In our last lecture, we explored the physics of motion, analyzing position, velocity, and acceleration. Today, we shift our perspective to the geometry of the path itself. We will answer fundamental questions like: "How long is a curved path?" and "How can we describe its orientation in space at any given moment?" This will lead us to the concepts of arc length and the foundational TNB frame, a moving coordinate system that travels along a curve.
How can we find the exact length of a winding curve in space? The core idea, as is common in calculus, is to approximate, sum, and take a limit. We can approximate the length of the curve by connecting a series of points along it with straight line segments, creating a polygonal path. The more segments we use, the better our approximation becomes.
We approximate the length of the curve by summing the lengths of many small, straight line segments. Move slider h to see how the line segment approximation gets better and better as h->0.
Let's focus on the length of one of these tiny segments, which we'll call $\Delta s$. This segment connects the point $\mathbf{r}(t)$ to a nearby point $\mathbf{r}(t + \Delta t)$. The vector representing this segment is $\Delta \mathbf{r} = \mathbf{r}(t + \Delta t) - \mathbf{r}(t)$. Its length is simply its magnitude, $|\Delta \mathbf{r}|$.
Here is where the derivative comes in. Recall the definition of the derivative of our vector function:
For a very small $\Delta t$, this means $\mathbf{r}'(t) \approx \frac{\Delta \mathbf{r}}{\Delta t}$, which we can rearrange to get $\Delta \mathbf{r} \approx \mathbf{r}'(t) \Delta t$. Taking the magnitude of both sides gives us the approximate length of our small segment:
To get the total length $L$, we sum up all these small lengths ($\sum \Delta s$) and take the limit as $\Delta t \to 0$. This process of summing infinitesimal pieces is precisely what a definite integral does.
The arc length $L$ of a curve defined by the vector function $\mathbf{r}(t)$ from $t=a$ to $t=b$ is given by the integral of its speed:
$$ L = \int_{a}^{b} |\mathbf{r}'(t)| \,dt = \int_{a}^{b} \sqrt{[x'(t)]^2 + [y'(t)]^2 + [z'(t)]^2} \,dt $$Find the length of one full turn of the helix with vector equation $\mathbf{r}(t) = \langle \cos(t), \sin(t), t \rangle$, starting from $t=0$.
Step 1: Find the derivative $\mathbf{r}'(t)$.
$$ \mathbf{r}'(t) = \langle -\sin(t), \cos(t), 1 \rangle $$
Step 2: Find the magnitude of the derivative (the speed).
$$ |\mathbf{r}'(t)| = \sqrt{(-\sin t)^2 + (\cos t)^2 + 1^2} = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt{1 + 1} = \sqrt{2} $$
Notice that the speed is constant! This makes the integration very simple.
Step 3: Set up and evaluate the integral. One full turn corresponds to the interval $0 \le t \le 2\pi$.
$$ L = \int_{0}^{2\pi} \sqrt{2} \,dt = [\sqrt{2}t]_{0}^{2\pi} = \sqrt{2}(2\pi) - 0 = 2\pi\sqrt{2} $$
The length of one turn of the helix is $2\pi\sqrt{2}$ units.
The blue curve shows one full turn of the helix from $t=0$ to $t=2\pi$. We just calculated its exact length.
Calculating the total length between two points is useful, but a more powerful idea is to create a function that tells us the distance traveled along a curve from a starting point $t=a$ to any arbitrary time $t$. This is the arc length function, $s(t)$.
The arc length function $s(t)$ measures the length of the curve from a starting time $a$ to any time $t$:
$$ s(t) = \int_{a}^{t} |\mathbf{r}'(u)| \,du $$(Note: We use $u$ as the variable of integration to avoid confusion with the upper limit $t$.)
As a direct result of this definition, the derivative of a function reparametrized by arc length always has a magnitude of 1. That is:
$$ |\mathbf{r}'(s)| = 1 $$This means that if you use distance $s$ as your parameter, you travel along the curve at a constant "unit speed" with respect to that parameter. This property is what makes arc length the most natural parameter for studying geometry.
Let's reparametrize the helix $\mathbf{r}(t) = \langle \cos(t), \sin(t), t \rangle$ with respect to arc length, starting from $t=0$.
Step 1: Find the arc length function $s(t)$.
We already know the speed is $|\mathbf{r}'(t)| = \sqrt{2}$. We'll use $a=0$ as our starting point.
$$ s(t) = \int_{0}^{t} |\mathbf{r}'(u)| \,du = \int_{0}^{t} \sqrt{2} \,du = \sqrt{2}t $$
So, we have a simple relationship: the distance traveled is $s = \sqrt{2}t$.
Step 2: Solve for $t$ in terms of $s$.
From $s = \sqrt{2}t$, we get $t = \frac{s}{\sqrt{2}}$. This equation allows us to convert from "distance traveled" back to "time elapsed."
Step 3: Substitute this expression for $t$ back into the original vector function.
Replace every $t$ in $\mathbf{r}(t)$ with $\frac{s}{\sqrt{2}}$ to get our new parametrization, $\mathbf{r}(s)$:
$$ \mathbf{r}(s) = \left\langle \cos\left(\frac{s}{\sqrt{2}}\right), \sin\left(\frac{s}{\sqrt{2}}\right), \frac{s}{\sqrt{2}} \right\rangle $$This is the same helix, but now the input parameter tells us how far we are along the curve instead of for how long we've been traveling.
Once we reparametrize and write $\mathbf{r}(s)$, we have a new derivative to consider: $\frac{d\mathbf{r}}{ds}$. Your intuition might tell you that since $\mathbf{r}$ is position and $s$ is distance, this derivative is measuring the "change in position with respect to distance." You might then ask, "Isn't the change in position the same as the change in distance?"
This is a fantastic question, and the answer lies in the crucial difference between vectors and scalars.
Imagine you drive from Washington, D.C. to Richmond, VA along the winding I-95 highway.
The key insight from calculus is that when we look at an infinitesimally small step along the curve, the tiny, winding piece of road becomes indistinguishable from a tiny, straight line. At that tiny scale, the length of the position change vector, $|d\mathbf{r}|$, becomes equal to the distance traveled, $ds$. So, we arrive at the crucial relationship: $|d\mathbf{r}| = ds$.
When we calculate the derivative $\frac{d\mathbf{r}}{ds}$, we are dividing a vector ($d\mathbf{r}$) by its own magnitude ($ds$). As we know, any vector divided by its magnitude results in a unit vector that points in the same direction. $$ \frac{d\mathbf{r}}{ds} = \text{a vector with a magnitude of exactly 1 pointing in the direction of travel} $$ This is the **unit tangent vector**, $\mathbf{T}$. By using $s$ as our parameter, we eliminate any changes in the math caused by how fast we are traveling. No matter how you drive from D.C. to Richmond—speeding up or slowing down—the odometer will always read 108 miles (assuming you take the same route). The geometry of the path is independent of your speed, and the arc length parameter $s$ captures this perfectly.
The velocity vector $\mathbf{v}(t) = \mathbf{r}'(t)$ tells us two things: the direction of motion and the speed. If we only care about the direction, we can create a unit vector by dividing the velocity vector by its magnitude (the speed). This gives us the Unit Tangent Vector, $\mathbf{T}(t)$.
The Unit Tangent Vector, $\mathbf{T}(t)$, is a vector of length 1 that points in the direction of motion along the curve.
$$ \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|} $$Find the unit tangent vector for the helix $\mathbf{r}(t) = \langle \cos(t), \sin(t), t \rangle$.
From our previous example, we already know $\mathbf{r}'(t) = \langle -\sin(t), \cos(t), 1 \rangle$ and $|\mathbf{r}'(t)| = \sqrt{2}$.
Therefore, we just need to divide the vector by its magnitude:
$$ \mathbf{T}(t) = \frac{\langle -\sin(t), \cos(t), 1 \rangle}{\sqrt{2}} = \left\langle -\frac{\sin t}{\sqrt{2}}, \frac{\cos t}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right\rangle $$Let's connect the last two topics. We just defined the unit tangent vector as $\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}$.
In the previous section, we showed that when we differentiate with respect to arc length, we get $\frac{d\mathbf{r}}{ds} = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}$.
This reveals a fundamental identity: the derivative of the position vector with respect to arc length is the unit tangent vector.
$$ \mathbf{r}'(s) = \mathbf{T} $$This is a core concept: traveling at "unit speed" with respect to distance means your velocity vector is always a unit vector pointing along the tangent.
The red vector is $\mathbf{T}(t)$. Drag the slider for 's' to move the point along the curve and observe how the tangent vector always points in the direction of travel.
Find the unit tangent vector $\mathbf{T}(t)$ for the curve $\mathbf{r}(t) = \langle t^2, 2t, \ln t \rangle$ at the point where $t=1$.
As a particle moves, its Unit direction vector $\mathbf{T}$ pivots to follow the path. To understand how it's changing at any instant, we look at its derivative, $\mathbf{T}'(t)$.
No, it is not. Think of the unit tangent vector $\mathbf{T}$ as the velocity vector "stripped" of its speed information; it only tells you the direction of travel. Because $\mathbf{T}$ only contains directional information, its derivative, $\mathbf{T}'$, can only tell us about the change in direction.
Acceleration, on the other hand, is the rate of change of the full velocity vector ($\mathbf{v}$), which has both speed and direction. Therefore, acceleration must account for changes in both speed and direction. This is why the product rule is essential. Since $\mathbf{v} = v\mathbf{T}$ (speed times direction), the acceleration is $\mathbf{a} = \mathbf{v}' = v'\mathbf{T} + v\mathbf{T}'$. The vector $\mathbf{T}'$ is a key ingredient, but it's not the whole recipe for acceleration.
We know that $\mathbf{T}$ is a unit vector, so its magnitude is always 1. Squaring this gives $|\mathbf{T}(t)|^2 = 1$.
A key property is that the square of a vector's magnitude is equal to the dot product of the vector with itself. So, we can write:
Now, we differentiate both sides of this equation with respect to $t$. We must use the product rule for dot products on the left side:
This implies that $\mathbf{T}(t) \cdot \mathbf{T}'(t) = 0$. Since their dot product is zero, the vectors $\mathbf{T}$ and $\mathbf{T}'$ are always orthogonal. This is a powerful result: the rate of change of any vector with constant magnitude is always orthogonal to the vector itself.
Because $\mathbf{T}'(t)$ gives the direction of the turn and is orthogonal to the direction of motion, we normalize it to create our Principal Unit Normal Vector.
The formula for curvature, $\kappa = \frac{|\mathbf{T}'(t)|}{|\mathbf{r}'(t)|}$, can be understood as a ratio that removes the influence of a particle's speed to describe a purely geometric property of the curve.
To see why, imagine two cars on the same circular track: Car A is driving fast and Car B is driving slow. The track itself has a single, constant curvature. However, Car A's direction vector T is changing very quickly, while Car B's is changing slowly. If we only used the "turn rate" ($|\mathbf{T}'(t)|$), we would incorrectly conclude the track has two different curvatures.
We must normalize by the speed ($|\mathbf{r}'(t)|$) to find the true curvature of the track:
$$\kappa_{\text{Track}} = \frac{\text{High Turn Rate (Car A)}}{\text{High Speed (Car A)}} = \frac{\text{Low Turn Rate (Car B)}}{\text{Low Speed (Car B)}}$$The ratio cancels out the "per time" aspect of the motion, leaving a pure geometric measure: the change in direction per unit of distance traveled. This is why a sharp corner has high curvature regardless of how fast you take it.
The Principal Unit Normal Vector, $\mathbf{N}(t)$, is a unit vector that points in the direction in which the curve is turning. It is always orthogonal to the unit tangent vector $\mathbf{T}(t)$.
$$ \mathbf{N}(t) = \frac{\mathbf{T}'(t)}{|\mathbf{T}'(t)|} $$At any point on a curve, there isn't just one vector orthogonal to the tangent vector $\mathbf{T}$—there's an entire plane of them, called the normal plane. So why is $\mathbf{N}$ considered the "principal" normal vector?
$\mathbf{N}$ is special because it is the unique vector in the normal plane that also points in the direction the curve is turning. It is defined by the geometry of the curve itself, not just by the condition of being orthogonal to $\mathbf{T}$.
The graph below shows the normal plane. Any vector lying in this plane is orthogonal to T. Move the slider 's' to rotate a sample vector around in the plane. Notice that only one vector (the blue N vector) correctly points in the direction of the curve's bend. You can also click on the expressions for vectors I and J to see other examples of vectors in this plane.
Find the principal unit normal vector for the helix $\mathbf{r}(t) = \langle \cos(t), \sin(t), t \rangle$.
Step 1: Differentiate $\mathbf{T}(t)$. From the previous example, $\mathbf{T}(t) = \langle -\frac{\sin t}{\sqrt{2}}, \frac{\cos t}{\sqrt{2}}, \frac{1}{\sqrt{2}} \rangle$.
$$ \mathbf{T}'(t) = \left\langle -\frac{\cos t}{\sqrt{2}}, -\frac{\sin t}{\sqrt{2}}, 0 \right\rangle $$
Step 2: Find the magnitude $|\mathbf{T}'(t)|$.
$$ |\mathbf{T}'(t)| = \sqrt{\left(-\frac{\cos t}{\sqrt{2}}\right)^2 + \left(-\frac{\sin t}{\sqrt{2}}\right)^2 + 0^2} = \sqrt{\frac{\cos^2 t}{2} + \frac{\sin^2 t}{2}} = \sqrt{\frac{1}{2}(\cos^2 t + \sin^2 t)} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} $$
Step 3: Calculate $\mathbf{N}(t) = \frac{\mathbf{T}'(t)}{|\mathbf{T}'(t)|}$.
$$ \mathbf{N}(t) = \frac{\langle -\frac{\cos t}{\sqrt{2}}, -\frac{\sin t}{\sqrt{2}}, 0 \rangle}{1/\sqrt{2}} = \langle -\cos t, -\sin t, 0 \rangle $$
Notice that for the helix, this vector always points directly toward the z-axis, which is the center of its circular path in the xy-plane.
The red vector is $\mathbf{T}(t)$ and the blue vector is $\mathbf{N}(t)$. Notice how $\mathbf{N}$ always points "inward" toward the center of the curve's bend.
Find the principal unit normal vector $\mathbf{N}(t)$ for the circular path $\mathbf{r}(t) = \langle 5\cos t, 5\sin t, 3 \rangle$.
We now have two orthogonal unit vectors, $\mathbf{T}$ and $\mathbf{N}$, that describe the plane in which the curve is turning at a particular moment. We can define a third vector, orthogonal to both, to complete a right-handed 3D coordinate system. This is achieved with the cross product.
The Binormal Vector, $\mathbf{B}(t)$, is defined as the cross product of the tangent and normal vectors.
$$ \mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t) $$Together, the set of mutually orthogonal unit vectors $\{\mathbf{T}, \mathbf{N}, \mathbf{B}\}$ is called the TNB Frame or Frenet-Serret frame. It's a moving coordinate system that provides a complete geometric description of the curve at every point.
Find the binormal vector for the helix $\mathbf{r}(t) = \langle \cos(t), \sin(t), t \rangle$.
We have already found:
$\mathbf{T}(t) = \langle -\frac{\sin t}{\sqrt{2}}, \frac{\cos t}{\sqrt{2}}, \frac{1}{\sqrt{2}} \rangle$
$\mathbf{N}(t) = \langle -\cos t, -\sin t, 0 \rangle$
Now we compute the cross product, $\mathbf{B} = \mathbf{T} \times \mathbf{N}$:
In astronautics, the TNB frame is mission-critical. A variation of it is often called the Local Vertical/Local Horizontal (LVLH) frame, and it's used constantly for orientation and maneuvering.
Putting it all together: When docking with the International Space Station, a visiting spacecraft must precisely align its own LVLH frame with the station's frame to ensure a safe connection.
The complete TNB Frame. Red is Tangent ($\mathbf{T}$), Blue is Normal ($\mathbf{N}$), and Green is Binormal ($\mathbf{B}$). Drag the slider for 's' to see how this coordinate system travels with the point.
Consider the curve $\mathbf{r}(t) = \langle t, 2t, t^2 \rangle$. Find the unit tangent vector $\mathbf{T}$ and the principal unit normal vector $\mathbf{N}$ at the point where $t=1$.
Find the arc length of the curve $\mathbf{r}(t) = \langle 2, t^2, \frac{1}{3}t^3 \rangle$ from $t=0$ to $t=\sqrt{5}$.
Today we moved from the physics of motion to its intrinsic geometry. We learned how to measure the length of a curve by integrating its speed. More profoundly, we constructed the TNB frame—a moving, right-handed coordinate system that gives us a complete local description of the curve.
This framework is the geometric foundation for describing not just motion, but the properties of any curve in three-dimensional space.