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.

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 $$t=1$$.

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

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.

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

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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 $$t=0$$.

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

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

2048 video

video by Krista King Math

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

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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 $$t=\pi/6$$.

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

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

2050 video

video by MIP4U

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

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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 $$t=1$$.

$$\langle 3/5,4/5 \rangle$$

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

2049 video

video by MIP4U

$$\langle 3/5,4/5 \rangle$$

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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 $$\vec{r}(t)=\langle 5t^2+1, -e^{-3t}, 2\sin(-3t) \rangle$$.

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

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

2051 video

video by MIP4U

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

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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 $$t=1$$.

$$\langle -\sqrt{2}/2, \sqrt{2}/2, 0 \rangle$$

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

2052 video

video by PatrickJMT

$$\langle -\sqrt{2}/2, \sqrt{2}/2, 0 \rangle$$

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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 $$t=0$$.

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

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

2053 video

video by David Lippman

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

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Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

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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