{\vec i: - 5 = {c_1}}\\& {\vec j:2 = - 4 + {c_2}}\\ &{\vec k: - 3 = {c_3}}\end{align*}\hspace{0.5in} \Rightarrow \hspace{0.25in}{c_1} = - 5,\,\,\,\,{c_2} = 6,\,\,\,\,{c_3} = - 3$

The position function is then,

\vec{r} (t) = (\frac{1}{2} t^{3} - 5) \vec{i} + (- 4 e^{- t} - 3 t + 6) \vec{j} + (t^{4} - 3 t - 3) \vec{k}

$\require{bbox} \bbox[2pt,border:1px solid black]{{\vec r\left( t \right) = \left( {\frac{1}{2}{t^3} - 5} \right)\,\vec i + \left( { - 4{{\bf{e}}^{ - t}} - 3t + 6} \right)\,\vec j + \left( {{t^4} - 3t - 3} \right)\vec k}}$