Arc Length and Curvature: Learn It 4

Giselle Miller

Arc Length and Curvature: Learn It 4

The Normal and Binormal Vectors

You’ve seen that the derivative [latex]{\bf{r}}'(t)[/latex] of a vector-valued function is a tangent vector to the curve, and the unit tangent vector [latex]{\bf{T}}(t)[/latex] can be calculated by dividing [latex]{\bf{r}}'(t)[/latex] by its magnitude. When studying motion in three dimensions, two other vectors are useful for describing the motion of a particle along a path in space: the principal unit vector and the binomial vector.

principal unit normal vector and binormal vector

Let [latex]C[/latex] be a three-dimensional smooth curve represented by [latex]{\bf{r}}[/latex] over an open interval [latex]I[/latex].

[latex]\\[/latex]

Principal unit normal vector: If [latex]{\bf{T}}'(t)\neq 0[/latex], then the principal unit normal vector at [latex]t[/latex] is:

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

Binormal vector: The binormal vector at [latex]t[/latex] is:

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

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

By definition, the binormal vector is orthogonal to both the unit tangent vector and the normal vector. Additionally, [latex]{\bf{B}}(t)[/latex] is always a unit vector.

We can verify that [latex]{\bf{B}}(t)[/latex] is a unit vector using the formula for the magnitude of a cross product:

[latex]\left\Vert{\bf{B}}(t)\right\Vert=\left\Vert{\bf{T}}(t)\times{\bf{N}}(t)\right\Vert=\left\Vert{\bf{T}}(t)\right\Vert\left\Vert{\bf{N}}(t)\right\Vert\sin\theta[/latex]

where [latex]\theta[/latex] is the angle between [latex]{\bf{T}}(t)[/latex] and [latex]{\bf{N}}(t)[/latex]. Since [latex]{\bf{N}}(t)[/latex] is the derivative of a unit vector, we know from earlier properties that [latex]{\bf{T}}(t)[/latex] and [latex]{\bf{N}}(t)[/latex] are orthogonal to each other, so [latex]\theta=\frac{\pi}{2}[/latex]. Both are unit vectors with magnitude [latex]1[/latex]. Therefore:

[latex]\left\Vert{\bf{T}}(t)\right\Vert\left\Vert{\bf{N}}(t)\right\Vert\sin\theta=(1)(1)\sin\big(\frac{\pi}{2}\big)=1[/latex]

This confirms that [latex]{\bf{B}}(t)[/latex] is a unit vector.

The principal unit normal vector can be challenging to calculate because the unit tangent vector involves a quotient, often with a square root in the denominator. In three dimensions, finding the cross product of the unit tangent vector and the unit normal vector can be even more cumbersome. Fortunately, alternative formulas exist for finding these vectors, which are presented presented in section on Motion in Space

For each of the following vector-valued functions, find the principal unit normal vector. Then, if possible, find the binormal vector.

[latex]{\bf{r}}(t)=4\cos{t{\bf{i}}}-4\sin{t{\bf{j}}}[/latex]
[latex]{\bf{r}}(t)=(6t+2){\bf{i}}+5t^{2}{\bf{j}}-8t{\bf{k}}[/latex]

This function describes a circle.

Figure 1.

To find the principal unit normal vector, we first must find the unit tangent vector [latex]{\bf{T}}(t)[/latex]:

[latex]\begin{array}{ccc}\hfill {{\bf{T}}(t)} &=\hfill&{\frac{{\bf{r}}'(t)}{\left\Vert{\bf{r}}'(t)\right\Vert}}\hfill \\ \hfill & =\hfill & {\frac{-4\sin{t{\bf{i}}}-4\cos{t{\bf{j}}}}{\sqrt{(-4\sin{t})^{2}+(-4\cos{t})^{2}}}} \hfill \\ \hfill & =\hfill & {\frac{-4\sin{t{\bf{i}}}-4\cos{t{\bf{j}}}}{\sqrt{16\sin{^{2}t}+16\cos{^{2}t}}}}\hfill \\ \hfill & =\hfill & {\frac{-4\sin{t{\bf{i}}}-4\cos{t{\bf{j}}}}{\sqrt{16(\sin{^{2}t}+\cos{^{2}t})}}}\hfill \\ \hfill & =\hfill & {\frac{-4\sin{t{\bf{i}}}-4\cos{t{\bf{j}}}}{4}}\hfill \\ \hfill & =\hfill & {-\sin{t{\bf{i}}}-\cos{t{\bf{j}}}.}\hfill \\ \hfill \end{array}[/latex]

Next, we use the first equation from the definition

[latex]\begin{array}{ccc}\hfill {{\bf{N}}(t)} &=\hfill&{\frac{{\bf{T}}'(t)}{\left\Vert{\bf{T}}'(t)\right\Vert}}\hfill \\ \hfill & =\hfill & {\frac{-\cos{t{\bf{i}}}+\sin{t{\bf{j}}}}{\sqrt{(-\cos{t})^{2}+(\sin{t})^{2}}}} \hfill \\ \hfill & =\hfill & {\frac{-\cos{t{\bf{i}}}+\sin{t{\bf{j}}}}{\sqrt{\cos{^{2}t}+\sin{^{2}t}}}}\hfill \\ \hfill & =\hfill & {-\cos{t{\bf{i}}}+\sin{t{\bf{j}}}}\hfill \\ \hfill \end{array}[/latex].

Notice that the unit tangent vector and the principal unit normal vector are orthogonal to each other for all rules of [latex]t[/latex]:

[latex]\begin{array}{ccc}\hfill {{\bf{T}}(t)\cdot{\bf{N}}(t)} &=\hfill&{\langle{-}\sin{t},-\cos{t}\rangle\cdot\langle{-}\cos{t},\sin{t}\rangle}\hfill \\ \hfill & =\hfill & {\sin{t}\cos{t}-\cos{t}\sin{t}} \hfill \\ \hfill & =\hfill & {0.}\hfill \\ \hfill \end{array}[/latex]

Furthermore, the principal unit normal vector points toward the center of the circle from every point on the circle. Since [latex]{\bf{r}}(t)[/latex] defines a curve in two dimensions, we cannot calculate the binormal vector

Figure 2.
This function looks like this:

Figure 3.

To find the principal unit normal vector, we first find the unit tangent vector [latex]{\bf{T}}(t)[/latex]:

[latex]\begin{array}{ccc}\hfill {{\bf{T}}(t)} &=\hfill & {\frac{{\bf{r}}'(t)}{\left\Vert{\bf{r}}'(t)\right\Vert}}\hfill \\ \hfill & =\hfill & {\frac{6{\bf{i}}+10t{\bf{j}}-8{\bf{k}}}{\sqrt{6^{2}+(10t)^{2}+(-8)^{2}}}} \hfill \\ \hfill & =\hfill & {\frac{6{\bf{i}}+10t{\bf{j}}-8{\bf{k}}}{\sqrt{36+100t^{2}+64}}}\hfill \\ \hfill & =\hfill & {\frac{6{\bf{i}}+10t{\bf{j}}-8{\bf{k}}}{\sqrt{100(t^{2}+1)}}}\hfill \\ \hfill & =\hfill & {\frac{3{\bf{i}}+5t{\bf{j}}-4{\bf{k}}}{5\sqrt{t^{2}+1}}}\hfill \\ \hfill & =\hfill & {\frac{3}{5}(t^{2}+1)^{-1/2}{\bf{i}}+t(t^{2}+1)^{-1/2}{\bf{j}}-\frac{4}{5}(t^{2}+1)^{-1/2}{\bf{k}}.}\hfill \\ \hfill\end{array}[/latex].

Next, we calculate [latex]{\bf{T}}'(t)[/latex] and [latex]\left\Vert{\bf{T}}'(t)\right\Vert[/latex]:

[latex]\begin{array}{ccc}\hfill {{\bf{T}}(t)} &=\hfill & {\frac{3}{5}(-\frac{1}{2})(t^{2}+1)^{-3/2}(2t){\bf{i}}+\big((t^{2}+1)^{-1/2}-t(\frac{1}{2})(t^{2}+1)^{-3/2}(2t)\big){\bf{j}}-\frac{4}{5}(-\frac{1}{2})(t^{2}+1)^{-3/2}(2t){\bf{k}}}\hfill \\ \hfill & =\hfill & {-\frac{3t}{5(t^{2}+1)^{3/2}}{\bf{i}}+\frac{1}{(t^{2}+1)^{3/2}}{\bf{j}}+\frac{4}{5(t^{2}+1)^{3/2}}{\bf{k}}} \hfill \\ \hfill {\left\Vert{\bf{T}}'(t)\right\Vert}& =\hfill & {\sqrt{\bigg(-\frac{3t}{5(t^{2}+1)^{3/2}}\bigg)^{2}+\bigg(-\frac{1}{(t^{2}+1)^{3/2}}\bigg)^{2}+\bigg(\frac{4t}{5(t^{2}+1)^{3/2}}\bigg)^{2}}}\hfill \\ \hfill & =\hfill & {\sqrt{\frac{9t^{2}}{25(t^{2}+1)^{3}}+\frac{1}{(t^{2}+1)^{3}}+\frac{16t^{2}}{25(t^{2}+1)^{3}}}}\hfill \\ \hfill & =\hfill & {\sqrt{\frac{25t^{2}+25}{25(t^{2}+1)^{3}}}}\hfill \\ \hfill & =\hfill & {\sqrt{\frac{1}{(t^{2}+1)^{2}}}}\hfill \\ \hfill & =\hfill & {\frac{1}{t^{2}+1}.}\hfill \\ \hfill\end{array}[/latex]

Therefore, according to the second equation from the above definition:

[latex]\begin{array}{ccc}\hfill {{\bf{N}}(t)} &=\hfill & {\frac{{\bf{T}}'(t)}{\left\Vert{\bf{T}}'(t)\right\Vert}}\hfill \\ \hfill & =\hfill & {\bigg(-\frac{3t}{5(t^{2}+1)^{3/2}}{\bf{i}}+\frac{1}{(t^{2}+1)^{3/2}}{\bf{j}}+\frac{4t}{5(t^{2}+1)^{3/2}}{\bf{k}}\bigg)(t^{2}+1)} \hfill \\ \hfill & =\hfill & {-\frac{3t}{5(t^{2}+1)^{1/2}}{\bf{i}}+\frac{1}{(t^{2}+1)^{1/2}}{\bf{j}}+\frac{4t}{5(t^{2}+1)^{1/2}}{\bf{k}}}\hfill \\ \hfill & =\hfill & {-\frac{3t{\bf{i}}-5{\bf{j}}-4t{\bf{k}}}{5\sqrt{t^{2}+1}}}\hfill \\ \hfill\end{array}[/latex]

Once again, the unit tangent vector and the principal unit normal vector are orthogonal to each other for all values of [latex]t[/latex]:

[latex]\begin{array}{ccc}\hfill {{\bf{T}}(t)\cdot{\bf{N}}(t)} &=\hfill & {\bigg(\frac{3{\bf{i}}+5t{\bf{j}}-4{\bf{k}}}{5\sqrt{t^{2}+1}}\bigg)\cdot\bigg(-\frac{3t{\bf{i}}-5{\bf{j}}-4t{\bf{k}}}{5\sqrt{t^{2}+1}}\bigg)}\hfill \\ \hfill & =\hfill & {\frac{3(-3)-5t(-5)-4(4t)}{5\sqrt{t^{2}+1}}} \hfill \\ \hfill & =\hfill & {\frac{-9t+25t-16t}{5\sqrt{t^{2}+1}}}\hfill \\ \hfill & =\hfill & {0.}\hfill \\ \hfill\end{array}[/latex]

Last, since [latex]{\bf{r}}(t)[/latex] represents a three-dimensional curve, we can calculate the binormal vector using the first equation from the above definition:

[latex]\begin{array}{ccc}\hfill {{\bf{B}}(t)} &=\hfill & {{\bf{T}}(t)\times{\bf{N}}(t)}\hfill \\ \hfill & =\hfill & {\begin{vmatrix}{\bf{i}}&{\bf{j}}&{\bf{k}}\\\frac{3}{5\sqrt{t^{2}+1}}&+\frac{5t}{5\sqrt{t^{2}+1}}&-\frac{4}{5\sqrt{t^{2}+1}}\\ -\frac{3t}{5\sqrt{t^{2}+1}}&+\frac{5}{5\sqrt{t^{2}+1}}&\frac{4t}{5\sqrt{t^{2}+1}}\end{vmatrix}} \hfill \\ \hfill & =\hfill & {\bigg(\Big(+\frac{5t}{5\sqrt{t^{2}+1}}\Big)\Big(\frac{4t}{5\sqrt{t^{2}+1}}\Big)-\Big(-\frac{4}{5\sqrt{t^{2}+1}}\Big)\Big(-\frac{5}{5\sqrt{t^{2}+1}}\Big)\bigg){\bf{i}} \\ +\bigg(\Big(\frac{3}{5\sqrt{t^{2}+1}}\Big)\Big(\frac{4t}{5\sqrt{t^{2}+1}}\Big)-\Big(-\frac{4}{5\sqrt{t^{2}+1}}\Big)\Big(-\frac{3t}{5\sqrt{t^{2}+1}}\Big)\bigg){\bf{j}} \\ +\bigg(\Big(\frac{3}{5\sqrt{t^{2}+1}}\Big)\Big(+\frac{5}{5\sqrt{t^{2}+1}}\Big)-\Big(+\frac{5t}{5\sqrt{t^{2}+1}}\Big)\Big(-\frac{3t}{5\sqrt{t^{2}+1}}\Big)\bigg){\bf{k}}}\hfill \\ \hfill & =\hfill & {\Big(\frac{20t^{2}+20}{25(t^{2}+1)}\Big){\bf{i}}+\Big(\frac{-15-15t^{2}}{25(t^{2}+1)}\Big){\bf{k}}}\hfill \\ \hfill & =\hfill & {20\Big(\frac{t^{2}+1}{25(t^{2}+1)}\Big){\bf{i}}-15\Big(\frac{t^{2}+1}{25(t^{2}+1)}\Big){\bf{k}}}\hfill \\ \hfill & =\hfill & {\frac{4}{5}{\bf{i}}-\frac{3}{5}{\bf{k}}.}\hfill \\ \hfill\end{array}[/latex]

The Frenet Frame and Osculating Plane

For any smooth curve in three dimensions defined by a vector-valued function, we now have formulas for three important vectors: the unit tangent vector [latex]{\bf{T}}[/latex], the unit normal vector [latex]{\bf{N}}[/latex], and the binormal vector [latex]{\bf{B}}[/latex].

The unit normal vector and the binormal vector form a plane perpendicular to the curve at any point, called the normal plane. These three vectors together form a frame of reference in three-dimensional space called the Frenet frame of reference (also called the TNB frame), shown in Figure 1. The plane determined by the vectors [latex]{\bf{T}}[/latex] and [latex]{\bf{N}}[/latex] forms the osculating plane of [latex]C[/latex] at any point [latex]P[/latex] on the curve.

This figure is the graph of a curve increasing and decreasing. Along the curve at 4 different points are 3 vectors at each point. The first vector is labeled “T” and is tangent to the curve at the point. The second vector is labeled “N” and is normal to the curve at the point. The third vector is labeled “B” and is orthogonal to T and N. — Figure 4. This figure depicts a Frenet frame of reference. At every point [latex]P[/latex] on a three-dimensional curve, the unit tangent, unit normal, and binormal vectors form a three-dimensional frame of reference.

The Osculating Circle

Suppose we form a circle in the osculating plane of [latex]C[/latex] at point [latex]P[/latex] on the curve. Assume the circle has the same curvature as the curve does at point [latex]P[/latex] and has radius [latex]r[/latex]. Then the curvature of the circle is [latex]\frac{1}{r}[/latex]. We call [latex]r[/latex] the radius of curvature of the curve—it equals the reciprocal of the curvature.

If this circle lies on the concave side of the curve and is tangent to the curve at point [latex]P[/latex], then this circle is called the osculating circle of [latex]C[/latex] at [latex]P[/latex], as shown in Figure 2.

This figure is the graph of a curve with a circle in the middle. The bottom of the circle is the same as part of the curve. Inside of the circle is a vector labeled “r”. It starts at point “P” on the circle and points towards the radius. There is also a line segment perpendicular to the radius and tangent to point P. — Figure 5. In this osculating circle, the circle is tangent to curve [latex]C[/latex] at point [latex]P[/latex] and shares the same curvature.

To find the equation of an osculating circle in two dimensions, you need only the center and radius of the circle. The radius is the reciprocal of the curvature at that point.

Find the equation of the osculating circle of the helix defined by the function [latex]y=x^{3}-3x+1[/latex] at [latex]x=1[/latex].