# 3.4 Motion in space

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• Describe the velocity and acceleration vectors of a particle moving in space.
• Explain the tangential and normal components of acceleration.
• State Kepler’s laws of planetary motion.

We have now seen how to describe curves in the plane and in space, and how to determine their properties, such as arc length and curvature. All of this leads to the main goal of this chapter, which is the description of motion along plane curves and space curves. We now have all the tools we need; in this section, we put these ideas together and look at how to use them.

## Motion vectors in the plane and in space

Our starting point is using vector-valued functions to represent the position of an object as a function of time. All of the following material can be applied either to curves in the plane or to space curves. For example, when we look at the orbit of the planets, the curves defining these orbits all lie in a plane because they are elliptical. However, a particle traveling along a helix moves on a curve in three dimensions.

## Definition

Let $\text{r}\left(t\right)$ be a twice-differentiable vector-valued function of the parameter t that represents the position of an object as a function of time. The velocity vector     $\text{v}\left(t\right)$ of the object is given by

$\text{Velocity}=\text{v}\left(t\right)={r}^{\prime }\left(t\right).$

The acceleration vector     $\text{a}\left(t\right)$ is defined to be

$\text{Acceleration}=\text{a}\left(t\right)={v}^{\prime }\left(t\right)=\text{r″}\left(t\right).$

The speed is defined to be

$\text{Speed}=v\left(t\right)=‖\text{v}\left(t\right)‖=‖{r}^{\prime }\left(t\right)‖=\frac{ds}{dt}.$

Since $\text{r}\left(t\right)$ can be in either two or three dimensions, these vector-valued functions can have either two or three components. In two dimensions, we define $\text{r}\left(t\right)=x\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{i}+y\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{j}$ and in three dimensions $\text{r}\left(t\right)=x\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{i}+y\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{j}+z\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{k}.$ Then the velocity, acceleration, and speed can be written as shown in the following table.

Formulas for position, velocity, acceleration, and speed
Quantity Two Dimensions Three Dimensions
Position $\text{r}\left(t\right)=x\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{i}+y\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{j}$ $\text{r}\left(t\right)=x\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{i}+y\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{j}+z\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{k}$
Velocity $\text{v}\left(t\right)={x}^{\prime }\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{i}+{y}^{\prime }\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{j}$ $\text{v}\left(t\right)={x}^{\prime }\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{i}+{y}^{\prime }\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{j}+{z}^{\prime }\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{k}$
Acceleration $\text{a}\left(t\right)=x\text{″}\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{i}+y\text{″}\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{j}$ $\text{a}\left(t\right)=x\text{″}\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{i}+y\text{″}\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{j}+z\text{″}\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{k}$
Speed $v\left(t\right)=\sqrt{{\left({x}^{\prime }\left(t\right)\right)}^{2}+{\left({y}^{\prime }\left(t\right)\right)}^{2}}$ $v\left(t\right)=\sqrt{{\left({x}^{\prime }\left(t\right)\right)}^{2}+{\left({y}^{\prime }\left(t\right)\right)}^{2}+{\left({z}^{\prime }\left(t\right)\right)}^{2}}$

## Studying motion along a parabola

A particle moves in a parabolic path defined by the vector-valued function $\text{r}\left(t\right)={t}^{2}\text{i}+\sqrt{5-{t}^{2}}\text{j},$ where t measures time in seconds.

1. Find the velocity, acceleration, and speed as functions of time.
2. Sketch the curve along with the velocity vector at time $t=1.$
$\begin{array}{ccc}\hfill \text{v}\left(t\right)& =\hfill & {r}^{\prime }\left(t\right)=2t\phantom{\rule{0.1em}{0ex}}\text{i}-\frac{t}{\sqrt{5-{t}^{2}}}\text{j}\hfill \\ \hfill \text{a}\left(t\right)& =\hfill & {v}^{\prime }\left(t\right)=2\text{i}-5{\left(5-{t}^{2}\right)}^{-3}{2}}\text{j}\hfill \\ \hfill v\left(t\right)& =\hfill & ‖{r}^{\prime }\left(t\right)‖\hfill \\ & =\hfill & \sqrt{{\left(2t\right)}^{2}+{\left(-\frac{t}{\sqrt{5-{t}^{2}}}\right)}^{2}}\hfill \\ & =\hfill & \sqrt{4{t}^{2}+\frac{{t}^{2}}{5-{t}^{2}}}\hfill \\ & =\hfill & \sqrt{\frac{21{t}^{2}-4{t}^{4}}{5-{t}^{2}}}.\hfill \end{array}$
2. The graph of $\text{r}\left(t\right)={t}^{2}\text{i}+\sqrt{5-{t}^{2}}\text{j}$ is a portion of a parabola ( [link] ). The velocity vector at $t=1$ is
$\text{v}\left(1\right)={r}^{\prime }\left(1\right)=2\left(1\right)\phantom{\rule{0.1em}{0ex}}\text{i}-\frac{1}{\sqrt{5-{\left(1\right)}^{2}}}\text{j}=2\text{i}-\frac{1}{2}\text{j}$

and the acceleration vector at $t=1$ is
$\text{a}\left(1\right)={v}^{\prime }\left(1\right)=2\text{i}-5{\left(5-{\left(1\right)}^{2}\right)}^{\text{−}3\text{/}2}\text{j}=2\text{i}-\frac{5}{8}\text{j}.$

Notice that the velocity vector is tangent to the path, as is always the case.

A particle moves in a path defined by the vector-valued function $\text{r}\left(t\right)=\left({t}^{2}-3t\right)\phantom{\rule{0.1em}{0ex}}\text{i}+\left(2t-4\right)\phantom{\rule{0.1em}{0ex}}\text{j}+\left(t+2\right)\text{k},$ where t measures time in seconds and where distance is measured in feet. Find the velocity, acceleration, and speed as functions of time.

$\begin{array}{c}\text{v}\left(t\right)={r}^{\prime }\left(t\right)=\left(2t-3\right)\phantom{\rule{0.1em}{0ex}}\text{i}+2\text{j}+\text{k}\hfill \\ \text{a}\left(t\right)={v}^{\prime }\left(t\right)=2\text{i}\hfill \\ v\left(t\right)=‖{r}^{\prime }\left(t\right)‖=\sqrt{{\left(2t-3\right)}^{2}+{2}^{2}+{1}^{2}}=\sqrt{4{t}^{2}-12t+14}\hfill \end{array}$

The units for velocity and speed are feet per second, and the units for acceleration are feet per second squared.

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