The Main Idea 

Arc length measures the distance along a curve. For vector-valued functions, the formula is straightforward: integrate the magnitude of the derivative vector.

The Core Formula

For any smooth curve defined by [latex]{\bf{r}}(t)[/latex] over [latex][a, b][/latex]:

[latex]s=\int_{a}^{b} \left\Vert{\bf{r}}'(t)\right\Vert dt[/latex]

Plane curves [latex]{\bf{r}}(t)=f(t){\bf{i}}+g(t){\bf{j}}[/latex]:

[latex]s=\int_{a}^{b} \sqrt{[f'(t)]^{2}+[g'(t)]^{2}}dt[/latex]

Space curves [latex]{\bf{r}}(t)=f(t){\bf{i}}+g(t){\bf{j}}+h(t){\bf{k}}[/latex]:

[latex]s=\int_{a}^{b} \sqrt{[f'(t)]^{2}+[g'(t)]^{2}+[h'(t)]^{2}}dt[/latex]

The only difference? Space curves include a third component.

Problem-Solving Strategy

1. Find [latex]{\bf{r}}'(t)[/latex] by differentiating each component
2. Calculate [latex]\left\Vert{\bf{r}}'(t)\right\Vert[/latex] using the square root of the sum of squares
3. Integrate [latex]\left\Vert{\bf{r}}'(t)\right\Vert[/latex] from [latex]a[/latex] to [latex]b[/latex]

The curve must be smooth: [latex]{\bf{r}}(t)[/latex] must be differentiable with a non-zero derivative. This guarantees no cusps or corners that break the formula.

Special Case: Helices

A helix with radius [latex]R[/latex], height [latex]h[/latex], and [latex]N[/latex] turns has arc length:

[latex]s=\sqrt{4{\pi}^{2}N^{2}R^{2}+h^{2}}[/latex]

This tells you the length of wire needed to form the helix.

The Main Idea 

Arc-length parameterization is a way to describe a curve where the parameter directly measures distance traveled along the curve. It’s like using mile markers instead of time to track position on a road.

Regular parameterization can trace a curve at any speed. Arc-length parameterization always traces at unit speed, making distance calculations straightforward.

The Arc-Length Function

For a smooth curve [latex]{\bf{r}}(t)[/latex] where [latex]t \geq a[/latex]:

[latex]s(t)=\int_{a}^{t} \left\Vert{\bf{r}}'(u)\right\Vert du[/latex]

This function [latex]s(t)[/latex] tells you the distance traveled from the starting point. Taking its derivative gives speed:

[latex]\frac{ds}{dt}=\left\Vert{\bf{r}}'(t)\right\Vert[/latex]

Special Case: Unit Speed

When [latex]\left\Vert{\bf{r}}'(t)\right\Vert=1[/latex] for all [latex]t[/latex], the parameter [latex]t[/latex] already measures arc length. The curve traces at unit speed.

Creating an Arc-Length Parameterization

1. Find the arc-length function: Calculate [latex]s(t)=\int_{a}^{t} \left\Vert{\bf{r}}'(u)\right\Vert du[/latex]
2. Solve for [latex]t[/latex]: Express [latex]t[/latex] as a function of [latex]s[/latex]
3. Substitute: Replace [latex]t[/latex] in [latex]{\bf{r}}(t)[/latex] with your expression in terms of [latex]s[/latex]
4. Adjust the domain: Convert the restriction on [latex]t[/latex] to a restriction on [latex]s[/latex]

Arc-length parameterization makes the parameter meaningful. The value of [latex]s[/latex] tells you exactly how far you’ve traveled along the curve. This simplifies formulas for curvature and line integrals.

The Main Idea 

Curvature measures how sharply a curve bends. A straight line has zero curvature. A tight turn has high curvature. A gentle curve has low curvature.

The Intuition

Picture driving on a road. A gentle curve needs barely any steering. A sharp turn requires turning the wheel more. Curvature quantifies this difference. For a circle, smaller radius means greater curvature.

The Definition

For a smooth curve [latex]{\bf{r}}(s)[/latex] with arc-length parameter [latex]s[/latex]:

[latex]\kappa=\left\Vert{\bf{T}}'(s)\right\Vert[/latex]

where [latex]{\bf{T}}[/latex] is the unit tangent vector.

Practical Formulas

The definition requires arc-length parameterization. These alternatives work with any parameterization:

• For any curve [latex]{\bf{r}}(t)[/latex]: [latex]\kappa=\frac{\left\Vert{\bf{T}}'(t)\right\Vert}{\left\Vert{\bf{r}}'(t)\right\Vert}[/latex]
• For space curves: [latex]\kappa=\frac{\left\Vert{\bf{r}}'(t)\times{\bf{r}}''(t)\right\Vert}{\left\Vert{\bf{r}}'(t)\right\Vert^{3}}[/latex]
• For graphs [latex]y=f(x)[/latex]: [latex]\kappa=\frac{|y''|}{[1+(y')^{2}]^{3/2}}[/latex]

Choosing Your Formula

• Use the first when you already have [latex]{\bf{T}}(t)[/latex] calculated
• Use the cross product formula for space curves—it only needs [latex]{\bf{r}}'(t)[/latex] and [latex]{\bf{r}}''(t)[/latex]
• Use the [latex]y=f(x)[/latex] formula for function graphs

Circles

For a circle of radius [latex]r[/latex], curvature equals [latex]\kappa = \frac{1}{r}[/latex]. Curvature is the reciprocal of radius.

The Main Idea 

Beyond the tangent vector, two more vectors describe motion along a three-dimensional curve: the principal unit normal vector and the binormal vector. Together with the tangent, they form a moving frame of reference.

The Principal Unit Normal Vector

The principal unit normal vector [latex]{\bf{N}}(t)[/latex] points in the direction the curve is turning:

[latex]{\bf{N}}(t)=\frac{{\bf{T}}'(t)}{\left\Vert{\bf{T}}'(t)\right\Vert}[/latex]

It’s always perpendicular to [latex]{\bf{T}}(t)[/latex] and has magnitude [latex]1[/latex].

The Binormal Vector

The binormal vector [latex]{\bf{B}}(t)[/latex] is perpendicular to both [latex]{\bf{T}}[/latex] and [latex]{\bf{N}}[/latex]:

[latex]{\bf{B}}(t)={\bf{T}}(t)\times{\bf{N}}(t)[/latex]

It’s automatically a unit vector.

The Frenet Frame

These three vectors—[latex]{\bf{T}}[/latex], [latex]{\bf{N}}[/latex], and [latex]{\bf{B}}[/latex]—form a frame of reference that moves along the curve. They’re mutually perpendicular and all have magnitude [latex]1[/latex].

Key Planes

• The osculating plane contains [latex]{\bf{T}}[/latex] and [latex]{\bf{N}}[/latex]. The curve lies in this plane locally.
• The normal plane contains [latex]{\bf{N}}[/latex] and [latex]{\bf{B}}[/latex]. It’s perpendicular to the curve.

The Osculating Circle

At any point on a curve, you can fit a circle that matches the curve’s curvature. This osculating circle has radius [latex]r = \frac{1}{\kappa}[/latex], called the radius of curvature. It lies in the osculating plane, is tangent to the curve, and sits on the concave side.

Problem-Solving Strategy

Find [latex]{\bf{T}}(t)[/latex] first. Differentiate to get [latex]{\bf{T}}'(t)[/latex]. Normalize it to get [latex]{\bf{N}}(t)[/latex]. Take the cross product [latex]{\bf{T}}\times{\bf{N}}[/latex] to get [latex]{\bf{B}}(t)[/latex].