Vector Functions and Space Curves

Vector Functions

In general, we describe the motion of a particle by a vector function
$$\vec{r}(t)=<f(t),g(t),h(t)>$$
where f, g, h are scalar-valued functions.

We define the notions of limits, continuity, differentiability.
$$li{m}_{t\to t_0}\vec{r}(t)=<li{m}_{t\to t_0}f(t),li{m}_{t\to t_0}g(t),li{m}_{t\to t_0}h(t)>$$
E.x.

Describe the curve $\vec{r}(t)=<1+t,2+5t,-1+6t>$.

sol:

At t=0, we are at the point A=(1,2,-1); let t vary from 0 to 1, then our particle will move from A to B=(2,7,5).

Another way to see it is to write it as $<1,2,-1>+t<1,5,6>$.

E.x.

Find a vector equation and parametric equation to describe the line segment between P=(1,3,-2) and Q=(2,-1,3)

sol:

We describe this line segment using
$$\begin{gathered} {\vec{r}}_0(1-t)+{\vec{r}}_{1}t \\ =(1-t)<1,3,-2>+t<2,-1,3> \\ =<1+t,3-4t,5t-2> \end{gathered}$$
For the parametric equation, just extract elements from the vector equation.

There are infinite number of parametric equations. One of them can be expressed as
$$\begin{gathered} {\vec{r}}_0(1-t^2)+{\vec{r}}_1{t}^2 \\ =<1+t^2,3-4t^2,-2+5t^2> \end{gathered}$$
E.x.

Describe the intersection of the cylinder $x^2+y^2=1$ and the plane $y+z=2$.

sol:

Because $\vec{r}(t)$ lies on the cylinder, we define
$$\begin{gathered} x=cos(t) \\ y=sin(t) \end{gathered}$$
Next, we use the information that it lies on the plane to get
$$z=2-sin(t)$$
Therefore, the curve is $<cos(t),sin(t),2-sin(t)>$.

Visualize Space Curves

Helix

curve $<cos(t),sin(t),t>$.

Derivatives and Integrals of Vector Functions

Derivatives of a Vector Function

We define the derivative of a vector function.
$${\vec{r}}^{\prime }(t)=<f^{\prime }(t),g^{\prime }(t),h^{\prime }(t)>$$
Proof:
$$\begin{gathered} li{m}_{t\to 0}\frac{\vec{r}(t+\Delta t)-\vec{r}(t)}{\Delta t} \\ =li{m}_{t\to 0}\frac{1}{\Delta t}<f(t+\Delta t)-f(t),g(t+\Delta t)-g(t),h(t+\Delta t)-h(t)> \\ =<f^{\prime }(t),g^{\prime }(t),h^{\prime }(t)> \end{gathered}$$

Geometric Meaning of Derivatives

We call ${\vec{r}}^{\prime }(t)$ the tangent vector of $\vec{r}(t)$ at $t$. When it is not a zero vector, we define
$$\vec{T}(t)=\frac{{\vec{r}}^{\prime }(t)}{|{\vec{r}}^{\prime }(t)|}$$
which is the unit tangent vector.

E.x.

We have a helix $<2cos(t),sin(t),t>$ . Find the tangent line at $(0,1,\pi /2)$.

sol:

We know the point corresponds to $t=\pi /2$ by looking at the z component. The tangent vector is ${\vec{r}}^{\prime }(t)=<-2sin(t),cos(t),1>$. Thus, at that point, ${\vec{r}}^{\prime }(\pi /2)=<-2sin(\pi /2),cos(\pi /2),1>=<-2,0,1>$. The tangent line is $<-2t,1,\pi /2+t>$.

Second Order Derivatives

We define the second order derivative of a vector function.
$${\vec{r}}^{\prime \prime }(t)=<f^{\prime \prime }(t),g^{\prime \prime }(t),h^{\prime \prime }(t)>$$
It distinguishes curves that are straight from those are curved.

Differentiation Rules

$$\begin{gathered} \frac{d}{dt}[\vec{u}(t)+\vec{v}(t)]={\vec{u}}^{\prime }(t)+{\vec{v}}^{\prime }(t) \\ \frac{d}{dt}[c\vec{u}(t)]=c{\vec{u}}^{\prime }(t) \\ \frac{d}{dt}[f(t)\vec{u}(t)]=f^{\prime }(t)\vec{u}(t)+f(t){\vec{u}}^{\prime }(t) \\ \frac{d}{dt}[\vec{u}(t)\vec{v}(t)]={\vec{u}}^{\prime }(t)\vec{v}(t)+\vec{u}(t){\vec{v}}^{\prime }(t) \\ \frac{d}{dt}[\vec{u}(t)\times \vec{v}(t)]={\vec{u}}^{\prime }(t)\times \vec{v}(t)+\vec{u}(t)\times {\vec{v}}^{\prime }(t) \\ \frac{d}{dt}[\vec{u}(f(t))]={\vec{u}}^{\prime }(f(t))f^{\prime }(t) \end{gathered}$$

E.x.

Show that if $|\vec{r}(t)|$ is constant with respect to t, then ${\vec{r}}^{\prime }(t)\perp \vec{r}(t)$ for every t.

sol:

Assume the length of the vector is $a$. We square both sides to get
$$|\vec{r}(t){|}^2=a^2=\vec{r}(t)\cdot \vec{r}(t)$$
We then compute the derivative on both sides.
$$0={\vec{r}}^{\prime }(t)\cdot \vec{r}(t)+\vec{r}(t)\cdot {\vec{r}}^{\prime }(t)$$

Arc Length and Curvature

Length

integrations along curves

1. path integral
2. line integral

For a spatial curve $\vec{r}(t)=<f(t),g(t),h(t)>$, its length can be computed as
$$\begin{gathered} {\int }_a^b|{\vec{r}}^{\prime }(t)|dt \\ ={\int }_a^b\sqrt{|f^{\prime }(t){|}^2+|g^{\prime }(t){|}^2+|h^{\prime }(t){|}^2}dt \end{gathered}$$

Parametrization and Re-parametrization

E.x.

We have a curve ${\vec{r}}_1(t)=<t,t^2,t^3>$ , and $t\in [1,2]$. Replace $t$ by $e^u$.

sol:

We can convert the domain to $u\in [0,ln2]$, and get the curve ${\vec{r}}_2(t)=<e^u,e^{2u},e^{3u}>$.

These two functions describe the same curve. The process is called the parametrization of the curve.

If we replace $t$ by $w(s)$ where its derivative is always positive/negative, then write the curve as $<w(s),w^2(s),w^3(s)>$. This process is called re-parametrization of the curve.

Arc Length Parametrization

E.x.

We have a curve $\vec{r}(t)=<cos(t),sin(t),t>$ , and $t\in [0,1]$.

sol:

We first compute the arc length
$$\begin{gathered} s(t)={\int }_0^t|{\vec{r}}^{\prime }(u)|du \\ ={\int }_0^t\sqrt{2}du \\ =\sqrt{2}t \end{gathered}$$
We solve t and express it by using s and get $t=\frac{s}{\sqrt{2}}$. Then, we replace t by s and obtain $<cos(\frac{s}{\sqrt{2}}),sin(\frac{s}{\sqrt{2}}),\frac{s}{\sqrt{2}}>$ while $s\in [0,\sqrt{2}]$. This gives the arc-length parametrization of a curve.

Curvature

Curvature means the the speed of change in tangent direction.

For a given curve $\vec{r}(t)$, we let $\vec{T}(t)$ be the unit tangent vector given by
$$\vec{T}(t)=\frac{{\vec{r}}^{\prime }(t)}{|{\vec{r}}^{\prime }(t)|}$$
The curvature of a curve is
$$\kappa =|\frac{d\vec{T}}{ds}|=\frac{|{\vec{T}}^{\prime }(t)|}{|{\vec{r}}^{\prime }(t)|}$$
where $s(t)$ is the arc-length function.

Another relevant formula
$$\kappa =\frac{{\vec{r}}^{\prime }(t)\times {\vec{r}}^{\prime \prime }(t)}{|{\vec{r}}^{\prime }(t){|}^3}$$
E.X.

Compute the curvature of a circle of radius a. $\vec{r}(t)=<acos(t),asin(t),0>$.

sol:
$$\begin{gathered} {\vec{r}}^{\prime }(t)=<-asin(t),acos(t),0> \\ |{\vec{r}}^{\prime }(t)|=a \\ \vec{T}(t)=<-sin(t),cos(t),0> \\ {\vec{T}}^{\prime }(t)=<-cos(t),-sin(t),0> \\ |{\vec{T}}^{\prime }(t)|=1 \\ \kappa =\frac{1}{a} \end{gathered}$$
The smaller the radius, the larger the curvature.

Normal Vector

The normal vector measures the direction that $\vec{T}(t)$ changes. The direction that ${\vec{T}}^{\prime }(t)$ points to is called the (principle) normal direction. We define the unit normal vector to be
$$\vec{N}(t)=\frac{{\vec{T}}^{\prime }(t)}{|{\vec{T}}^{\prime }(t)|}$$
We know that $\vec{T}(t)\cdot {\vec{T}}^{\prime }(t)=0$, and $\vec{N}(t)$ has the same direction with ${\vec{T}}^{\prime }(t)$. Therefore, $\vec{T}(t)\cdot \vec{N}(t)=0$.

Binormal Vector

The binormal vector is perpendicular to both the normal vector and the tangent vector.
$$\vec{B}(t)=\vec{T}(t)\times \vec{N}(t)$$
E.X.

Consider the helix $\vec{r}(t)=<cos(t),sin(t),t>$. Compute the tangent, normal, binormal vec.

sol:
$$\begin{gathered} {\vec{r}}^{\prime }(t)=<-sin(t),cos(t),1> \\ |{\vec{r}}^{\prime }(t)|=\sqrt{2} \\ \vec{T}(t)=\frac{1}{\sqrt{2}}<-sin(t),cos(t),1> \\ {\vec{T}}^{\prime }(t)=\frac{1}{\sqrt{2}}<-cos(t),-sin(t),0> \\ \vec{N}(t)=<-cos(t),-sin(t),0> \\ \vec{B}(t)=\vec{T}(t)\times \vec{N}(t) \end{gathered}$$

Normal Planes and Osculating Planes

The normal plane is determined by the normal vector and binormal vector, passing through the curve.

The osculating plane is determined by the normal vector and tangent vector, passing through the curve.

The osculating circle is the circle that best approximates the curve around $t$. The circle is tangent to the curve and the tangent vector, and its radius is $\frac{1}{\kappa (t)}$.

Motion in Space

Velocity and Acceleration

The velocity vector is the tangent vector. The speed is the magnitude of the velocity.

$$\begin{gathered} \vec{v}(t)={\vec{r}}^{\prime }(t) \\ v(t)=|\vec{v}(t)| \\ \vec{v}(t)=v(t)\vec{T}(t) \end{gathered}$$

The acceleration is defined to be the second derivative of the curve.
$$\begin{gathered} \vec{a}(t)={\vec{r}}^{\prime \prime }(t)={\vec{v}}^{\prime }(t) \\ \vec{a}(t)=v^{\prime }(t)\vec{T}(t)+v(t){\vec{T}}^{\prime }(t) \end{gathered}$$
Recall that curvature and normal vector
$$\begin{gathered} \kappa (t)=\frac{|{\vec{T}}^{\prime }(t)|}{|{\vec{r}}^{\prime }(t)|}=\frac{{\vec{T}}^{\prime }(t)}{v(t)} \\ \vec{N}(t)=\frac{{\vec{T}}^{\prime }(t)}{|{\vec{T}}^{\prime }(t)|} \\ \vec{a}(t)=v^{\prime }(t)\vec{T}(t)+\kappa v(t)\vec{N}(t) \end{gathered}$$
E.X.

Consider the position curve at a single point $\vec{r}(0)=<1,0,0>$ with initial velocity $\vec{v}(0)=<1,-1,1>$. We know that $\vec{a}(t)=<4t,6t,1>$. Compute the position curve.

sol:
$$\begin{gathered} \vec{v}(t)-\vec{v}(0)={\int }_0^t\vec{a}(u)du=<2t^2,3t^2,t> \\ \vec{v}(t)=<2t^2+1,3t^2-1,t+1> \\ \vec{r}(t)-\vec{r}(0)={\int }_0^t\vec{v}(u)du \end{gathered}$$

Newton’s Law

E.X.

An object with mass $m$ moves in a circular path with constant angular speed $\omega$. The position vector is $\vec{r}(t)=<acos(\omega t),asin(\omega t)>$. Find the force.

sol:
$$\begin{gathered} \vec{F}(t)=m\cdot \vec{a}(t) \\ \vec{a}(t)=<-a{\omega }^{2}cos(\omega t),-a{\omega }^{2}sin(\omega t)> \end{gathered}$$

Optimization Problem

E.X.

Assume that the only force that acts on the object is gravity. We know its initial velocity, with its angle to the ground as $\alpha$. Predict where the object will land.

sol:
$$\begin{gathered} \vec{F}(t)=m\cdot \vec{a}(t)=-mg\vec{j} \\ \vec{v}(t)-\vec{v}(0)={\int }_0^t\vec{a}(u)du=<0,-gt> \\ \vec{r}(t)={\int }_0^t\vec{v}(u)du=\vec{v}(0)t+<0,-\frac{1}{2}g{t}^2> \\ \vec{v}(0)t=<|\vec{v}(0)|cos\alpha ,|\vec{v}(0)|sin\alpha > \\ \vec{r}(t)=<|\vec{v}(0)|tcos\alpha ,|\vec{v}(0)|tsin\alpha -\frac{1}{2}g{t}^2> \end{gathered}$$
We set the second component to be 0 to solve for $t$ that is non-zero. Then, we can get the final answer.

Reference

• Multivariable_Calculus_8th_Edition (13.1-13.4), James Stewart