Path Length Slides Arc Length Re-mapping
Lesson 1: Parameterization by Distance
The Road Trip Problem
Time vs. Distance
Usually, we describe a path \( \mathbf{r}(t) \) based on time .
But if two cars drive the same road at different speeds, their equations look different. How do we describe the geometry of the road itself?
We use mile markers.Arc Length (s) .
The Arc Length Function
The distance \( s \) along a curve from a starting point \( a \) to time \( t \) is:
\[ s(t) = \int_{a}^{t} \|\mathbf{r}'(u)\| \, du \]
\( \mathbf{r}'(u) \) is the velocity vector.
\( \|\mathbf{r}'(u)\| \) is the speed.
The "Unit Speed" Advantage
Standardized
Every curve parameterized by \( s \) is traversed at exactly speed 1.
Intrinsic
The geometry depends only on the shape, not how fast you move through it.
Simplified
Calculations for curvature and TNB frames become much more elegant.
The 3-Step Process
01
Integrate to find \( s(t) \)
Solve the integral of speed from \( a \) to \( t \).
02
Invert to find \( t(s) \)
Solve the equation for \( t \) in terms of \( s \).
03
Substitute
Replace every \( t \) in your original \( \mathbf{r}(t) \) with your result for \( t(s) \).
Arc Length Worksheet Surveyor's Notebook
Arc Length Parameterization
Surveyor:
Log Date:
Objective: Transform a time-dependent path \( \mathbf{r}(t) \) into a distance-dependent path \( \mathbf{r}(s) \). This ensures the velocity vector always has a magnitude of 1 (unit speed).
1. The Transformation Integral
Recall the formula for arc length \( s \) as a function of time \( t \):
\[ s(t) = \int_{0}^{t} \| \mathbf{r}'(u) \| \, du \]
In your own words, explain why we use the magnitude of the derivative inside the integral to calculate the total length of the curve:
2. Guided Field Procedure
Example Problem: The Circular Helix
\[ \mathbf{r}(t) = \langle 3\cos(t), 3\sin(t), 4t \rangle \]
Step 1: Calculate \( \mathbf{r}'(t) \)
Step 2: Calculate speed \( \| \mathbf{r}'(t) \| \)
Step 3: Integrate to find \( s(t) \)
Step 4: Solve for \( t \) in terms of \( s \)
Step 5: Write the new \( \mathbf{r}(s) \)
3. Independent Practice
Re-parameterize the following curve with respect to arc length \( s \) measured from \( t = 0 \):
\[ \mathbf{r}(t) = \langle e^t \cos(t), e^t \sin(t), e^t \rangle \]
Calculation Area:
Final Verification:
Calculate \( \| \frac{d\mathbf{r}}{ds} \| \) for your resulting function. It should equal 1.
Unit Tangent Slides The Direction Vector
Lesson 2: Isolating the Heading with T(t)
Speed vs. Heading
The velocity vector \( \mathbf{r}'(t) \) contains two pieces of information:
"To analyze the geometry, we must strip away the speed."
A unit vector tells us only the where , never the how fast .
Defining T(t)
Unit Tangent Vector
\[ \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\| \mathbf{r}'(t) \|} \]
Dividing a vector by its own magnitude "normalizes" it, resulting in a vector of length 1 that points in the same direction.
Visualizing the Heading
At any point on the curve, \( \mathbf{T}(t) \) is the unit vector tangent to the curve.
Key Properties:
Always length 1
Points in the direction of motion
Undefined if speed is 0 (singular point)
T(t) points "down the road"
Calculation Demo
For the helix \( \mathbf{r}(t) = \langle \cos t, \sin t, t \rangle \):
1. Velocity:
\( \mathbf{r}'(t) = \langle -\sin t, \cos t, 1 \rangle \)
2. Speed:
\( \| \mathbf{r}'(t) \| = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt{2} \)
\( \mathbf{T}(t) = \left\langle -\frac{1}{\sqrt{2}}\sin t, \frac{1}{\sqrt{2}}\cos t, \frac{1}{\sqrt{2}} \right\rangle \)
Unit Tangent Worksheet Heading Analysis
Unit Tangent Vector T(t)
Navigator:
Course Date:
Standard Operating Procedure:
The velocity vector \( \mathbf{r}'(t) \) captures both speed and direction. To isolate the heading , we normalize the velocity vector. This provides a "unit speed" description of the path's orientation.
01 Normalization Routine
Calculate the Unit Tangent Vector \( \mathbf{T}(t) \) for the following position function:
\[ \mathbf{r}(t) = \langle t, t^2, \frac{2}{3}t^3 \rangle \]
A. Velocity Vector \( \mathbf{r}'(t) \)
B. Magnitude \( \| \mathbf{r}'(t) \| \)
C. Normalized Unit Tangent \( \mathbf{T}(t) \)
02 Interpretation
Consider a particle moving along a straight line path \( \mathbf{r}(t) = \mathbf{a} + t\mathbf{v} \).
What do you expect the unit tangent vector \( \mathbf{T}(t) \) to be? Explain your reasoning using the definition of normalization.
03 The Twisted Heading
Find \( \mathbf{T}(t) \) at the specific point \( t = \pi \) for the helix:
\[ \mathbf{r}(t) = \langle 4\cos(t), 4\sin(t), 3t \rangle \]
Critical Thinking
If a particle is traveling at constant speed, is the magnitude of \( \mathbf{T}'(t) \) necessarily zero? Explain.
Curvature Slides Bending and Curvature
Lesson 3: Measuring the Sharpness of a Turn
Quantifying "Curvy"
How quickly does your direction change as you move along a path?
"Curvature is the magnitude of the rate of change of the unit tangent vector with respect to arc length."
LOW
Large Highway Ramp (Gentle Bend)
HIGH
Tight Roundabout (Sharp Bend)
Defining \(\kappa\) (Kappa)
Theoretical Definition
\[ \kappa = \left\| \frac{d\mathbf{T}}{ds} \right\| \]
Change in tangent per unit of distance.
Computational Formula
\[ \kappa(t) = \frac{\| \mathbf{T}'(t) \|}{\| \mathbf{r}'(t) \|} \]
Normalize the rate of change by the speed.
The Intuition Check
For a circle of radius \( a \), the curvature is constant :
\[ \kappa = \frac{1}{a} \]
As radius \( a \) gets HUGE , curvature goes to 0 (a straight line).
As radius \( a \) gets tiny , curvature goes to infinity (a sharp point).
The "Kissing" Circle
The Osculating Circle at a point is the circle that best fits the curve at that specific location.
Its radius is \( \rho = \frac{1}{\kappa} \) (Radius of Curvature).
It shares the same tangent and curvature as the path.
"Osculare" = To Kiss
Curvature Worksheet Kappa Analytics
Curvature and Osculating Circles
Analyst:
Timestamp:
Curvature Formula
\[ \kappa(t) = \frac{\| \mathbf{T}'(t) \|}{\| \mathbf{r}'(t) \|} \]
Measures how fast the heading changes per unit distance.
Radius of Curvature
\[ \rho = \frac{1}{\kappa} \]
The radius of the circle that best fits the curve at a point.
1. Straight-Line Geometry
Consider the linear path \( \mathbf{r}(t) = \langle 3+2t, 5-t, 4t \rangle \). Show mathematically that the curvature \( \kappa(t) = 0 \) for all \( t \).
2. Constant Bend
Find the curvature \( \kappa \) of a circle in the xy-plane defined by \( \mathbf{r}(t) = \langle R\cos t, R\sin t \rangle \). Your answer should be in terms of the radius \( R \).
3. Local Sharpness
Calculate the curvature \( \kappa \) of the parabola \( y = x^2 \) at the origin \( (0,0) \).
Hint: Parameterize as \( \mathbf{r}(t) = \langle t, t^2, 0 \rangle \).
Finding the Center:
Based on your curvature at \( (0,0) \), what is the center of the osculating circle for this parabola at the origin?
Geometric Investigation
Where does the function \( y = \ln(x) \) have its maximum curvature? Does it occur where the slope is greatest, or elsewhere? Use your intuition (or some quick sketches) to form a hypothesis.
Principal Normal Slides The Turning Vector
Lesson 4: Defining the Principal Normal N(t)
Where is the Turn?
The Unit Tangent \( \mathbf{T}(t) \) tells us where we are going .
But it is constantly changing. The direction of that change tells us where we are turning .
"If T(t) is velocity of direction, N(t) is the acceleration of direction."
N(t) points toward the center of the bend.
Defining N(t)
Principal Unit Normal
\[ \mathbf{N}(t) = \frac{\mathbf{T}'(t)}{\| \mathbf{T}'(t) \|} \]
Just like we normalized velocity to get T, we normalize the derivative of T to get N.
A Geometric Guarantee
If \| \mathbf{T}(t) \| is constant...
\[ \mathbf{T}(t) \cdot \mathbf{T}'(t) = 0 \]
Therefore: \( \mathbf{T} \perp \mathbf{N} \)
The direction of change is always perpendicular to the direction of travel.
The Osculating Plane
The vectors T and N define a plane known as the Osculating Plane .
Geometric Fact:
This is the plane that contains the curve's velocity and acceleration vectors. It's the plane in which the curve is "currently turning."
Osculating Plane
T N
Principal Normal Worksheet Inward Force
The Principal Normal Vector N(t)
Navigator:
Course Date:
Standard Operating Procedure:
The Principal Normal Vector \( \mathbf{N}(t) \) points in the direction that the path is turning. It is always orthogonal to the unit tangent vector \( \mathbf{T}(t) \).
\[ \mathbf{N}(t) = \frac{\mathbf{T}'(t)}{\| \mathbf{T}'(t) \|} \]
01 The Property of Orthogonality
Explain why the dot product \( \mathbf{T}(t) \cdot \mathbf{T}'(t) \) must equal zero. Hint: Start with the fact that \( \mathbf{T}(t) \cdot \mathbf{T}(t) = 1 \) and differentiate both sides with respect to \( t \).
02 Normal Calculation
Find \( \mathbf{N}(t) \) for the circular path:
\[ \mathbf{r}(t) = \langle \cos(3t), \sin(3t) \rangle \]
Step A: Calculate \( \mathbf{T}(t) \)
Step B: Calculate \( \mathbf{T}'(t) \)
Step C: Final Normalized \( \mathbf{N}(t) \)
Does this vector point toward the origin (center of the circle)? Explain.
03 The Direction of the Turn
On the graph of the parabola \( y = x^2 \), sketch a representative **T** and **N** vector at the point \( (1, 1) \).
Mathematically verify your sketch by finding \( \mathbf{N}(t) \) for \( \mathbf{r}(t) = \langle t, t^2 \rangle \) at \( t = 1 \).
Binormal Torsion Slides Twist and Torsion
Lesson 5: Completing the TNB Frame
Completing the Basis
The **Binormal Vector** \( \mathbf{B}(t) \) is the final vector in our intrinsic coordinate system.
\[ \mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t) \]
It is a unit vector perpendicular to both the Tangent and the Normal.
A Right-Handed System
What is Torsion?
While **curvature** (\( \kappa \)) measures how a curve bends *within* its plane...
**Torsion** (\( \tau \)) measures how a curve twists *out* of its plane.
Planar Curves have \( \tau = 0 \)
Example: Circles, Parabolas
Defining \(\tau\) (Tau)
Rate of Twist
\[ \frac{d\mathbf{B}}{ds} = -\tau \mathbf{N} \]
Torsion is the speed at which the binormal vector (and thus the osculating plane) is rotating.
Computational Formula:
\[ \tau = \frac{(\mathbf{r}' \times \mathbf{r}'') \cdot \mathbf{r}'''}{\| \mathbf{r}' \times \mathbf{r}'' \|^2} \]
The Frenet-Serret Formulas
Change in T
\[ \frac{d\mathbf{T}}{ds} = \kappa \mathbf{N} \]
Change in N
\[ \frac{d\mathbf{N}}{ds} = -\kappa \mathbf{T} + \tau \mathbf{B} \]
Change in B
\[ \frac{d\mathbf{B}}{ds} = -\tau \mathbf{N} \]
"These equations describe how the frame moves through space, completely defining the curve's geometry."
Binormal Torsion Worksheet The Third Axis
Binormal Vectors & Torsion
Observer:
Timestamp:
The Binormal B(t)
Defines the orientation of the osculating plane. Perpendicular to both T and N.
\[ \mathbf{B} = \mathbf{T} \times \mathbf{N} \]
Torsion \(\tau\)
Measures the rate at which the curve twists out of its plane of motion.
\[ \frac{d\mathbf{B}}{ds} = -\tau \mathbf{N} \]
1. Completing the Frame
A curve has the following vectors at a specific point \( P \):
\( \mathbf{T} = \langle 1, 0, 0 \rangle \) \( \mathbf{N} = \langle 0, 1, 0 \rangle \)
Calculate the Binormal Vector \( \mathbf{B} \):
2. Twisted Paths
Determine if the following curves will have zero or non-zero torsion based on their geometry:
A circle in the xy-plane.
Zero Non-Zero
Justification:
A circular helix rising in z.
Zero Non-Zero
Justification:
3. Full TNB Calculation
For the helix \( \mathbf{r}(t) = \langle \cos t, \sin t, t \rangle \), we have already found:
\( \mathbf{T}(t) = \frac{1}{\sqrt{2}}\langle -\sin t, \cos t, 1 \rangle \)
Find \( \mathbf{B}(t) \):
Find \( \mathbf{B}'(t) \):
Solving for Torsion (\(\tau\)):
Using the relation \( \mathbf{B}'(s) = -\tau \mathbf{N} \). Note that \( \mathbf{B}'(s) = \frac{\mathbf{B}'(t)}{\| \mathbf{r}'(t) \|} \).
