\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\units}[1]{\,\text{#1}} \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus Vector Functions - Unit Tangent Vector

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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 \(\displaystyle{ \vec{T}(t) = \frac{\vec{v}(t)}{ \| \vec{v}(t) \| } }\).

Notation - As we mentioned on the unit vectors page, many times books and instructors will use the 'hat' notation to indicate that a vector is a unit vector. So you may see the unit tangent vector written as \( \hat{T} \). Check with your instructor to see what they expect. Your textbook will also give you an indication of the preferred notation in class.

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.

How to Read and Do Proofs: An Introduction to Mathematical Thought Processes

Practice

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

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

Problem Statement

Calculate the unit tangent vector for the vector function \( \vec{r}(t) = t\vhat{i} + (1/t)\vhat{j} \) at the point \(t=1\).

Final Answer

\(\displaystyle{ \vec{T}(1) = \frac{\vhat{i} - \vhat{j}}{\sqrt{2}} }\)

Problem Statement

Calculate the unit tangent vector for the vector function \( \vec{r}(t) = t\vhat{i} + (1/t)\vhat{j} \) at the point \(t=1\).

Solution

\( \vec{r}'(t) = \vhat{i} - (1/t^2)\vhat{j} \)
\( \vec{r}'(1) = \vhat{i} - \vhat{j} \)
\(\displaystyle{ \vec{T}(1) = \frac{\vec{r}'(1)}{\| \vec{r}'(1) \|} = \frac{\vhat{i}-\vhat{j}}{\sqrt{2}} }\)
Here is a plot of the solution. The black line is the curve \( \vec{r}(t) \) with the black arrow indicating that it is being traced out left to right (or down the curve). (We have plotted only a section of the curve from x=1/4 to x=4.) The red vector is the unit tangent vector. Notice that it is pointing in the direction that the curve is being traced.

Final Answer

\(\displaystyle{ \vec{T}(1) = \frac{\vhat{i} - \vhat{j}}{\sqrt{2}} }\)

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

Problem Statement

Calculate the unit tangent vector for the vector function \( \vec{r}(t) = \cos t \vhat{i} + 3t\vhat{j} + \) \( 2\sin 2t \vhat{k} \) at the point \(t=0\).

Final Answer

\( \vec{T}(0) = (3/5)\vhat{j} + (4/5)\vhat{k} \)

Problem Statement

Calculate the unit tangent vector for the vector function \( \vec{r}(t) = \cos t \vhat{i} + 3t\vhat{j} + \) \( 2\sin 2t \vhat{k} \) at the point \(t=0\).

Solution

Krista King Math - 2048 video solution

video by Krista King Math

Final Answer

\( \vec{T}(0) = (3/5)\vhat{j} + (4/5)\vhat{k} \)

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

Problem Statement

Calculate the unit tangent vector for the vector function \( \vec{r}(t) = \langle 2\sin(t), 4\cos(t), 4\sin^2(t) \rangle \) at the point \(t=\pi/6\).

Final Answer

\(\displaystyle{ \frac{1}{\sqrt{19}} \langle \sqrt{3}, -2, 2\sqrt{3} \rangle }\)

Problem Statement

Calculate the unit tangent vector for the vector function \( \vec{r}(t) = \langle 2\sin(t), 4\cos(t), 4\sin^2(t) \rangle \) at the point \(t=\pi/6\).

Solution

MIP4U - 2050 video solution

video by MIP4U

Final Answer

\(\displaystyle{ \frac{1}{\sqrt{19}} \langle \sqrt{3}, -2, 2\sqrt{3} \rangle }\)

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

Problem Statement

Calculate the unit tangent vector for the vector function \(\vec{r}(t) = \langle t^3,2t^2 \rangle \) at the point \(t=1\).

Final Answer

\( \langle 3/5,4/5 \rangle \)

Problem Statement

Calculate the unit tangent vector for the vector function \(\vec{r}(t) = \langle t^3,2t^2 \rangle \) at the point \(t=1\).

Solution

MIP4U - 2049 video solution

video by MIP4U

Final Answer

\( \langle 3/5,4/5 \rangle \)

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

Problem Statement

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

Final Answer

\(\displaystyle{ \vec{T}(t) = \frac{\langle 10t, 3e^{-3t}, -6\cos(-3t) \rangle}{\sqrt{100t^2+9e^{-6t}+36\cos^2(-3t)}} }\)
\( \vec{T}(0) = \langle 0,1/\sqrt{5},-2/\sqrt{5} \rangle \)

Problem Statement

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

Solution

MIP4U - 2051 video solution

video by MIP4U

Final Answer

\(\displaystyle{ \vec{T}(t) = \frac{\langle 10t, 3e^{-3t}, -6\cos(-3t) \rangle}{\sqrt{100t^2+9e^{-6t}+36\cos^2(-3t)}} }\)
\( \vec{T}(0) = \langle 0,1/\sqrt{5},-2/\sqrt{5} \rangle \)

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

Problem Statement

Calculate the unit tangent vector for the vector function \( \vec{r}(t) = (-t^3+t)\vhat{i} + \) \( (\ln(t^2))\vhat{j} + (\cos(\pi t))\vhat{k} \) at the point \( t=1 \).

Final Answer

\( \langle -\sqrt{2}/2, \sqrt{2}/2, 0 \rangle \)

Problem Statement

Calculate the unit tangent vector for the vector function \( \vec{r}(t) = (-t^3+t)\vhat{i} + \) \( (\ln(t^2))\vhat{j} + (\cos(\pi t))\vhat{k} \) at the point \( t=1 \).

Solution

PatrickJMT - 2052 video solution

video by PatrickJMT

Final Answer

\( \langle -\sqrt{2}/2, \sqrt{2}/2, 0 \rangle \)

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

Problem Statement

Calculate the unit tangent vector for the vector function \( \vec{r}(t) = \langle t\sqrt{2}, e^t, e^{-t} \rangle \) at the point \( t=0 \).

Final Answer

\( \vec{T}(0) = \langle \sqrt{2}/2, 1/2, -1/2 \rangle \)

Problem Statement

Calculate the unit tangent vector for the vector function \( \vec{r}(t) = \langle t\sqrt{2}, e^t, e^{-t} \rangle \) at the point \( t=0 \).

Solution

David Lippman - 2053 video solution

video by David Lippman

Final Answer

\( \vec{T}(0) = \langle \sqrt{2}/2, 1/2, -1/2 \rangle \)

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The curves \(\vec{r}_1(t) = \langle 5t+2, \ln t, \sqrt{4-2t-t^2} \rangle\) and \(\vec{r}_2(t) = \langle 7, s^{3/2}\sqrt{2}, e^{s^2-2s} \rangle\) intersect at a point. Find \(\cos\theta\) where \(\theta\) is the angle of intersection (the angle between their respective tangent lines).

Problem Statement

The curves \(\vec{r}_1(t) = \langle 5t+2, \ln t, \sqrt{4-2t-t^2} \rangle\) and \(\vec{r}_2(t) = \langle 7, s^{3/2}\sqrt{2}, e^{s^2-2s} \rangle\) intersect at a point. Find \(\cos\theta\) where \(\theta\) is the angle of intersection (the angle between their respective tangent lines).

Hint

Remember the angle between two vectors \(\vec{u}\) and \(\vec{v}\) can be found using \(\displaystyle{ \cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{ \norm{\vec{u}} \norm{\vec{v}} } }\)

Problem Statement

The curves \(\vec{r}_1(t) = \langle 5t+2, \ln t, \sqrt{4-2t-t^2} \rangle\) and \(\vec{r}_2(t) = \langle 7, s^{3/2}\sqrt{2}, e^{s^2-2s} \rangle\) intersect at a point. Find \(\cos\theta\) where \(\theta\) is the angle of intersection (the angle between their respective tangent lines).

Hint

Remember the angle between two vectors \(\vec{u}\) and \(\vec{v}\) can be found using \(\displaystyle{ \cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{ \norm{\vec{u}} \norm{\vec{v}} } }\)

Solution

Steve Butler - 4374 video solution

video by Steve Butler

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Practice Instructions

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

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