# 8. Properties of Curves

## e. Tangent, Normal, and Binormal Vectors

In the examples on the previous page the acceleration was perpendicular to the velocity, which made it easy to compute the unit normal vector. We now derive formulas which are more convenient even when the acceleration is not perpendicular to the velocity.

## 2. Computational Formulas

### Unit Tangent Vector, $$\hat T$$

Given the curve, $$\vec{r}(t)$$, it is easy to compute the velocity, $$\vec{v}=\dfrac{d\vec{r}}{dt}$$. Then $$\hat{T}$$ is the unit vector in the direction of $$\vec{v}$$.

$\hat{T}=\hat{v}=\dfrac{\vec{v}}{|\vec{v}|}$

### Unit Normal Vector, $$\hat N$$

Given the velocity, $$\vec{v}=\dfrac{d\vec{r}}{dt}$$, it is easy to compute the acceleration $$\vec{a}=\dfrac{d\vec{v}}{dt}$$. Recall from the discussion of projections that the projection of $$\vec a$$ perpendicular to $$\vec v$$ is $\text{proj}_{\bot \vec{v}}\vec{a} =\vec{a}-\,\dfrac{\vec{a}\cdot\vec{v}}{|\vec{v}|^2}\vec{v}$ Then $$\hat{N}$$ is the unit vector in the direction of $$\text{proj}_{\bot \vec{v}}\vec{a}$$: $\hat{N} =\dfrac{\text{proj}_{\bot\vec{v}}\vec{a}} {\left|\text{proj}_{\bot\vec{v}}\vec{a}\right|} =\dfrac{|\vec{v}|}{|\vec{v}\times\vec{a}|} \left(\vec{a}-\,\dfrac{\vec{a}\cdot\vec{v}}{|\vec{v}|^2}\vec{v}\right)$ where the second formula used the equation $$\left|\text{proj}_{\bot \vec{v}}\vec{a}\right| =\dfrac{|\vec{v}\times\vec{a}|}{|\vec{v}|}$$.

\begin{aligned} \left|\text{proj}_{\bot \vec{v}}\vec{a}\right| &=\sqrt{\left(\vec{a}-\,\dfrac{\vec{a}\cdot\vec{v}}{|\vec{v}|^2}\vec{v}\right) \cdot\left(\vec{a}-\,\dfrac{\vec{a}\cdot\vec{v}}{|\vec{v}|^2}\vec{v}\right)} \\ &=\sqrt{\vec{a}\cdot\vec{a} -2\dfrac{\vec{a}\cdot\vec{v}}{|\vec{v}|^2}\vec{v}\cdot\vec{a} +\dfrac{\vec{a}\cdot\vec{v}}{|\vec{v}|^2}\vec{v} \cdot\dfrac{\vec{a}\cdot\vec{v}}{|\vec{v}|^2}\vec{v}} \\ &=\sqrt{\vec{a}\cdot\vec{a}-\,\dfrac{(\vec{a}\cdot\vec{v})^2}{|\vec{v}|^2}} =\dfrac{\sqrt{|\vec{a}|^2|\vec{v}|^2-(\vec{v}\cdot\vec{a})^2}}{|\vec{v}|} =\dfrac{|\vec{v}\times\vec{a}|}{|\vec{v}|} \end{aligned} In the last step we used the Pythagorean Identity for Dot and Cross Products, $$|\vec{a}|^2|\vec{v}|^2-(\vec{v}\cdot\vec{a})^2 =|\vec{v}\times\vec{a}|^2$$.

Besides this definition, there are two simpler formulas which are useful in different situations. The first is useful when the formula for $$\hat{T}$$ is simple. The second is useful if we compute $$\hat{B}$$ before $$\hat{N}$$.

$\hat{N} =\dfrac{\dfrac{d\hat{T}}{dt}}{\;\left|\dfrac{d\hat{T}}{dt}\right|\;} =\dfrac{\dfrac{d\hat{T}}{ds}}{\;\left|\dfrac{d\hat{T}}{ds}\right|\;} =\hat{B}\times\hat{T}$

We have two things to prove:
(1) $$\dfrac{d\hat{T}}{dt}$$ lies in the plane of $$\vec{v}$$ and $$\vec{a}$$ and is on the same side of $$\vec{v}$$ as $$\vec{a}$$.
(2) $$\dfrac{d\hat{T}}{dt}$$ is perpendicular to $$\vec{v}$$.
Then, $$\left.\dfrac{d\hat{T}}{dt}\middle/\left|\dfrac{d\hat{T}}{dt}\right|\right.$$ will be the unit vector perpendicular to $$\vec{v}$$ in the plane of $$\vec{v}$$ and $$\vec{a}$$ on the same side of $$\vec{v}$$ as $$\vec{a}$$, which is the definition of $$\hat{N}$$.

First, we compute the derivative of $$\hat T=\dfrac{1}{|\vec{v}|}\vec{v}$$ using the product rule: \begin{aligned} \dfrac{d\hat{T}}{dt} &=\dfrac{1}{|\vec{v}|}\dfrac{d\vec{v}}{dt} +\dfrac{d}{dt}\left(\dfrac{1}{|\vec{v}|}\right)\vec{v} \\ &=\dfrac{1}{|\vec{v}|}\vec{a} +\dfrac{d}{dt}\left(\dfrac{1}{|\vec{v}|}\right)\vec{v} \end{aligned} So $$\dfrac{d\hat{T}}{dt}$$ lies in the plane of $$\vec{v}$$ and $$\vec{a}$$ and is on the same side of $$\vec{v}$$ as $$\vec{a}$$ since the coefficient of $$\vec{a}$$ is positive.

Second, since $$\hat T$$ is a unit vector, we have $$\hat T\cdot\hat T=1$$. We take the derivative using the product rule: $\dfrac{d\hat{T}}{dt}\cdot\hat T+\hat T\cdot\dfrac{d\hat{T}}{dt}=0$ So $$\dfrac{d\hat{T}}{dt}\cdot\hat T=0$$ and $$\dfrac{d\hat{T}}{dt}$$ is perpendicular to $$\hat T$$, and hence $$\vec v$$.

Converting the derivatives from $$t$$ to $$s$$ is just the Chain Rule: $=\dfrac{\dfrac{d\hat{T}}{dt}}{\;\left|\dfrac{d\hat{T}}{dt}\right|\;} =\dfrac{\dfrac{ds}{dt}\dfrac{d\hat{T}}{ds}}{\;\left|\dfrac{ds}{dt}\dfrac{d\hat{T}}{ds}\right|\;} =\dfrac{|\vec v|\dfrac{d\hat{T}}{ds}}{\;|\vec v|\left|\dfrac{d\hat{T}}{ds}\right|\;} =\dfrac{\dfrac{d\hat{T}}{ds}}{\;\left|\dfrac{d\hat{T}}{ds}\right|\;}$

We note that $$\hat{T}$$, $$\hat{N}$$ and $$\hat{B}$$ are a right handed triplet of unit vectors. So $$\hat{N}=\hat{B}\times\hat{T}$$.

The formula $$\hat{N}=\hat{B}\times\hat{T}$$ is much simpler than the definition $$\hat{N}=\dfrac{\text{proj}_{\bot \vec{v}}\vec{a}} {\left|\text{proj}_{\bot\vec{v}}\vec{a}\right|}$$. However, you should remember that $$\hat{N}$$ is defined before $$\hat{B}$$ and is the second vector in the basis triplet $$\hat{T}$$, $$\hat{N}$$ and $$\hat{B}$$ which are related by the right hand rule.

### Unit Binormal Vector, $$\hat B$$

By definition, $\hat{B}=\hat{T}\times\hat{N}$ However, there is an easier formula for computing $$\hat{B}$$, which does not depend on first computing $$\hat{N}$$.

$\hat{B}=\dfrac{\vec{v}\times\vec{a}}{|\vec{v}\times\vec{a}|}$

Since $$\hat{T}$$ and $$\hat{N}$$ define the same plane as $$\vec{v}$$ and $$\vec{a}$$ and $$\hat{N}$$ is on the same side of $$\vec{v}$$ as $$\vec{a}$$, the cross product $$\hat{T}\times\hat{N}$$ is in the same direction as $$\vec{v}\times\vec{a}$$. However, $$\hat{B}=\hat{T}\times\hat{N}$$ is a unit vector. So we make $$\vec{v}\times\vec{a}$$ into a unit vector and get the formula: $\hat{B}=\dfrac{\vec{v}\times\vec{a}}{|\vec{v}\times\vec{a}|}$

Find the unit tangent vector, $$\hat{T}$$, the unit normal vector $$\hat{N}$$, and the unit binormal vector $$\hat{B}$$ of the twisted cubic $$\vec{r}=\left\langle t,t^2,\dfrac{2}{3}t^{3}\right\rangle$$ for general $$t$$ and at $$t=2$$.

The velocity and acceleration are: $\vec{v} =\langle 1,2t,2t^2\rangle \qquad \vec{a} =\langle 0,2,4t\rangle$ The magnitude of the velocity is: (Notice the quantity in the square root is a perfect square. ) \begin{aligned} |\vec{v}| &=\sqrt{1^2+(2t)^2+(2t^2)^2} \\ &=\sqrt{1+4t^2+4t^{4}} =1+2t^2 \end{aligned}

This was why we chose to put the $$\dfrac{2}{3}$$ in the version of the twisted cubic in this exercise.

Then the unit tangent vector is \begin{aligned} \hat{T} &=\dfrac{\vec{v}}{|\vec{v}|} =\dfrac{1}{1+2t^2}\langle 1,2t,2t^2\rangle \\ &=\left\langle \dfrac{1}{1+2t^2},\dfrac{2t}{1+2t^2}, \dfrac{2t^2}{1+2t^2}\right\rangle \end{aligned} To find the unit binormal vector, we first find $$\vec{v}\times\vec{a}$$ and its length: \begin{aligned} \vec{v}\times\vec{a} &=\begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & 2t & 2t^2 \\ 0 & 2 & 4t \end{vmatrix} =\hat{\imath}(8t^2-4t^2)-\hat{\jmath}(4t-0)+\hat{k}(2-0) \\ &=\langle 4t^2,-4t,2\rangle \\ |\vec{v}\times\vec{a}| &=\sqrt{(4t^2)^2+(4t)^2+(2)^2} =2\sqrt{4t^{4}+4t^2+1} \\ &=2(1+2t^2) \end{aligned} where we again identify the quantity in the square root as a perfect square. So the unit binormal vector is \begin{aligned} \hat{B} &=\dfrac{\vec{v}\times\vec{a}}{|\vec{v}\times\vec{a}|} =\dfrac{1}{2(1+2t^2)}\langle 4t^2,-4t,2\rangle \\ &=\dfrac{1}{1+2t^2}\langle 2t^2,-2t,1\rangle =\left\langle \dfrac{2t^2}{1+2t^2},\dfrac{-2t}{1+2t^2}, \dfrac{1}{1+2t^2}\right\rangle \end{aligned} Finally we compute the unit normal vector as $$\hat{N}=\hat{B}\times\hat{T}$$. In doing this, we simplify our work by using one of the properties of the cross product: $$(a\vec{u})\times(b\vec{v})=ab(\vec{u}\times\vec{v})$$. \begin{aligned} \hat{N} &=\hat{B}\times\hat{T} =\dfrac{1}{1+2t^2}\langle 2t^2,-2t,1\rangle\times \dfrac{1}{1+2t^2}\langle 1,2t,2t^2\rangle \\ &=\dfrac{1}{(1+2t^2)^2}\langle 2t^2,-2t,1\rangle\times\langle 1,2t,2t^2\rangle \\ &=\dfrac{1}{(1+2t^2)^2} \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2t^2 & -2t & 1 \\ 1 & 2t & 2t^2 \end{vmatrix} \\ &=\dfrac{1}{(1+2t^2)^2} \left(\hat{\imath}(-4t^{3}-2t) -\hat{\jmath}(4t^{4}-1) +\hat{k}(4t^{3}+2t)\right) \\ &=\dfrac{1}{(1+2t^2)^2}\langle -4t^{3}-2t,1-4t^{4},4t^{3}+2t\rangle \end{aligned} We now factor each term and cancel one factor of $$1+2t^2$$: \begin{aligned} \hat{N} &=\dfrac{1}{1+2t^2}\langle -2t,1-2t^2,2t\rangle \\ &=\left\langle \dfrac{-2t}{1+2t^2},\dfrac{1-2t^2}{1+2t^2}, \dfrac{2t}{1+2t^2}\right\rangle \end{aligned}

The figure shows $$\hat{T}$$, $$\hat{N}$$ and $$\hat{B}$$ moving along the curve. In particular, at $$t=2$$ we have \begin{aligned} \hat{T}&=\left\langle \dfrac{1}{9},\dfrac{4}{9},\dfrac{8}{9}\right\rangle \\ \hat{B}&=\left\langle \dfrac{8}{9},\dfrac{-4}{9},\dfrac{1}{9}\right\rangle \\ \hat{N}&=\left\langle \dfrac{-4}{9},\dfrac{-7}{9},\dfrac{4}{9}\right\rangle \end{aligned} It is easy to check these are $$3$$ mutually perpendicular unit vectors.

That was a bit long, but still straightforward.

Find the unit tangent vector, $$\hat{T}$$, the unit normal vector $$\hat{N}$$, and the unit binormal vector $$\hat{B}$$ for the twisted quartic $$\vec{r}=\left\langle t^2,\dfrac{2}{3}t^3,\dfrac{1}{4}t^4\right\rangle$$ for general $$t$$ and at $$t=1$$.

The quantities in square roots are again perfect squares.

Generally: \begin{aligned} \hat{T}&=\left\langle\dfrac{2}{2+t^2},\dfrac{2t}{2+t^2},\dfrac{t^2}{2+t^2}\right\rangle \\ \hat{B}&=\left\langle \dfrac{t^2}{2+t^2},\dfrac{-2t}{2+t^2},\dfrac{2}{2+t^2}\right\rangle \\ \hat{N}&=\left\langle \dfrac{-2t}{2+t^2},\dfrac{2-t^2}{2+t^2},\dfrac{2t}{2+t^2}\right\rangle \end{aligned} For $$t=1$$: \begin{aligned} \hat{T} &=\left\langle\dfrac{2}{3},\dfrac{2}{3},\dfrac{1}{3}\right\rangle \\ \hat{B} &=\left\langle \dfrac{1}{3},\dfrac{-2}{3},\dfrac{2}{3}\right\rangle \\ \hat{N} &=\left\langle \dfrac{-2}{3},\dfrac{1}{3},\dfrac{2}{3}\right\rangle \end{aligned}

The velocity and acceleration are: $\vec{v} =\left\langle2t,2t^2,t^3\right\rangle \qquad \vec{a} =\left\langle2,4t,3t^2\right\rangle$ The magnitude of the velocity is: (Notice the quantity in the square root is a perfect square.) \begin{aligned} |\vec{v}| &=\sqrt{4t^2+4t^4+t^6} =t\sqrt{4+4t^2+t^4} \\ &=t\sqrt{(2+t^2)^2} =t(2+t^2)=2t+t^3 \end{aligned} Then the unit tangent vector is \begin{aligned} \hat{T} &=\dfrac{\vec{v}}{|\vec{v}|} =\dfrac{1}{t(2+t^2)}\left\langle2t,2t^2,t^3\right\rangle \\ &=\left\langle\dfrac{2}{2+t^2},\dfrac{2t}{2+t^2},\dfrac{t^2}{2+t^2}\right\rangle \end{aligned} To find the unit binormal vector, we first find $$\vec{v}\times\vec{a}$$ and its length: \begin{aligned} \vec{v}\times\vec{a} &=\begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2t & 2t^2 & t^3 \\ 2 & 4t & 3t^2 \end{vmatrix} \\ &=\hat{\imath}(6t^4-4t^4)-\hat{\jmath}(6t^3-2t^3)+\hat{k}(8t^2-4t^2) \\ &=\langle 2t^4,-4t^3,4t^2\rangle \\ |\vec{v}\times\vec{a}| &=\sqrt{4t^8+16t^6+16t^4} =2t^2\sqrt{t^4+4t^2+4} \\ &=2t^2\sqrt{(t^2+2)^2)} =2t^2(2+t^2) =4t^2+2t^4 \end{aligned} where we again identify the quantity in the square root as a perfect square. So the unit binormal vector is \begin{aligned} \hat{B} &=\dfrac{\vec{v}\times\vec{a}}{|\vec{v}\times\vec{a}|} =\dfrac{1}{2t^2(2+t^2)}\langle 2t^4,-4t^3,4t^2\rangle \\ &=\left\langle\dfrac{t^2}{2+t^2},\dfrac{-2t}{2+t^2}, \dfrac{2}{2+t^2}\right\rangle \end{aligned} Finally we compute the unit normal vector as $$\hat{N}=\hat{B}\times\hat{T}$$. In doing this, we simplify our work by using the property: $$(a\vec{u})\times(b\vec{v})=ab(\vec{u}\times\vec{v})$$. \begin{aligned} \hat{N} &=\hat{B}\times\hat{T} =\dfrac{1}{2+t^2}\langle t^2,-2t,2\rangle\times \dfrac{1}{2+t^2}\left\langle2,2t,t^2\right\rangle \\ &=\dfrac{1}{(2+t^2)^2} \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ t^2 & -2t & 2 \\ 2 & 2t & t^2 \end{vmatrix} \\ &=\dfrac{1}{(2+t^2)^2} \left(\hat{\imath}(-2t^3-4t) -\hat{\jmath}(t^4-4) +\hat{k}(2t^3+4t)\right) \\ &=\dfrac{1}{(2+t^2)^2}\langle -2t^3-4t,4-t^4,2t^3+4t\rangle \end{aligned} We now factor each term and cancel one factor of $$2+t^2$$: \begin{aligned} \hat{N} &=\dfrac{1}{2+t^2}\langle -2t,1-2t^2,2t\rangle \\ &=\left\langle \dfrac{-2t}{2+t^2},\dfrac{2-t^2}{2+t^2}, \dfrac{2t}{2+t^2}\right\rangle \end{aligned}

The figure shows $$\hat{T}$$, $$\hat{N}$$ and $$\hat{B}$$ moving along the curve. In particular, for $$t=1$$ we have \begin{aligned} \hat{T}&=\left\langle\dfrac{2}{3},\dfrac{2}{3},\dfrac{1}{3}\right\rangle \\ \hat{B}&=\left\langle \dfrac{1}{3},\dfrac{-2}{3},\dfrac{2}{3}\right\rangle \\ \hat{N}&=\left\langle \dfrac{-2}{3},\dfrac{1}{3},\dfrac{2}{3}\right\rangle \end{aligned} It is easy to check these are $$3$$ mutually perpendicular unit vectors.

Supported in part by NSF Grant #1123255 