Vector Functions

Introduction

Throughout the rest of this course, we'll use vector-valued functions extensively. We've actually already seen an example of a vector function, when we described a line in R3\mathbb{R}^3 with parametric equations ⟨x0,y0,z0⟩+t⟨a,b,c⟩\langle x_0,y_0,z_0 \rangle +t\langle a,b,c \rangle

If we write it like r⃗(t)=⟨x0+at,y0+bt,z0+ct⟩,\vec{r}(t) = \langle x_0+at,y_0+bt,z_0+ct \rangle, it may be clearer that it's a vector valued function. Basically, for each value of tt, a vector-valued function describes a vector, and if we draw these vectors from the origin, the tips of the vectors trace out a path in space (a straight line for the function above).

Vector valued function

In general, a vector-valued function has the form (in R3\mathbb{R}^3) of r⃗(t)=⟨f(t),g(t),h(t)⟩\vec{r}(t) = \langle f(t),g(t),h(t) \rangle or r⃗(t)=⟨x(t),y(t),z(t)⟩.\vec{r}(t) = \langle x(t),y(t),z(t) \rangle.

Example

For example, consider the function r⃗(t)=⟨t,t2,2t+1⟩\vec{r}(t) = \langle t,t^2,2t+1 \rangle (note that x(t)=tx(t)=t, y(t)=t2y(t)=t^2, and z(t)=2t+1z(t)=2t+1). The domain of this function is all tt in R\mathbb{R}, but the range is not quite so simple (we'd need to see a graph to get an idea of it).

If we pick a few representative values for tt, we can start to build a table of vectors: t⟨x,y,z⟩0⟨0,0,1⟩1⟨1,1,3⟩2⟨2,4,5⟩⋮⋮\begin{array}{c | c} t & \langle x,y,z \rangle\ \hline 0 & \langle 0,0,1 \rangle\ 1 & \langle 1,1,3 \rangle\ 2 & \langle 2,4,5 \rangle\ \vdots & \vdots \end{array}

We can graph this space curve with Matlab using the following command:

>> ezplot3('t','t^2','2*t+1',[-6*pi,6*pi])

Vector valued function

Another Example

Graph r⃗(t)=⟨cos⁡t,sin⁡t,tln⁡t⟩\vec{r}(t)=\langle \cos t,\sin t,t \ln t \rangle:

>> ezplot3('cos(t)','sin(t)','t*log(t)',[-6*pi,6*pi])

Vector valued function

Another Example

Graph r⃗(t)=⟨4cos⁡t,t2+2t,4sin⁡t⟩\vec{r}(t)=\langle 4 \cos t,t^2+2t,4 \sin t \rangle:

>> ezplot3('4*cos(t)','t^2+2*t','4*sin(t)',[-6*pi,6*pi])

alt text

Limits With Vector-Valued Functions

If we have a vector function, the limit of this function as tt approaches some value is just what you would expect (and hope): it is equal to the limit of each component function.

Find lim⁡t→0+⟨cos⁡t,sin⁡t,t ln⁡t⟩\displaystyle\lim_{t \to 0^+} \langle \cos t, \sin t, t\ \ln t \rangle

Solution

Find the limit of each component function:

lim⁡t→0+cos⁡t=1lim⁡t→0+sin⁡t=0lim⁡t→0+t ln⁡t=0⋅(−∞) (indeterminate: use L’Hopital’s Rule)\begin{aligned} \lim_{t \to 0^+} \cos t &= 1\ \lim_{t \to 0^+} \sin t &= 0\ \lim_{t \to 0^+} t\ \ln t &= 0 \cdot (-\infty) \textrm{ (indeterminate: use L'Hopital's Rule)}\ \end{aligned}

Using L'Hopital's Rule on the third limit: lim⁡t→0+t ln⁡t=lim⁡t→0+ln⁡t1/t=(L)lim⁡t→0+1/t−1/t2=lim⁡t→0+−t=0\lim_{t \to 0^+} t\ \ln t = \lim_{t \to 0^+} \dfrac{\ln t}{1/t} = \textrm{(L)} \lim_{t \to 0^+} \dfrac{1/t}{-1/t^2} = \lim_{t \to 0^+} -t = 0

Therefore, the limit is lim⁡t→0+⟨cos⁡t,sin⁡t,t ln⁡t⟩=⟨1,0,0⟩\ans{\lim_{t \to 0^+} \langle \cos t, \sin t, t\ \ln t \rangle = \langle 1,0,0 \rangle}

  1. Find lim⁡t→2cos⁡πti^+sin⁡πtj^+e−tk^\displaystyle\lim_{t \to 2} \cos \pi t \hat{i} + \sin \pi t \hat{j} + e^{-t} \hat{k}

    2. i^+e−2k^\hat{i}+e^{-2}\hat{k}

  2. Find lim⁡t→∞cos⁡πti^+sin⁡πtj^+e−tk^\displaystyle\lim_{t \to \infty} \cos \pi t \hat{i} + \sin \pi t \hat{j} + e^{-t} \hat{k}

    3. Does not exist (DNE)

Derivatives of Vector-Valued Functions

Just like with limits, derivatives of vector functions work as you would hope: we can differentiate each component function individually.

r⃗ ′(t)=lim⁡Δt→0ΔrΔt=lim⁡Δt→0r⃗(t+Δt)−r⃗(t)Δt=lim⁡Δt→0(x(t+Δt)−x(t)Δti^+y(t+Δt)−y(t)Δtj^+z(t+Δt)−z(t)Δtk^)=(lim⁡Δt→0x(t+Δt)−x(t)Δt)i^+(lim⁡Δt→0y(t+Δt)−y(t)Δt)j^+(lim⁡Δt→0z(t+Δt)−z(t)Δt)k^=x′(t)i^+y′(t)j^+z′(t)k^\begin{aligned} \vec{r}\ '(t) &= \lim_{\Delta t \to 0} \dfrac{\Delta r}{\Delta t} = \lim_{\Delta t \to 0} \dfrac{\vec{r}(t+\Delta t) - \vec{r}(t)}{\Delta t}\ &= \lim_{\Delta t \to 0} \left(\dfrac{x(t+\Delta t)-x(t)}{\Delta t} \hat{i} + \dfrac{y(t+\Delta t)-y(t)}{\Delta t} \hat{j} + \dfrac{z(t+\Delta t)-z(t)}{\Delta t} \hat{k}\right)\ &= \left(\lim_{\Delta t \to 0} \dfrac{x(t+\Delta t)-x(t)}{\Delta t}\right) \hat{i} + \left(\lim_{\Delta t \to 0} \dfrac{y(t+\Delta t)-y(t)}{\Delta t}\right) \hat{j} + \left(\lim_{\Delta t \to 0} \dfrac{z(t+\Delta t)-z(t)}{\Delta t}\right) \hat{k}\ &= x'(t) \hat{i} + y'(t) \hat{j} + z'(t) \hat{k} \end{aligned}

Summary

r⃗ ′(t)=⟨x′(t),y′(t),z′(t)⟩\vec{r}\ '(t) = \langle x'(t),y'(t),z'(t) \rangle

If r⃗ ′ = 0⃗\vec{r}\ '\ \cancel=\ \vec{0} (the zero vector), then r⃗′(t)r⃗ '(t) is a tangent vector to the space curve.

Find the first and second derivatives of r⃗(t)=⟨t,t2,2t+1⟩\vec{r}(t) = \langle t,t^2,2t+1 \rangle.

Solution

r⃗ ′(t)=⟨1,2t,2⟩r⃗ ′′(t)=⟨0,2,0⟩\ans{\begin{aligned} \vec{r}\ '(t) &= \langle 1,2t,2 \rangle\ \vec{r}\ ''(t) &= \langle 0,2,0 \rangle \end{aligned}}

Unit Tangent Vector

This vector will come back later; to find the unit tangent vector, simply divide the tangent vector by its magnitude.

The unit tangent vector to r⃗(t)r⃗ (t) is T⃗(t)=r⃗ ′(t)∣r⃗ ′(t)∣\vec{T}(t) = \dfrac{\vec{r}\ '(t)}{|\vec{r}\ '(t)|}

Find the unit tangent vector for r⃗(t)=⟨t2,4t,4ln⁡t⟩,    t>0\vec{r}(t) = \langle t^2,4t,4 \ln t \rangle,\ \ \ \ t > 0.

Solution

First, find r⃗′(t)r⃗ '(t): r⃗ ′(t)=⟨2t,4,4t⟩\vec{r}\ '(t) = \left\langle 2t,4,\dfrac{4}{t} \right\rangle

Then divide this by its magnitude to find the unit tangent vector: r⃗ ′(t)∣r⃗ ′(t)∣=⟨2t,4,4/t⟩4t2+16+16/t2=⟨2t,4,4/t⟩(2t+4/t)2  ⟹  T⃗(t)=12t+4t⟨2t,4,4t⟩\begin{aligned} \dfrac{\vec{r}\ '(t)}{|\vec{r}\ '(t)|} &= \dfrac{\langle 2t,4,4/t \rangle}{\sqrt{4t^2+16+16/t^2}}\ &= \dfrac{\langle 2t,4,4/t \rangle}{\sqrt{(2t+4/t)^2}}\ &\implies \ans{\vec{T}(t) = \dfrac{1}{2t+\dfrac{4}{t}} \left\langle 2t,4,\dfrac{4}{t} \right\rangle} \end{aligned}

  1. Find the unit tangent vector for r⃗(t)=⟨10,3cos⁡t,3sin⁡t⟩\vec{r} (t) = \langle 10,3\cos t, 3\sin t \rangle.

    2. T⃗(t)=⟨0,−sin⁡t,cos⁡t⟩\vec{T}(t) = \langle 0,-\sin t, \cos t \rangle

Let r⃗(t)=⟨t5,t4,t3⟩\vec{r}(t) = \langle t^5,t^4,t^3 \rangle.

  1. Find the unit tangent vector to this curve at t=1t=1.
  2. Find parametric equations for the tangent line to r⃗(t)r⃗ (t) at t=1t=1.

Solution

  1. The unit tangent vector to this curve at t=1t=1: r⃗ ′(t)=⟨5t4,4t3,3t2⟩r⃗ ′(1)=⟨5,4,3⟩r⃗ ′(1)∣r⃗ ′(1)∣=⟨5,4,3⟩25+16+9=152⟨5,4,3⟩=⟨22,225,3210⟩\begin{aligned} \vec{r}\ '(t) &= \langle 5t^4,4t^3,3t^2 \rangle\ \vec{r}\ '(1) &= \langle 5,4,3 \rangle\ \dfrac{\vec{r}\ '(1)}{|\vec{r}\ '(1)|} &= \dfrac{\langle 5,4,3 \rangle}{\sqrt{25+16+9}} = \dfrac{1}{5\sqrt{2}} \langle 5,4,3 \rangle\ &= \ans{\left\langle \dfrac{\sqrt{2}}{2}, \dfrac{2\sqrt{2}}{5}, \dfrac{3\sqrt{2}}{10} \right\rangle} \end{aligned}

  2. Parametric equations for the tangent line to r⃗(t)r⃗ (t) at t=1t=1: r⃗ ′(1)=⟨5,4,3⟩=⟨a,b,c⟩⟨x0,y0,z0⟩=⟨x(1),y(1),z(1)⟩=⟨1,1,1⟩\begin{aligned} \vec{r}\ '(1) &= \langle 5,4,3 \rangle = \langle a,b,c \rangle\ \langle x_0,y_0,z_0 \rangle &= \langle x(1),y(1),z(1) \rangle = \ans{\langle 1,1,1 \rangle} \end{aligned}

Putting the starting point and the direction together: x=1+5ty=1+4tz=1+3t\ans{\begin{aligned} x &= 1+5t\ y &= 1+4t\ z &= 1+3t \end{aligned}}

  1. Find parametric equations for the tangent line to r⃗(t)=⟨10,3cos⁡t,3sin⁡t⟩\vec{r}(t) = \langle 10,3\cos t,3\sin t \rangle at t=0t=0.

    2. x=10y=3z=3t\begin{aligned} x &= 10\ y &= 3\ z &= 3t \end{aligned}

Find the angle between the following two curves at the point where t=1t=1 and s=2s=2. r⃗1(t)=⟨t,1−t,3+t2⟩r⃗2(s)=⟨3−s,s−2,s2⟩\begin{aligned} \vec{r}_1(t) &= \langle t,1-t,3+t^2 \rangle\ \vec{r}_2(s) &= \langle 3-s,s-2,s^2 \rangle \end{aligned}

Solution

Find the angle between their tangent vectors: r⃗1 ′(t)=⟨1,−1,2t⟩⟶r⃗1 ′(1)=⟨1,−1,2⟩r⃗2 ′(s)=⟨−1,1,2s⟩⟶r⃗2 ′(2)=⟨−1,1,4⟩\begin{aligned} \vec{r}_1\ '(t) &= \langle 1,-1,2t \rangle \longrightarrow \vec{r}_1\ '(1) = \langle 1,-1,2 \rangle\ \vec{r}_2\ '(s) &= \langle -1,1,2s \rangle \longrightarrow \vec{r}_2\ '(2) = \langle -1,1,4 \rangle \end{aligned}

Take the dot product to find the angle between them:

⟨1,−1,2⟩⋅⟨−1,1,4⟩=−1−1+8=6cos⁡θ=61+1+4 1+1+16⟶θ=54.74∘\begin{aligned} \langle 1,-1,2 \rangle \cdot \langle -1,1,4 \rangle &= -1-1+8 = 6\ \cos \theta = \dfrac{6}{\sqrt{1+1+4}\ \sqrt{1+1+16}} &\longrightarrow \ans{\theta = 54.74^{\circ}} \end{aligned}

Most of the derivative rules are what you would expect (the derivative of a sum/difference, constant multiple, etc.), but here are a few extensions of basic derivative rules, the Chain Rule and Product Rule:

Chain Rule

ddt(u⃗(f(t)))=u⃗ ′(f(t))f′(t)\dfrac{d}{dt} \left(\vec{u}(f(t))\right) = \vec{u}\ '\left(f(t)\right) f'(t)

Dot Product Rule

ddt(u⃗(t)⋅v⃗(t))=u⃗ ′(t)⋅v⃗(t)+u⃗(t)⋅v⃗ ′(t)\dfrac{d}{dt} \left(\vec{u}(t) \cdot \vec{v}(t)\right) = \vec{u}\ '(t) \cdot \vec{v}(t) + \vec{u}(t) \cdot \vec{v}\ '(t)

Cross Product Rule

ddt(u⃗(t)×v⃗(t))=u⃗ ′(t)×v⃗(t)+u⃗(t)×v⃗ ′(t)\dfrac{d}{dt} \left(\vec{u}(t) \times \vec{v}(t)\right) = \vec{u}\ '(t) \times \vec{v}(t) + \vec{u}(t) \times \vec{v}\ '(t)

Integrals of Vector-Valued Functions

By now it shouldn't come as a surprise that to integrate a vector-valued function, we can integrate the component functions individually. If r⃗(t)=⟨x(t),y(t),z(t)⟩\vec{r}(t) = \langle x(t),y(t),z(t) \rangle, then ∫r⃗(t) dt=(∫x(t) dt)i^+(∫y(t) dt)j^+(∫z(t) dt)k^+C⃗\int \vec{r}(t)\ dt = \left(\int x(t)\ dt\right) \hat{i} + \left(\int y(t)\ dt\right) \hat{j} + \left(\int z(t)\ dt\right) \hat{k} + \vec{C}

where C⃗\vec{C} is a constant vector ⟨c1,c2,c3⟩⟨c_1,c_2,c_3⟩. For a definite integral, ∫abr⃗(t) dt=(∫abx(t) dt)i^+(∫aby(t) dt)j^+(∫abz(t) dt)k^\int_a^b \vec{r}(t)\ dt = \left(\int_a^b x(t)\ dt\right) \hat{i} + \left(\int_a^b y(t)\ dt\right) \hat{j} + \left(\int_a^b z(t)\ dt\right) \hat{k}

If r⃗(t)=2t i^+sin⁡t j^+t2 k^\vec{r}(t) = 2t\ \hat{i} + \sin t\ \hat{j} + t^2\ \hat{k}, find ∫r⃗(t) dt\displaystyle\int \vec{r}(t)\ dt.

Solution

∫r⃗(t) dt=(t2+c1)i^+(−cos⁡t+c2)j^+(13t3+c3)k^=⟨t2,−cos⁡t,13t3⟩+⟨c1,c2,c3⟩\ans{\begin{aligned} \int \vec{r}(t)\ dt &= (t^2+c_1) \hat{i} + (-\cos t+c_2) \hat{j} + \left(\dfrac{1}{3}t^3+c_3\right) \hat{k}\ &= \left\langle t^2,-\cos t,\dfrac{1}{3}t^3 \right\rangle + \langle c_1,c_2,c_3 \rangle \end{aligned}}

If r⃗(t)=⟨2t+1,t2,t−3⟩\vec{r}(t) = \langle 2t+1,t^2,t-3 \rangle, find ∫12r⃗(t) dt\displaystyle\int_1^2 \vec{r}(t)\ dt.

Solution

∫12r⃗(t) dt=(t2+t∣12)i^+(13t3∣12)j^+(12t2−3t∣12)k^=4i^+73j^−32k^\begin{aligned} \int_1^2 \vec{r}(t)\ dt &= \left(t^2+t \bigg|_1^2\right) \hat{i} + \left(\dfrac{1}{3}t^3 \bigg|_1^2\right) \hat{j} + \left(\dfrac{1}{2}t^2-3t \bigg|_1^2\right) \hat{k}\ &= \ans{4\hat{i} + \dfrac{7}{3}\hat{j} - \dfrac{3}{2}\hat{k}} \end{aligned}

  1. If r⃗(t)=⟨tt2+2,e−3t,sin⁡(4t)+1⟩\vec{r}(t) = \left\langle \dfrac{t}{\sqrt{t^2+2}},e^{-3t},\sin (4t) + 1 \right\rangle, find ∫r⃗(t) dt\displaystyle\int \vec{r}(t)\ dt.

    2. ∫r⃗(t) dt=⟨t2+2,−13e−3t,t−14cos⁡4t⟩+C⃗\displaystyle\int \vec{r}(t)\ dt = \left\langle \sqrt{t^2+2},-\dfrac{1}{3}e^{-3t},t-\dfrac{1}{4}\cos 4t \right\rangle + \vec{C}

  2. If r⃗(t)=⟨1,3cos⁡t2,4t⟩\vec{r}(t) = \left\langle 1,3\cos \dfrac{t}{2},4t \right\rangle, find ∫0πr⃗(t) dt\displaystyle\int_0^\pi \vec{r}(t)\ dt

    3. ∫0πr⃗(t) dt=⟨π,6,2π2⟩\displaystyle\int_0^\pi \vec{r}(t)\ dt = \left\langle \pi,6,2\pi^2 \right\rangle