\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \)
\( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \)
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17calculus > vector functions > unit tangent vector

Vector Functions - Unit Tangent Vector

In order to discuss curvature and a few other topics, we need to define a special vector called the unit tangent vector. As the name indicates, the unit tangent vector is a vector that is tangent to the curve and it's length is one.

What may not be obvious is that there is only one unit tangent vector and it points in the direction of motion. Given the vector function \(\vec{r}(t)\), the most basic equation we use to find the unit tangent vector is

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

The vector function \( \vec{r}(t) \) is often a position vector. As you know from basic calculus, the derivative of the position is velocity. So you will often see \(\vec{v}(t)=\vec{r}'(t)\) where \(\vec{v}(t)\) is referred to as the velocity vector. This allows us to write the unit tangent vector as \(\displaystyle{ \vec{T}(t) = \frac{\vec{v}(t)}{ \| \vec{v}(t) \| } }\).

This unit tangent vector is used a lot when calculating the principal unit normal vector, acceleration vector components and curvature. So take a few minutes to work some practice problems before going on to the next topic.

next: principal unit normal vector →

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Practice Problems

Instructions - - Unless otherwise instructed, find the unit tangent vector for the given vector function at the given point. If no point is given, find the general unit tangent vector \(\vec{T}(t)\).

Level A - Basic

Practice A01

\(\vec{r}(t)=t\vhat{i}+(1/t)\vhat{j}\), \(t=1\)

answer

solution

Practice A02

\(\vec{r}(t)=\cos t\vhat{i}+3t\vhat{j}+2\sin 2t\vhat{k}\), \(t=0\)

answer

solution

Practice A03

\(\vec{r}(t)=\langle 2\sin(t),4\cos(t),4\sin^2(t)\rangle\), \(t=\pi/6\)

answer

solution

Practice A04

\(\vec{r}(t)=\langle t^3,2t^2\rangle\), \(t=1\)

answer

solution

Practice A05

Find \(\vec{T}(t)\) and \(\vec{T}(0)\) for \(\vec{r}(t)=\langle 5t^2+1,-e^{-3t},2\sin(-3t)\rangle\).

answer

solution

Practice A06

\(\vec{r}(t)=(-t^3+t)\vhat{i}+(\ln(t^2))\vhat{j}+(\cos(\pi t))\vhat{k}\), \(t=1\)

answer

solution

Practice A07

\(\vec{r}(t)=\langle t\sqrt{2},e^t,e^{-t}\rangle\), \(t=0\)

answer

solution

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