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\centerline
{
\bf Dr. Z's Math251 Handout \#13.4 [Motion in Space:
Velocity and Acceleration]
}

By Doron Zeilberger

{\bf Problem Type 13.4a}: Find the velocity, acceleration,
and speed of a particle with the given position function.
$$
{\bf r}(t)= x(t) \, {\bf i}+y(t) \, {\bf j}+z(t) \, {\bf k} \quad.
$$
{\bf Example Problem 13.4a}: 
Find the velocity, acceleration,
and speed of a particle with the given position function.
$$
{\bf r}(t)= t^2 \, {\bf i}+ \ln t \, {\bf j}+t \, {\bf k} \quad.
$$
%%Specific Problem


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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} The velocity is ${\bf r}'(t)$ and
the acceleration is ${\bf r}''(t)$.
&
{\bf 1.} 
${\bf v}(t)={\bf r}'(t)= (t^2)' \, {\bf i}+ (\ln t)' \, {\bf j}+t' \, {\bf k}$
$= 2t \, {\bf i}+ 1/t \, {\bf j}+ \, {\bf k} $.

${\bf a}(t)= {\bf r}''(t)= (2t)' \, {\bf i}+ (1/t)' \, {\bf j}+ 1'\, {\bf k}$
$=2 \, {\bf i} -(1/t^2) \, {\bf j}$ \quad .
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{\bf 2.} To find the speed, take the magnitude of the
velocity, i.e. compute $\vert {\bf v}(t)\vert$
&
{\bf 2.} the speed is
$\vert {\bf v}(t) \vert= \vert 2t \, {\bf i}+ 1/t \, {\bf j}+ \, {\bf k} \vert$
$=\sqrt{(2t)^2+(1/t)^2+1^2}=\sqrt{4t^2+1/t^2+1} \,\,$  .
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{\bf Problem Type 13.4b}: A force with magnitute
$F N$ acts on a body of mass $m$ in the directon $\langle d_1,d_2,d_3 \rangle$.
The object starts at the $(x_0,y_0,z_0)$ with initial
velocity ${\bf v} (0)= \langle v_1,v_2,v_3 \rangle$.
Find its position function and its speed at time $t$.

{\bf Example Problem 13.4b}: A force with magnitute
$300 N$ acts on a body of mass $100$ kg
in the directon $\langle 1,2,2 \rangle$.
The object starts at the $(1,2,3)$ with initial
velocity ${\bf v} (0)= \langle 0,1,-1 \rangle$.
Find its position function and its speed at time $t$.

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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Find the {\bf unit direction vector} 
by dividing $\langle d_1,d_2,d_3 \rangle$
by its length. To get the {\bf force} vector,
multiply this vector by the magnitude of the force $F$.

&
{\bf 1.} $\vert \langle 1,2,2 \rangle \vert =\sqrt{1^2+2^2+2^2}=3$.
So the direction of the force is 
$(1/3)\langle 1,2,2 \rangle$=
$ \langle 1/3,2/3,2/3 \rangle$.
The {\bf force is} ${\bf F}=300\langle 1/3,2/3,2/3 \rangle$, so
${\bf F}=\langle 100,200,200 \rangle$.
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{\bf 2.} Set-up {\bf Newton's Second Law} 
$$
{\bf F}=m {\bf r}''(t) \quad.
$$
&
{\bf 2.} 
$$
\langle 100,200,200 \rangle= 100 {\bf r}''(t) \quad.
$$
so
$$
{\bf r}''(t) =\langle 1,2,2 \rangle \quad.
$$
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{\bf 3.} Integrate with respect to $t$ to get
${\bf v}(t)={\bf r}'(t)$, and don't forget
the {\bf arbitrary constant} vector. Then
plug-in $t=0$ to get it.
&

{\bf 3.}
$$
{\bf v}(t)={\bf r}'(t) =\int \langle 1,2,2 \rangle \, dt=
\langle t,2t,2t \rangle + {\bf C} \quad .
$$
But ${\bf v}(0)=\langle 0,1,-1 \rangle$,
so ${\bf C}=\langle 0,1,-1 \rangle$ and we get
$$
{\bf v}(t)={\bf r}'(t) =\langle t,2t,2t \rangle +
\langle 0,1,-1 \rangle=
\langle t,2t+1,2t-1 \rangle \quad .
$$
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{\bf 4.} To get the position vector ${\bf r}(t)$
integrate the velocity vector ${\bf v}(t)$ that
you found in step 2, once again not
forgetting the arbitrary constant vector.
&

{\bf 4.}
$$
{\bf r}(t)=\int \langle t,2t+1,2t-1 \rangle \, dt=
\langle t^2/2,t^2+t,t^2-t \rangle + {\bf C}
$$
When $t=0$, ${\bf r}(0)=\langle 1,2,3 \rangle$, so
${\bf C}=\langle 1,2,3 \rangle$, and
$$
{\bf r}(t)=
\langle t^2/2,t^2+t,t^2-t \rangle + \langle 1,2,3 \rangle= 
\langle t^2/2+1,t^2+t+2,t^2-t+3 \rangle \quad .
$$
This is the {\bf position function}.

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{\bf 5.} To get the {\bf speed}, compute $\vert {\bf v}(t) \vert$,
an expression in $t$.
&
{\bf 5.}
$$
\vert {\bf v} (t) \vert = \vert \langle t,2t+1,2t-1 \rangle \vert 
$$
$$
=\sqrt{t^2+(2t+1)^2+(2t-1)^2}=
\sqrt{9t^2+2} \quad .
$$

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