Wize University Calculus 3 Textbook > Vector Functions

Tangent, Normal, Binormal Vectors - Applications of Vector Functions

Tangent, Normal, Binormal Vectors

The derivative, $f'(x)$ , of a function $f(x)$ represented the slope of the tangent line in single variable calculus.

In multivariable calculus, vector functions, $\vec{r(t)}$ , have a derivative, $\vec{r'(t)},~$ called the tangent vector.

Therefore, the tangent line to $\vec{r(t)}~$ at $P(x_0,y_0,z_0)$ is the line passing through the point $P$ and is parallel to the tangent vector.

Suppose that $\vec{r(t)}~$ is a vector and

$|\vec{r(t)}|=k,~k\in\Re$
$\vec{r'(t)}~$ is orthogonal to $\vec{r(t)}$

The unit tangent vector is defined as:

\boxed{\vec{T}=\frac{\vec{r'(t)}}{|\vec{r'(t)|}}}

The unit normal vector is defined as:

\boxed{\vec{N}=\frac{\vec{T'(t)}}{|\vec{T'(t)|}}}

The binormal vector is defined as:

\boxed{\vec{B(t)}=\vec{T(t)}\times\vec{N(t)}}

Let us take a look:

Image illustrating Tangent, Normal, Binormal Vectors

Example

Find the vector equation of the tangent line to the curve given by $\vec{r(t)}=3t^3\vec{i}+4\cos{t}\vec{j}+4\sin{t}\vec{k}$ and $t=\frac{\pi}{6}$ .

We need a point at $t=\frac{\pi}{6}~$ and we need a "slope" (i.e. a vector tangent to $\vec{r(t)}~$ which is $\vec{r'(t)}$ at $t=\frac{\pi}{6}.$

The point:

\vec{r(\frac{\pi}{6})}=\Bigg<3(\frac{\pi}{6})^3+4cos(\frac{\pi}{6})+4sin(\frac{\pi}{6})\Bigg>\newline{}\newline{} \vec{r(\frac{\pi}{6})}=\big<\frac{\pi^{3}}{72},~2\sqrt{3},~2\big>

The vector tangent $\vec{r'(t)}$ :

\vec{r'(t)}=<9t^2,~-4sint,~4cost>\newline \vec{r'(\frac{\pi}{6})}=\Big<9(\frac{\pi^2}{36}),~-4sin(\frac{\pi}{6}),~4cos(\frac{\pi}{6})\Big>\newline{}\newline{}\vec{r'(\frac{\pi}{6})}=\big<\frac{\pi^2}{4},~-2,~2\sqrt{3}\big>

The equation of the tangent line is:

L_1:~\big<\frac{\pi^{3}}{72},~2\sqrt{3},~2\big>+t\big<\frac{\pi^2}{4},~-2,~2\sqrt{3}\big>

Example

Find the tangent line, $\vec{A(t)},$ to the function $\vec{r(t)}=<9e^{t+5},~2\ln{t},~-\sqrt{t}+1>$ at $t=1$ .

We need a point at t =1 and a tangent vector at t = 1. First, find the point:

\begin{array}{rcl} \vec{r(1)}&=&<9e^{1+5},~2ln1,~-\sqrt{1}+1>\\\\ &=&<9e^6,0,0> \end{array}

The tangent vector:

\begin{array}{rcl} \vec{r'(t)}&=&\Big<9e^{t+5},\frac{2}{t},-\frac{1}{2\sqrt{t}}\Big>\\\\ \vec{r'(1)}&=&\Big<9e^{1+5},\frac{2}{1},-\frac{1}{2\sqrt{1}}\Big>\\\\ &=&<9e^6,~2,-\frac{1}{2}> \end{array}

So, the tangent line is:

\vec{A(t)}=~<9e^6,0,0>+~t\Big<9e^6,2,-\dfrac{1}{2}\Big>

Practice

Find the equation to the tangent line to the curve given by $\vec{r(t)}=\dfrac{1}{2}t^2\vec{i}+81\cos(\frac{t}{9})\vec{j}+81\sin(\frac{t}{9})\vec{k}$ at $t=3\pi$ .

$\vec{r_{tangent}}(t)=\ \ \ $ (Enter your answer as $<a,b,c>+t<p,q,r>$ with no spaces in between)

I don't know

Extra Practice

For the following curve, find the unit tangent, principle unit normal, binormal, curvature, radius of curvature, torsion, and the tangential and normal components of acceleration at time $t = \pi.$

\vec r(t) = (\cos t) \vec i + (\sin t) \vec j + (\cos^2 t) \vec k

Tangent, Normal, Binormal Vectors - Applications of Vector Functions

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