17Calculus 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.
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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.
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 |
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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 |
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\(\displaystyle{ \vec{T}(1) = \frac{\vhat{i} - \vhat{j}}{\sqrt{2}} }\)
Problem Statement |
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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 |
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\( \vec{r}'(t) = \vhat{i} - (1/t^2)\vhat{j} \)
\( \vec{r}'(1) = \vhat{i} - \vhat{j} \)
\(\displaystyle{ \vec{T}(1) = \frac{\vec{r}'(1)}{\| \vec{r}'(1) \|} = \frac{\vhat{i}-\vhat{j}}{\sqrt{2}} }\)
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 |
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\(\displaystyle{ \vec{T}(1) = \frac{\vhat{i} - \vhat{j}}{\sqrt{2}} }\) |
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\( \vec{r}(t) = \cos t \vhat{i} + 3t\vhat{j} + \) \( 2\sin 2t \vhat{k} \), \(t=0\).
Problem Statement |
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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 |
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\( \vec{T}(0) = (3/5)\vhat{j} + (4/5)\vhat{k} \)
Problem Statement |
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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 |
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Final Answer |
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\( \vec{T}(0) = (3/5)\vhat{j} + (4/5)\vhat{k} \) |
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\( \vec{r}(t) = \langle 2\sin(t), 4\cos(t), 4\sin^2(t) \rangle \), \(t=\pi/6\).
Problem Statement |
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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 |
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\(\displaystyle{ \frac{1}{\sqrt{19}} \langle \sqrt{3}, -2, 2\sqrt{3} \rangle }\)
Problem Statement |
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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 |
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Final Answer |
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\(\displaystyle{ \frac{1}{\sqrt{19}} \langle \sqrt{3}, -2, 2\sqrt{3} \rangle }\) |
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\(\vec{r}(t) = \langle t^3,2t^2 \rangle \), \(t=1\).
Problem Statement |
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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 |
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\( \langle 3/5,4/5 \rangle \)
Problem Statement |
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Calculate the unit tangent vector for the vector function \(\vec{r}(t) = \langle t^3,2t^2 \rangle \) at the point \(t=1\).
Solution |
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video by MIP4U |
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Final Answer |
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\( \langle 3/5,4/5 \rangle \) |
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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 |
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Find \(\vec{T}(t)\) and \(\vec{T}(0)\) for \(\vec{r}(t)=\langle 5t^2+1, -e^{-3t}, 2\sin(-3t) \rangle \).
Final Answer |
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\(\displaystyle{ \vec{T}(t) = \frac{\langle 10t, 3e^{-3t}, -6\cos(-3t) \rangle}{\sqrt{100t^2+9e^{-6t}+36\cos^2(-3t)}} }\)
\( \vec{T}(0) = \langle 0,1/\sqrt{5},-2/\sqrt{5} \rangle \)
Problem Statement |
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Find \(\vec{T}(t)\) and \(\vec{T}(0)\) for \(\vec{r}(t)=\langle 5t^2+1, -e^{-3t}, 2\sin(-3t) \rangle \).
Solution |
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2051 video
video by MIP4U |
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Final Answer |
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\(\displaystyle{ \vec{T}(t) = \frac{\langle 10t, 3e^{-3t}, -6\cos(-3t) \rangle}{\sqrt{100t^2+9e^{-6t}+36\cos^2(-3t)}} }\) |
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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 |
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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 |
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\( \langle -\sqrt{2}/2, \sqrt{2}/2, 0 \rangle \)
Problem Statement |
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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 |
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Final Answer |
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\( \langle -\sqrt{2}/2, \sqrt{2}/2, 0 \rangle \) |
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\( \vec{r}(t) = \langle t\sqrt{2}, e^t, e^{-t} \rangle \), \( t=0 \).
Problem Statement |
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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 |
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\( \vec{T}(0) = \langle \sqrt{2}/2, 1/2, -1/2 \rangle \)
Problem Statement |
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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 |
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video by David Lippman |
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Final Answer |
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\( \vec{T}(0) = \langle \sqrt{2}/2, 1/2, -1/2 \rangle \) |
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You CAN Ace Calculus
Topics You Need To Understand For This Page
Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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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 \( \vec{T}(t) \).
