17Calculus Vector Functions - Principal Unit Normal Vector
On this page, we explain how to find a normal vector to a curve. In general, there are many normal vectors. However, one of them, the principal unit normal vector, is the one we focus on since this normal vector tells in what direction the curve is 'curving'. When we get to curvature, we will use this vector to also determine how 'tight' the curve is.
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Conceptual Understanding
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Plot 1 |
Before we get into the details of how to calculate the principal unit normal vector, let's look at an example to get a conceptual understanding of what we are talking about.
Plot 1 shows a curve (in black), the unit tangent vector (in green) and a normal vector (in blue) at the point \((1,2)\). Let's discuss each curve individually.
Basic Curve - - The basic curve (in black) is traced out starting at \((0,0)\) and going up and to the right, as shown by the arrow on the curve.
Unit Tangent Vector - - The unit tangent vector (in green) is tangent to the curve at the point of interest. Of course, as you know from basic calculus, there are two vectors tangent to the curve at each point but we are only interested in the unit tangent vector pointing in the direction of motion.
Principal Unit Normal Vector - - A normal vector (in blue) is shown in Plot 1. There are actually two normal vectors, the one we show and another in the opposite direction (not shown). We are interested only in the principal unit normal vector, which is the normal vector of length one that points to the inside of the curve.
How To Calculate the Principal Unit Normal Vector
To calculate the principal unit normal vector, we use the unit tangent vector \(\vec{T}(t)\). The equation is
\(\displaystyle{ \vec{N}(t) = \frac{d\vec{T}/dt}{ \| d\vec{T}/dt \| } }\)
We normally use \(\vec{N}\) to specify this vector within the context of vector functions. Here is quick video proving that \(\vec{T}\) and \(\vec{N}\) are orthogonal.
MIP4U - Proving the Unit Normal Vector Formula [3mins-41secs]
video by MIP4U |
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Shortcut in 2-dim
Remember from vector algebra that if you have a vector, you can get a vector normal to it by switching the x and y components and changing the sign of one of them. For example, for the vector \(\vec{v} = \langle a,b \rangle\), vectors \(\vec{w} = \langle -b,a \rangle\) and \(\vec{z} = \langle b,-a \rangle\) are both normal to \(\vec{v}\). (To check this, take the dot products and make sure you get zero.) However, there are two vectors and it is not possible to tell which is the principal normal vector. But, if you are allowed to, you can usually graph the original function and the vectors and pull the information off the graph.
Notes - -
1. This works only in 2-dim, i.e. for space curves (3-dim) you must use the vector equation given above for \(\vec{N}\).
2. As with everything on this site, make sure to check with your instructor to see if they allow you to use this shortcut.
Here is a quick video clip discussing the principal unit normal vector and this shortcut.
MIP4U - Determining the Unit Normal Vector [2mins-41secs]
video by MIP4U |
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Okay, time for some practice problems. After that, your next logical topic is how to describe the acceleration vector in terms of the unit tangent and principal unit normal vector.
Practice
Unless otherwise instructed, find the principal unit normal vector of the curve. If a point is given, also find the principal unit normal vector at that point.
\( \vec{r}(t)=\langle \cos(t),\sin(t),t \rangle \), \( t=\pi/4 \)
Problem Statement |
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Calculate the principal unit normal vector for the vector function \( \vec{r}(t)=\langle \cos(t),\sin(t),t \rangle \) at the point \( t=\pi/4 \).
Final Answer |
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\( \vec{N}(t) = \langle -\cos(t),-\sin(t),0 \rangle \); \( \vec{N}(\pi/4) = \langle -\sqrt{2}/2,-\sqrt{2}/2,0 \rangle \)
Problem Statement |
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Calculate the principal unit normal vector for the vector function \( \vec{r}(t)=\langle \cos(t),\sin(t),t \rangle \) at the point \( t=\pi/4 \).
Solution |
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2061 video
video by MIP4U |
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Final Answer |
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\( \vec{N}(t) = \langle -\cos(t),-\sin(t),0 \rangle \); \( \vec{N}(\pi/4) = \langle -\sqrt{2}/2,-\sqrt{2}/2,0 \rangle \) |
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\( \vec{r}(t) = \langle t^3,2t^2 \rangle \), \( t=1 \)
Problem Statement |
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Calculate the principal unit normal 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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\( \displaystyle{ \vec{N}(t) = \frac{\langle 4,-3t\rangle}{\sqrt{9t^2+16}}}\); \( \vec{N}(1) = \langle 4/5,-3/5\rangle \)
Problem Statement |
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Calculate the principal unit normal vector for the vector function \( \vec{r}(t) = \langle t^3,2t^2 \rangle \) at the point \( t=1 \).
Solution |
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Although not required for this problem, here is the graph of the vector function and the unit tangent and principal unit normal vectors.
2062 video
video by MIP4U |
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Final Answer |
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\( \displaystyle{ \vec{N}(t) = \frac{\langle 4,-3t\rangle}{\sqrt{9t^2+16}}}\); \( \vec{N}(1) = \langle 4/5,-3/5\rangle \) |
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Calculate \(\vec{T}(0)\) and \(\vec{N}(0)\) for \( \vec{r}(t) = \langle t\sqrt{2}, e^t, e^{-t} \rangle \).
Problem Statement |
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Calculate \(\vec{T}(0)\) and \(\vec{N}(0)\) for \( \vec{r}(t) = \langle t\sqrt{2}, e^t, e^{-t} \rangle \).
Final Answer |
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\( \vec{T}(0) = \langle \sqrt{2}/2,1/2,-1/2\rangle \); \( \vec{N}(0) = \langle 0, \sqrt{2}/2, \sqrt{2}/2 \rangle \)
Problem Statement |
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Calculate \(\vec{T}(0)\) and \(\vec{N}(0)\) for \( \vec{r}(t) = \langle t\sqrt{2}, e^t, e^{-t} \rangle \).
Solution |
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2063 video
video by David Lippman |
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Final Answer |
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\( \vec{T}(0) = \langle \sqrt{2}/2,1/2,-1/2\rangle \); \( \vec{N}(0) = \langle 0, \sqrt{2}/2, \sqrt{2}/2 \rangle \) |
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\( \vec{r}(t)=\langle t,3\cos(t),3\sin(t)\rangle \)
Problem Statement |
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Calculate the principal unit normal vector for the vector function \( \vec{r}(t)=\langle t,3\cos(t),3\sin(t)\rangle \).
Final Answer |
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\(\displaystyle{ \vec{T}(t) = \frac{\langle 1,-3\sin(t),3\cos(t)\rangle}{\sqrt{10}} }\); \( \vec{N}(t) = \langle 0,-\cos(t),-\sin(t) \rangle \)
Problem Statement |
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Calculate the principal unit normal vector for the vector function \( \vec{r}(t)=\langle t,3\cos(t),3\sin(t)\rangle \).
Solution |
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2064 video
video by Krista King Math |
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Final Answer |
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\(\displaystyle{ \vec{T}(t) = \frac{\langle 1,-3\sin(t),3\cos(t)\rangle}{\sqrt{10}} }\); \( \vec{N}(t) = \langle 0,-\cos(t),-\sin(t) \rangle \) |
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principal unit normal vector 17calculus youtube playlist
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, find the principal unit normal vector of the curve. If a point is given, also find the principal unit normal vector at that point.
