Section 2.4 Curvature

2.4 Curvature

In this section, we learn alternative formulas for curvature, and normal and binormal vectors, two important vectors that have geometry meaning for a curve. in space. We first recall the definition of curvature.

Definition

Let [latex]C[/latex] be a smooth curve in the plane or in space given by [latex]\overrightarrow{r(s)}[/latex], where [latex]s[/latex] is the arc-length parameter. The curvature [latex]\mathit{k}[/latex] at [latex]s[/latex] is \[k=||\frac{\overrightarrow{T(s)}}{ds}||=||\overrightarrow{T'(s)}||.\]

Theorem: Alternative Formulas for Curvature

Let [latex]C[/latex] be a smooth curve in the plane or in space given by [latex]\overrightarrow{r(t)}[/latex], where [latex]t[/latex] is the parameter. The curvature [latex]\mathit{k}[/latex] at [latex]t[/latex] is

(a)  \[k=\frac{||\overrightarrow{T'(t)}||}{||\overrightarrow{r'(t)}||}\]

(b)  \[k=\frac{||\overrightarrow{r'(t)}\times\overrightarrow{r”(t)}||}{||\overrightarrow{r'(t)}||^{3}}\]

Proof of (b) use (a) and [latex]\frac{\overrightarrow{r'(t)}}{||\overrightarrow{r'(t)}||}=\overrightarrow{T(t)}[/latex]

Example 1: Find the curvature of [latex]\overrightarrow{r(t)}=\lt \text{cos}(t),\text{sin}(t),t\gt[/latex].

Exercise 1: Find the curvature of [latex]\overrightarrow{r(t)}=\lt t,\text{cos}(t),\text{sin}(t)\gt[/latex].

Example 2: Find the curvature of [latex]\overrightarrow{r(t)}=\lt t\text{ln}(t),t,t^{2}\gt[/latex] at the point [latex](0,1,1)[/latex].

Exercise 2: Find the curvature of [latex]\overrightarrow{r(t)}=\lt e^{t},2t-1,e^{2t}\gt[/latex] at the point [latex](1,-1,1)[/latex].

When studying motion in three dimensions, two other vectors are useful in describing the motion of a particle along a path in space: the principal unit normal vector and the binormal vector.

Definition: The Normal and Binormal Vectors

Let [latex]\overrightarrow{r(t)}[/latex] be a smooth curve in space and let [latex]\overrightarrow{T(t)}[/latex] be its unit tangent vector, if [latex]\overrightarrow{T'(t)}\neq\overrightarrow{0}[/latex], then the unit normal vector at [latex]t[/latex] is defined to be \[\overrightarrow{N(t)}=\frac{\overrightarrow{T'(t)}}{||\overrightarrow{T'(t)}||}.\]

The binormal vector at [latex]t[/latex] is defined as  \[\overrightarrow{B(t)}=\overrightarrow{T(t)}\times\overrightarrow{N(t)}.\]

Notice that the unit normal vector and the binomial vector are unit vectors. The unit normal vector is a vector that is orthogonal to the tangent vector point to the center of the curve if the curvature is not zero. The binomial vector is a vector that is orthogonal to both [latex]\overrightarrow{T(t)}[/latex] and [latex]\overrightarrow{N(t)}[/latex].

Example 3: Find the unit normal and binormal vectors of [latex]\overrightarrow{r(t)}=\lt \text{cos}(2t),\text{sin}(2t),t\gt .[/latex] 

Exercise 3: Find the unit normal and binormal vectors of [latex]\overrightarrow{r(t)}=\lt t,\text{cos}(3t),\text{sin}(3t)\gt .[/latex] 

Definition:

The unit normal vector [latex]\overrightarrow{N(t)}[/latex] and the binormal vector [latex]\overrightarrow{B(t)}[/latex] of a curve [latex]\overrightarrow{r(t)}[/latex] form a plane that is perpendicular to the curve at any point on the curve, called the normal plane. Plane determined by the vectors [latex]\overrightarrow{T(t)}[/latex] and [latex]\overrightarrow{N(t)}[/latex] forms the osculating plane of [latex]\overrightarrow{r(t)}[/latex] at any point [latex]P[/latex] on the curve. The circle that lies in the osculating plane of [latex]\overrightarrow{r(t)}[/latex] at [latex]P[/latex], has the same tangent as [latex]\overrightarrow{r(t)}[/latex] at [latex]P[/latex], lies on the concave side of curve, and has radius [latex]\rho=\frac{1}{k}[/latex] (the reciprocal of the curvature) is called the osculating circle (or the circle of curvature) of the curve at [latex]P[/latex].

Notice that the normal plane has [latex]\overrightarrow{T(t)}[/latex] as its normal vector and the osculating plane has [latex]\overrightarrow{B(t)}[/latex] as its normal vector.

Example 4: The equations of normal plane and osculating plane of [latex]\overrightarrow{r(t)}=\lt \text{cos}(2t),\text{sin}(2t),4t\gt[/latex] at [latex]P=(-1,0,2\pi)[/latex].

Exercise 4: The equations of normal plane and osculating plane of [latex]\overrightarrow{r(t)}=\lt 6t,\text{cos}(3t),-\text{sin}(3t)\gt[/latex] at [latex]P=(3\pi,0,1)[/latex].

Example 5: Find the unit normal and binormal vectors of [latex]\overrightarrow{r(t)}=\lt 2\text{sin}(t),2\sqrt{2}\text{cos}(t),-2\text{sin}(t)\gt[/latex].

Group work:

1.Find the curvature of [latex]\overrightarrow{r(t)}=\lt 2\text{sin}(t),2\sqrt{2}\text{cos}(t),-2\text{sin}(t)\gt[/latex]

2. Find the unit normal and binormal vectors of [latex]\overrightarrow{r(t)}=\lt -3\sqrt{2}\text{sin}(t),3\text{cos}(t),3\text{cos}(t)\gt[/latex]. 

3. Find the equations of normal plane and osculating plane of [latex]\overrightarrow{r(t)}=\lt 2\text{cos}(3t),2\text{sin}(3t),t\gt[/latex]

at [latex]P=(-2,0,\pi)[/latex].