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Wize University Calculus 2 Textbook > Vector Functions

Vector Functions

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Vector-Valued Functions

Vector-valued functions (or vector functions) take in real numbers and outputs vectors
r⃗(t)=⟨f(t), g(t), h(t)⟩\vec r\left(t\right)=\left\langle f\left(t\right),\ g\left(t\right),\ h\left(t\right)\right\rangle or r⃗(t)=f(t)i+g(t)j+h(t)k\vec{r}\left(t\right)=f\left(t\right)\bold{i}+g\left(t\right)\bold{j}+h\left(t\right)\bold{k}
where f, g, hf,\ g,\ h are called component functions


Limits & continuity

The limit is defined by lim⁡t→ar⃗(t)=⟨lim⁡t→af(t), lim⁡t→ag(t), lim⁡t→ah(t)⟩\displaystyle \lim_{t\to a}\vec r\left(t\right)=\left\langle\lim_{t\to a}f\left(t\right),\ \lim_{t\to a}g\left(t\right),\ \lim_{t\to a}h\left(t\right)\right\rangle

A vector function is continuous at t=at=a if lim⁡t→ar⃗(t)=r⃗(a)\displaystyle \lim_{t\to a}\vec r\left(t\right)=\vec r\left(a\right)
(i.e. r(t)r(t) is continuous at t=at=a if and only if all of the component functions are continuous at t=at=a)


Space Curves

A space curve traces out the set of points ⟨x, y, z⟩= ⟨f(t), g(t), h(t)⟩\left\langle x,\ y,\ z\right\rangle=\ \left\langle f\left(t\right),\ g\left(t\right),\ h\left(t\right)\right\rangle, often for tt values within a certain range a≤t≤ba\le t\le b.


Sometimes the question will give you two different versions of the same vector function, or they will ask kyou to rewrite the vector function using different component functions. This is called reparametrization of the curve

Example
r⃗(t)=⟨t, 2t, 3t⟩\vec{r}\left(t\right)=\left\langle t,\ 2t,\ 3t\right\rangle, 0≤t≤10\le t\le1 and r⃗(s)=⟨s14, 2s14, 3s14⟩\vec{r}\left(s\right)=\left\langle\frac{s}{\sqrt{14}},\ \frac{2s}{\sqrt{14}},\ \frac{3s}{\sqrt{14}}\right\rangle, 0≤s≤140\le s\le\sqrt{14} are two parametrizations of the same space curve.
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Derivatives

The derivative or tangent vector to the curve at the point tt is defined by
dr⃗dt or r⃗′(t)=lim⁡h→0r⃗(t+h)−r⃗(t)h=⟨f′(t), g′(t), h′(t)⟩\displaystyle \frac{d\vec r}{dt}\ \text{or}\ \vec r'\left(t\right)=\lim_{h\to0}\frac{\vec r\left(t+h\right)-\vec r\left(t\right)}{h}=\langle f'(t),\ g'(t),\ h'(t)\rangle


Derivative Rules

  • ddt[u⃗(t)±v⃗(t)]=u⃗′(t)+v⃗′(t)\displaystyle \frac{d}{dt}\left[\vec{u}\left(t\right)\pm\vec{v}\left(t\right)\right]=\vec{u}'\left(t\right)+\vec{v}'\left(t\right)
  • ddt[cu⃗(t)]=cu⃗′(t)\displaystyle \frac{d}{dt}\left[c\vec{u}\left(t\right)\right]=c\vec{u}'\left(t\right)
  • ddt[u⃗(t) v⃗(t)]=u⃗′(t) v⃗(t)+u⃗(t) v⃗′(t)\displaystyle \frac{d}{dt}\left[\vec{u}\left(t\right)\ \vec{v}\left(t\right)\right]=\vec{u}'\left(t\right)\ \vec{v}\left(t\right)+\vec{u}\left(t\right)\ \vec{v}'\left(t\right) *this rule is true for dot and cross products as well
  • ddt[u⃗(f(t))]=u⃗′(f(t)) f′(t)\displaystyle \frac{d}{dt}\left[\vec{u}\left(f\left(t\right)\right)\right]=\vec{u}'\left(f\left(t\right)\right)\ f'\left(t\right)


Integrals

The indefinite integral is defined as
∫r⃗(t)dt=⟨∫f(t)dt,  ∫g(t)dt,  ∫h(t)dt⟩\displaystyle \int \vec{r}\left(t\right)dt=\left\langle \int f\left(t\right)dt,\ \ \int g\left(t\right)dt,\ \ \int h\left(t\right)dt\right\rangle

*There will be a constant vector C⃗\displaystyle \vec C at the end

The definite integral is defined as ∫abr⃗(t)dt=⟨∫abf(t)dt,  ∫abg(t)dt,  ∫abh(t)dt⟩\displaystyle \int _a^b\vec{r}\left(t\right)dt=\left\langle \int _a^bf\left(t\right)dt,\ \ \int _a^bg\left(t\right)dt,\ \ \int _a^bh\left(t\right)dt\right\rangle

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Arc Length

The length of a space curved defined by r⃗(t)=⟨f(t), g(t), h(t)⟩\displaystyle \vec{r}(t)=\left\langle f\left(t\right),\ g\left(t\right),\ h\left(t\right)\right\rangle where a≤t≤ba\le t\le b is
L=∫ab∣r⃗ ′(t)∣dt\displaystyle L=\int_a^b|\vec{r}\ '(t)|dt

The arc length function s(t)s(t) defines the length of the space curve CC from the starting point t=at=a to any point tt, and is given by
s(t)=∫at∣r⃗ ′(u)∣du\displaystyle s(t)=\int_a^t\left|\vec r\ '(u)\right|du

*By FTC I, we see that dsdt=∣r⃗′(t)∣\frac{ds}{dt}=\left|\vec{r}'\left(t\right)\right|

Reparametrizing a curve w/ respect to arc length

If a vector function and it's initial position is given, we can reparametrize the curve using the following method
  1. Determine the tt value
  2. Find the arc length function of the curve s(t)=∫0t∣r⃗ ′(u)∣dus\left(t\right)=\int_0^t\left|\vec{r}\ '\left(u\right)\right|du
  3. Solve for tt in terms of ss
  4. Replace tt in the vector function with this new ss expression, and change the initial bounds on tt to bounds on ss
Example
Reparamaterize the curve r⃗(t)=etcos⁡ti⃗+etsin⁡tj⃗+k\vec{r}\left(t\right)=e^t\cos t\vec{i}+e^t\sin t\vec{j}+k with respect to the arc length starting from the point (1, 0, 1)\left(1,\ 0,\ 1\right)
1. The point (1, 0 ,1)\left(1,\ 0\ ,1\right) tells us that
  • etcos⁡t=1e^t\cos t=1
  • etsin⁡t=0e^t\sin t=0
  • 1=11=1
Meaning that our curve starts at t=0t=0.

2. We need the arc length function:
s(t)=∫0t∣r⃗ ′(u)∣dus\left(t\right)=\int_0^t\left|\vec{r}\ '\left(u\right)\right|du
s(t)=∫0t∣⟨eucos⁡u−eusin⁡u, eusin⁡u+eucos⁡u, 0⟩∣ dus\left(t\right)=\int_0^t\left|\left\langle e^u\cos u-e^u\sin u,\ e^u\sin u+e^u\cos u,\ 0\right\rangle\right|\ du
s(t)=∫0t(eucos⁡u−eusin⁡u)2+(eusin⁡u+eucos⁡u)2+(0)2dus(t)=\int_0^t\sqrt{\left(e^u\cos u-e^u\sin u\right)^2+\left(e^u\sin u+e^u\cos u\right)^2+\left(0\right)^2}du
s(t)=∫0te2u(1−2cos⁡usin⁡u)+e2u(1+2sin⁡ucos⁡u)dus\left(t\right)=\int_0^t\sqrt{e^{2u}\left(1-2\cos u\sin u\right)+e^{2u}\left(1+2\sin u\cos u\right)}du
s(t)=∫0t2e2udus\left(t\right)=\int_0^t\sqrt{2e^{2u}}du
s(t)=∫0t2eu dus(t)=\int_0^t\sqrt 2e^u\ du
s(t)=2et−2s(t)=\sqrt 2e^t-\sqrt 2

3. Solving for tt:
s+22=et\frac{s+\sqrt{2}}{\sqrt{2}}=e^t
ln⁡∣s+22∣=t\ln\left|\frac{s+\sqrt{2}}{\sqrt{2}}\right|=t

4. Therefore,
r⃗(t(s))=eln⁡∣s+22∣cos⁡(ln⁡∣s+22∣)i⃗+eln⁡∣s+22∣sin⁡(ln⁡∣s+22∣)j⃗+k⃗\vec{r}\left(t\left(s\right)\right)=e^{\ln\left|\frac{s+\sqrt{2}}{\sqrt{2}}\right|}\cos\left(\ln\left|\frac{s+\sqrt{2}}{\sqrt{2}}\right|\right)\vec{i}+e^{\ln\left|\frac{s+\sqrt{2}}{\sqrt{2}}\right|}\sin\left(\ln\left|\frac{s+\sqrt{2}}{\sqrt{2}}\right|\right)\vec{j}+\vec{k}
r⃗(t(s))=s+22cos⁡(ln⁡∣s+22∣)i⃗+s+22sin⁡(ln⁡∣s+22∣)j⃗+k⃗\vec{r}\left(t\left(s\right)\right)=\frac{s+\sqrt{2}}{\sqrt{2}}\cos\left(\ln\left|\frac{s+\sqrt{2}}{\sqrt{2}}\right|\right)\vec{i}+\frac{s+\sqrt{2}}{\sqrt{2}}\sin\left(\ln\left|\frac{s+\sqrt{2}}{\sqrt{2}}\right|\right)\vec{j}+\vec{k}
where ss starts at 00
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Curvature

A curve is said to be smooth if it's parametrization r⃗(t)\vec{r}\left(t\right) is such that r⃗ ′(t)\vec{r}\ '\left(t\right) is continuous and r⃗ ′(t)≠0\vec{r}\ '\left(t\right)\ne0 -- the curve has no sharp corners or cusps.

Suppose that a smooth curve C is defined by the vector function r⃗(t)\vec{r}\left(t\right), then we have the following:
Unit Tangent VectorT⃗(t)=r⃗ ′(t)∣r⃗ ′(t)∣Tells us the direction of the cruve at a given pointUnit Normal VectorN⃗(t)=T⃗ ′(t)∣T⃗ ′(t)∣Tells us the direction the curve is turning at a given point*Orthogonal to the unit tangent vectorCurvatureκ=∣T⃗ ′(t)∣∣r⃗ ′(t)∣=∣r⃗ ′(t)×r⃗ ′′(t)∣∣r⃗ ′(t)∣3Tells us how fast the curve changes direction at a given pointBinomial VectorB⃗(t)=T⃗(t)×N⃗(t)Orthogonal to both unit tangent and unit normal vectors\begin{array}{|c|c|c|} \hline \textcolor{orange}{\text{Unit Tangent Vector}}&\displaystyle\vec T(t)=\frac{\vec r\ '(t)}{|\vec r\ '(t)|}&\text{Tells us the direction of the cruve at a given point}\\\\ \hline \textcolor{orange}{\text{Unit Normal Vector}}&\displaystyle\vec N(t)=\frac{\vec T\ '(t)}{|\vec T\ '(t)|}&\begin{array}{c}\text{Tells us the direction the curve is turning at a given point}\\\text{*Orthogonal to the unit tangent vector}\end{array}\\\\ \hline \textcolor{orange}{\text{Curvature}}&\displaystyle \kappa =\frac{|\vec T\ '(t)|}{|\vec r\ '(t)|}=\frac{|\vec r\ '(t)\times\vec r\ ''(t)|}{|\vec r\ '(t)|^3}&\text{Tells us how fast the curve changes direction at a given point}\\\\ \hline \textcolor{orange}{\text{Binomial Vector}}&\displaystyle \vec B(t)=\vec T(t)\times\vec N(t)&\text{Orthogonal to both unit tangent and unit normal vectors}\\\\\hline \end{array}
  • Normal plane at a given point: the plane determined by N⃗(t)\vec{N}\left(t\right) and B⃗(t)\vec{B}\left(t\right)
  • Osculating plane at a given point: the plane determined by T⃗(t)\vec{T}\left(t\right) and N⃗(t)\vec{N}\left(t\right)
  • Osculating circle (or the circle of curvature) at a given point: best describes how the curve behaves near that point, and shares the same tangent, normal, and curvature at the point

Example
Find the unit tangent vector, curvature, and unit normal vector of the curve r⃗(t)=⟨3sin⁡t,3cos⁡t,4t⟩\vec r(t)=\left\langle3\sin t,3\cos t,4t\right\rangle at the point (0,3,0)\left(0,3,0\right)

Unit Tangent Vector
At the point (0,3,0)(0,3,0), we can solve for t and get t=0t=0
  • r⃗′(t)=⟨3cos⁡t,−3sin⁡t,4⟩\vec{r}'\left(t\right)=\left\langle3\cos t,-3\sin t,4\right\rangle
  • ∣r⃗′(t)∣=9cos⁡2t+9sin⁡2t+16=25=5\left|\vec{r}'\left(t\right)\right|=\sqrt{9\cos^2t+9\sin^2t+16}=\sqrt{25}=5
So, the unit tangent vector is T⃗(t)=⟨3cos⁡t,−3sin⁡t,4⟩5 → T⃗(0)= ⟨35,0,45⟩\vec{T}\left(t\right)=\frac{\left\langle3\cos t,-3\sin t,4\right\rangle}{5}\ \to\ \vec{T}\left(0\right)=\ \left\langle\frac{3}{5},0,\frac{4}{5}\right\rangle

Curvature
  • ∣T⃗′(t)∣=∣⟨−35sin⁡t, −35cos⁡t, 0⟩∣=925sin⁡2t+925cos⁡2t=35\left|\vec{T}'\left(t\right)\right|=\left|\left\langle-\frac{3}{5}\sin t,\ -\frac{3}{5}\cos t,\ 0\right\rangle\right|=\sqrt{\frac{9}{25}\sin^2t+\frac{9}{25}\cos^2t}=\frac{3}{5}
  • ∣r⃗′(t)∣=5\left|\vec{r}'\left(t\right)\right|=5
So, the curvature is κ=355=325\kappa=\frac{\frac{3}{5}}{5}=\frac{3}{25}

Unit Normal Vector
  • T⃗′(t)=⟨−35sin⁡t,−35cos⁡t,0⟩\vec{T}'\left(t\right)=\left\langle-\frac{3}{5}\sin t,-\frac{3}{5}\cos t,0\right\rangle
  • ∣T⃗′(t)∣=35\left|\vec{T}'\left(t\right)\right|=\frac{3}{5}
So, the unit normal vector is N⃗(t)=⟨−35sin⁡t,−35cos⁡t,0⟩35 → N⃗(0)=⟨0,−1,0⟩\vec{N}\left(t\right)=\frac{\left\langle-\frac{3}{5}\sin t,-\frac{3}{5}\cos t,0\right\rangle}{\frac{3}{5}}\ \to\ \vec{N}\left(0\right)=\left\langle0,-1,0\right\rangle
Extra Practice
Given the following vector equation, sketch out the curve that it generates.
r(t)=⟨cos⁡(t),t⟩r(t) = \langle \cos (t), t \rangle
3 The position of an object is given by
r⃗(t)=ln⁡(sin⁡t)i⃗−tk⃗\vec r(t) = \ln(\sin t)\vec i -t\vec k
for 0<t<π. 0 < t < \pi. Determine the minimum speed of the object.
2 The position of an object is given by
r⃗(t)=ln⁡(sin⁡t)i⃗−tk⃗\vec r(t) = \ln(\sin t)\vec i -t\vec k
Determine the velocity, acceleration, and speed at time t.
1 The position of an object is given by
r⃗(t)=etsin⁡ti⃗+etcos⁡tj⃗−etk⃗\vec r(t) = e^t\sin t\vec i + e^t\cos t\vec j-e^t\vec k
Determine the velocity, acceleration, and speed at time t.
Find the Cartesian equation of the tangent line to the following curve at the indicated point:
r⃗(t)=ln⁡(cos⁡t)i⃗+sin⁡(2t)j⃗−cos⁡2tk⃗ at t=π/6\vec r(t) = \ln(\cos t) \vec i + \sin(2t) \vec j- \cos^2 t \vec k\quad \text{ at } t = \pi/6
Practice: Domain of a vector-valued function 2
Find the domain of the following function. Express the domain as an interval, or a union of intervals. i.e. t∈[−3,3)t \in [-3, 3) or t∈(−∞,π]∪(13,∞)t \in (-\infty, \pi] \cup ( 13, \infty) .
v(x)=⟨x9x2−x4,−log⁡(−x3)⟩v(x) = \left \langle \frac{x}{\sqrt{9x^2 - x^4}}, - \log \left ( - \frac{x}{3} \right )\right \rangle
The following curves intersect when t=0t = 0 . What is the angle between these two curves:
r1(t)=⟨t,1+t2,cos⁡(t)+t⟩r2(t)=⟨sin⁡(t),cos⁡(t),et⟩r_1(t) = \langle t, 1 + t^2, \cos(t) + t \rangle \quad\quad\quad r_2(t) = \langle \sin(t), \cos(t), e^t \rangle
Let's say that a vector function r(t)r(t) that describes a curve CC has the property that for all tt , r(t)r(t) is perpendicular to its tangent vector. Demonstrate that the curve must be completely contained on the surface of a sphere centered at the origin.
Give at least one point where the following vector function
r(t)=⟨−sin⁡t,2et,−t⟩r(t) = \langle -\sin t, 2e^t, -t\rangle
has tangent vectors normal to the plane whose equation is