1.4: Review of Vectors

If you have ever sat through physics class, you must have heard your instructor talk about position/velocity/ acceleration "vectors". If it has been a while since your last physics lecture, you may need to be refreshed about these concepts. In this section, we will be discussing the basics about vectors and the correlation between position, velocity, and acceleration vectors.

Introduction

A particles position during time interval, , described in parametric equations is defined as:

\[x=f(t), \;\;\; y=g(t), \;\;\; z=h(t), \;\;\; t\in I \nonumber \]

where the particle's path is composed of points:

\[(x,y,z) = (f(t),g(t),h(t)), t\in I . \nonumber \]

Written as vector form, the particle's position is modeled by the vector:

\[\mathbf{r}(t)=\underset{OP}{\rightarrow}=f(t)\hat{\textbf{i}}+g(t)\hat{\textbf{j}}+h(t)\hat{\textbf{j}}\nonumber \]

where \(f,g,h\) are component functions of the vector and \( \hat{\textbf{i}}, \hat{\textbf{j}}, \hat{\textbf{k}}\) represent the \(x,y,z\) components.

Definition: Limit

Let

\[\mathbf{r}(t)=f(t)\hat{\textbf{i}}+g(t) \hat{\mathbf{j}} +h(t) \hat{\mathbf{k}} \nonumber \]

be a vector function with domain \(D\), and \(\textbf{L}\)a vector. We say that \(\textbf{r}\) has a limit \(\textbf{L}\) as \(t\) approaches \(t_0\) and write

\[\lim_{t\rightarrow t_0}\textbf{r}(t)=\textbf{L}\nonumber \]

if for every number \(\epsilon > 0\) there exists a corresponding number \(\delta >0\) such that for all \(t\in D\),

\[\left | \mathbf{r}(t)-\mathbf{L} \right |<\epsilon\nonumber \]

whenever

\[0<\left | t-t_0 \right |<\delta.\nonumber \]

Definition: Continuous

A vector function \(\textbf{r}(t)\) is continuous at a point \(t=t_0\) in its domain if

\[\lim_{t \to t_{0}}\mathbf{r}(t)=\mathbf{r}(t_0).\nonumber \]

The function is continuous if it is continuous at every point in its domain.

The above definition sets the boundaries for continuity for a vector function. As we can see, it is very similar to the that of a scalar function.

Definition: Derivative

The vector function,

\[\textbf{r}(t)=f(t)\hat{\textbf{i}}+g(t) \hat{\textbf{j}} +h(t) \hat{\textbf{k}} \nonumber \]

has a derivative (is differentiable) at \(t\) if\(f\), \(g\), and \(h\) have derivatives at \(t\). The derivative is the vector function:

\[\textbf{r}'(t)=\dfrac{d\mathbf{r}}{dt}=\lim_{\bigtriangleup t\rightarrow 0}\dfrac{\mathbf{r}(t+\bigtriangleup t)-\mathbf{r}(t)}{\bigtriangleup t}=\dfrac{df}{dt} \hat{\textbf{i}} +\dfrac{dg}{dt} \hat{\textbf{j}} +\dfrac{dh}{dt} \hat{\textbf{k}}. \nonumber \]

Definition

If r is the position vector of a particle moving along a smooth curve in space, then

\[\mathbf{v}(t)=\dfrac{d\mathbf{r}}{dt}\nonumber \]

is the particle's velocity vector, tangent to the curve. At any time \(t\), the direction of v is the direction of motion, the magnitude of v is the particle's speed, and the derivative \(\textbf{a}=\dfrac{d\textbf{v}}{dt}\), when it exists, is the particle's acceleration vector. In summary,

1. Velocity is the derivative of position: \[\mathbf{v}=\dfrac{d\mathbf{r}}{dt},\nonumber \]
2. Speed is the magnitude of velocity: \[speed= \left|\mathbf{v}\right|,\nonumber \]
3. Acceleration is the derivative of velocity: \[a=\dfrac{d\mathbf{v}}{dt}=\dfrac{d^{2}\mathbf{r}}{dt},\nonumber \]
4. The unit vector, \[\dfrac{\mathbf{v}}{\left | \mathbf{v} \right |}\nonumber \], is the direction of motion at time \(t\).

Differential Rules for Vector Function

Let u and v be clifferentiable vector functions of t, C a constant vector, c any scalar, and f any differentiable scalar function.

1. Constant Function Rule \[\dfrac{d}{dt}\mathbf{C}=\mathbf{0},\nonumber \]
2. Scalar Multiple Rules \[\dfrac{d}{dt}[c\mathbf{u}(t))]=c\mathbf{u}'(t)\nonumber \] \[\dfrac{d}{dt}[f(t)\mathbf{u}(t)]=f'(t)\mathbf{u}(t)+f(t)\mathbf{u}'(t),\nonumber \]
3. Sum Rule: \[\dfrac{d}{dt}[\mathbf{u}(t)+\mathbf{v}(t)]=\mathbf{u}'(t)+\mathbf{v}'(t),\nonumber \]
4. Difference Rule: \[\dfrac{d}{dt}[\mathbf{u}(t)-\mathbf{v}(t)]=\mathbf{u}'(t)-\mathbf{v}'(t),\nonumber \]
5. Dot Product Rule: \[\dfrac{d}{dt}[\mathbf{u}(t)\cdot \mathbf{v}(t)]=\mathbf{u}'(t)\mathbf{v}(t)\cdot\mathbf{u}(t)\mathbf{v}'(t),\nonumber \]
6. Cross Product Rule: \[\dfrac{d}{dt}[\mathbf{u}(t)\times \mathbf{v}(t)]=\mathbf{u}'(t)\mathbf{v}(t)\times\mathbf{u}(t)\mathbf{v}'(t),\nonumber \]
7. Chain Rule: \[\dfrac{d}{dt}[\mathbf{u}(f(t))]=f'(t)\mathbf{u}'(f(t)).\nonumber \]

If r is a differentiable vector function of t of constant length, then

\[\mathbf{r}\cdot\dfrac{d\mathbf{r}}{dt}=0.\nonumber \]

Example 1

Prove the vector function \[\mathbf{r}(t)=(\cos(t)) \hat{\textbf{i}} +(\sin(t)) \hat{\textbf{j}} +t \hat{\textbf{k}} \nonumber \] is continuous at \(t=\dfrac{\pi}{4}\).

Solution

To prove the vector function is continuous, we need to prove that:

\[\mathbf{r} \left(\dfrac{\pi}{4}\right)=\lim_{t \to \frac{\pi}{4}}\mathbf{r}(t).\nonumber \]

To find the limit at \( t= \pi/4 \), we must compute:

\[\begin{align*} \lim_{t \to \frac{ \pi}{4}} \mathbf{r}(t) &= (\lim_{t \to \frac{\pi}{4}} f(t)) \hat{\textbf{i}} +(\lim_{t \to \frac{\pi}{4}} g(t)) \hat{\textbf{j}} +(\lim_{t \to \frac{\pi}{4}} h(t)) \hat{\textbf{k}} \\&= (\lim_{t \to \frac{\pi}{4}} \sin(t)) \hat{\textbf{i}} +(\lim_{t \to \frac{\pi}{4}} \cos(t)) \hat{\textbf{j}} +(\lim_{t \to \frac{\pi}{4}} t) \hat{\textbf{k}} \end{align*}\nonumber \]

\[\lim_{t \to \frac{\pi}{4}}\mathbf{r}(t)= \dfrac{\sqrt{2}}{2} \hat{\textbf{i}} +\dfrac{\sqrt{2}}{2} \hat{\textbf{j}} +\dfrac{\pi}{4} \hat{\textbf{k}}. \nonumber \]

Now we need to find out what \( \textbf{r}( \pi/4) \) is equal to:

\[\mathbf{r}\left(\dfrac{\pi}{4}\right)=\left(\sin(\dfrac{\pi}{4})\right) \hat{\textbf{i}} +\left(\cos(\dfrac{\pi}{4})\right) \hat{\textbf{j}} +\left(\dfrac{\pi}{4}\right) \hat{\textbf{k}} \nonumber \]

\[= \dfrac{\sqrt{2}}{2} \hat{\textbf{i}} +\dfrac{\sqrt{2}}{2} \hat{\textbf{j}} +\dfrac{\pi}{4} \hat{\textbf{k}}. \nonumber \]

Since

\[\mathbf{r}(\dfrac{\pi}{4})=\lim_{t \to \frac{\pi}{4}}\mathbf{r}(t)= \dfrac{\sqrt{2}}{2} \hat{\textbf{i}} +\dfrac{\sqrt{2}}{2} \hat{\textbf{j}} +\dfrac{\pi}{4} \hat{\textbf{k}}, \nonumber \]

the vector function is continuous at \(t=\dfrac{\pi}{4}\).

Example 2

Given the vector function, find the velocity and acceleration acceleration vectors, and speed when \(t=1\):

\[\mathbf{r}(t)=(t+1) \hat{\textbf{i}} +\dfrac{t^2}{\sqrt(2)} \hat{\textbf{j}} +\dfrac{t^3}{3} \hat{\textbf{k}} \nonumber \]

The find the velocity, we must take the first derivative of the vector function.

\[\mathbf{v}(t)=\dfrac{d\mathbf{r}}{dt}\nonumber \]

\[\mathbf{v}=\mathbf{r}'(t)= \hat{\textbf{i}} +\dfrac{2t}{\sqrt(2)} \hat{\textbf{j}} +t^2 \hat{\textbf{k}} \nonumber \]

To find the acceleration, we must take the first derivative of the velocity vector and second of the position vector.

\[\mathbf{a}=\dfrac{d\mathbf{v}}{dt}=\dfrac{d^2\mathbf{r}}{dt^2}\nonumber \]

\[\mathbf{a}(t)=\mathbf{v}'(t)=\mathbf{r}''(t)=\dfrac{2}{\sqrt{2}} \hat{\textbf{j}} +2t \hat{\textbf{k}} \nonumber \]

The speed at \(t=1\) is found by taking the magnitude of the velocity at \(t=1\).

\[\mathbf{v}(1)= \hat{\textbf{i}} + \dfrac{2}{\sqrt{2}} \hat{\textbf{j}}+t^2 \hat{\textbf{k}} \nonumber \]

\[|\mathbf{v}(1)|= \sqrt{(1^2)+ (\dfrac{2}{\sqrt{2}})^2+ (1^2)} =2\nonumber \]

References

1. Weir, Maurice D., Joel Hass, and George B. Thomas. Thomas' Calculus: Early Transcendentals. Boston: Addison-Wesley, 2010. Print.

Contributors and Attributions

• Shilpa Vasuki (UCD)

• Integrated by Justin Marshall.