
# 12.4: Motion 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: Speed, Velocity, and Acceleration

Let $$\vec{\mathbf r}(t)$$ 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 $$\vec{\mathbf v}(t)$$ of the object is given by

$\mathrm{Velocity}=\vec{\mathbf v}(t)=\vec{\mathbf r}′(t).$

The acceleration vector $$\vec{\mathbf a}(t)$$ is defined to be

$\mathrm{Acceleration}=\vec{\mathbf a}(t)=\vec{\mathbf v}′(t)=\vec{\mathbf r}″(t).$

The speed is defined to be

$\mathrm{Speed}=\vec{\mathbf v}(t)=‖\vec{\mathbf v}(t)‖=‖\vec{\mathbf r}′(t)‖=\dfrac{ds}{dt}.$

Since $$\vec{\mathbf r}(t)$$ can be in either two or three dimensions, these vector-valued functions can have either two or three components. In two dimensions, we define $$\vec{\mathbf r}(t)=x(t) \hat{\mathbf i}+y(t) \hat{\mathbf j}$$ and in three dimensions $$\vec{\mathbf r}(t)=x(t) \hat{\mathbf i}+y(t) \hat{\mathbf j}+z(t) \hat{\mathbf k}$$. 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 $$\vec{\mathbf r}(t)=x(t) \hat{\mathbf i}+y(t) \hat{\mathbf j}$$ $$\vec{\mathbf r}(t)=x(t) \hat{\mathbf i}+y(t) \hat{\mathbf j}+z(t) \hat{\mathbf k}$$
Velocity $$\vec{\mathbf v}(t)=x′(t) \hat{\mathbf i}+y′(t) \hat{\mathbf j}$$ $$\vec{\mathbf v}(t)=x′(t) \hat{\mathbf i}+y′(t) \hat{\mathbf j}+z′(t) \hat{\mathbf k}$$
Acceleration $$\vec{\mathbf a}(t)=x″(t) \hat{\mathbf i}+y″(t) \hat{\mathbf j}$$ $$\vec{\mathbf a}(t)=x″(t) \hat{\mathbf i}+y″(t) \hat{\mathbf j}+z″(t) \hat{\mathbf k}$$
Speed $$\|\vec{\mathbf v}(t)\|= \sqrt{(x′(t))^2+(y′(t))^2}$$ $$\|\vec{\mathbf v}(t)\|=\sqrt{(x′(t))^2+(y′(t))^2+(z′(t))^2}$$

Example $$\PageIndex{1}$$: Studying Motion Along a Parabola

A particle moves in a parabolic path defined by the vector-valued function $$\vec{\mathbf r}(t)=t^2 \hat{\mathbf i}+ \sqrt{5−t^2} \hat{\mathbf j}$$, 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 $$t=1$$.

Solution

1. We use EquationEquation, and Equation:

\begin{align*} \vec{\mathbf v}(t) & = \vec{\mathbf r}′(t)=2t\hat{\mathbf i}−\dfrac{t}{\sqrt{5-t^2}}\hat{\mathbf j} \\ \vec{\mathbf a}(t) & =\vec{\mathbf v}′(t)=2\hat{\mathbf i}−5(5−t^2)^{\frac{3}{2}}\hat{\mathbf j} \\ ||\vec{\mathbf v}(t)||&=||\vec{\mathbf r}′(t)|| \\ &=(2t)^2+\left(-\dfrac{t}{\sqrt{5-t^2}}\right)^2 \\ &=\sqrt{4t^2+\dfrac{t^2}{5-t^2}} \\ &=\sqrt{\dfrac{21t^2-4t^4}{5-t^2}}. \end{align*}

2. The graph of $$\vec{\mathbf r}(t)=t^2 \hat{\mathbf i}+ \sqrt{5−t^2} \hat{\mathbf j}$$ is a portion of a parabola (Figure). The velocity vector at $$t=1$$ is

$\vec{\mathbf v}(1)=\vec{\mathbf r}′(1)=2(1)\hat{\mathbf i}−15−(1)2\hat{\mathbf j}=2\hat{\mathbf i}−12\hat{\mathbf j}$

and the acceleration vector at $$t=1$$ is

$\vec{\mathbf a}(1)=\vec{\mathbf v}′(1)=2\hat{\mathbf i}−5(5−(1)2)−3/2\hat{\mathbf j}=2\hat{\mathbf i}−58\hat{\mathbf j}.$

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

This graph depicts the velocity vector at time $$t=1$$ for a particle moving in a parabolic path.

Exercise $$\PageIndex{1}$$

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

Hint

Use EquationEquation, and Equation.

\begin{align*}\mathrm{v}(t)=\mathrm{r}'(t)&=(2t-3)\mathrm{i}+2\mathrm{j}+\mathrm{k}\\ a(t)&=v′(t) =2\mathrm{i}v(t) \end{align*}
$||\mathrm{r}′(t)||=\sqrt{(2t-3)^2+2^2+1^2} =\sqrt{4t^2-12t+14}$
Figure $$\PageIndex{2}$$: The dashed line represents the trajectory of an object (a car, for example). The acceleration vector points toward the inside of the turn at all times.