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

17Calculus Vector Functions - Curvature

Coordinate Systems

Vectors

Using Vectors

Applications

Vector Functions

Partial Derivatives

Partial Integrals

Double Integrals - 2Int

Triple Integrals - 3Int

Practice

Vector Fields

Calculus Tools

Learning Tools

Articles

For the following discussion, we will consider a parameterized curve defined by the vector function \(\displaystyle{ \vec{r}(t) = \langle x(t), y(t), z(t) \rangle }\) which is traversed once on the continuous interval \( a \leq t \leq b \).
Some books use the Greek letter \(\kappa\) (kappa) for curvature. We use a capital K.

The curvature of a smooth curve is a measure of how 'tight' or 'sharp' the curve is. If we have a smooth curve \(\vec{r}\) and we have a function s which is the arc length function, the curvature is defined to be
\(\displaystyle{ K(s) = \left\| \frac{d\vec{T}}{ds} \right\|}\).

This equation for the curvature is not particularly useful for calculations. So we have several other ways to write the equation of the curvature. But first notice, that the curvature is a scalar function, not a vector function. And since it is the norm of a vector, the curvature will always be positive.

Curvature Formula #1

For our first equation to use when calculating the curvature, we will use the chain rule to write \(\displaystyle{ \frac{d\vec{T}}{dt} = \frac{d\vec{T}}{ds} \cdot \frac{ds}{dt} }\). We can solve for \(d\vec{T}/ds\) to get

\(\displaystyle{ K = \left\| \frac{d\vec{T}}{ds} \right\| = \frac{\| d\vec{T}/dt \|}{ \| ds/dt \| } = \frac{1}{\| \vec{v} \|} \left\| \frac{d\vec{T}}{dt} \right\| }\)

Notice in the previous equation, we used \(ds/dt = \|\vec{v}\|\) to simplify the equation somewhat.
Now we can write the first curvature formula in a form that we can use for calculations. \[\displaystyle{ K(t) = \frac{1}{\|\vec{v}\|} \left\| \frac{d\vec{T}}{dt} \right\| = \frac{\| \vec{T}'(t) \|}{\|\vec{r}'(t)\|} }\]

Curvature Formula #2

If we define a vector \(\vec{a} = d\vec{v}/dt\) as an acceleration vector, a second curvature formula is \[\displaystyle{ K=\frac{\|\vec{v} \times \vec{a} \|}{\|\vec{v}\|^3} }\] Before working some practice problems, here is a quick video clip for you that should help you understand curvature a bit better.

MIP4U - Determining Curvature of a Curve Defined by a Vector Valued Function [3mins-9secs]

video by MIP4U

Practice

Unless otherwise instructed, calculate the curvature of these vector functions. If a point is given, calculate the curvature at that point also. Give your answers in exact, completely factored form.

\( \vec{r}(t) = 3t\vhat{i} + 4\sin t\vhat{j} + \) \( 4\cos t\vhat{k} \)

Problem Statement

Calculate the curvature of the vector function \( \vec{r}(t) = 3t\vhat{i} + 4\sin t\vhat{j} + \) \( 4\cos t\vhat{k} \). Give your answer in exact, completely factored form.

Final Answer

\( K(t) = 4/25 \)

Problem Statement

Calculate the curvature of the vector function \( \vec{r}(t) = 3t\vhat{i} + 4\sin t\vhat{j} + \) \( 4\cos t\vhat{k} \). Give your answer in exact, completely factored form.

Solution

2073 video

video by Krista King Math

Final Answer

\( K(t) = 4/25 \)

close solution

Log in to rate this practice problem and to see it's current rating.

\( \vec{r}(t) = \langle 2\cos(t), 2\sin(t) \rangle \)

Problem Statement

Calculate the curvature of the vector function \( \vec{r}(t) = \langle 2\cos(t), 2\sin(t) \rangle \). Give your answer in exact, completely factored form.

Final Answer

\( K(t) = 1/2 \)

Problem Statement

Calculate the curvature of the vector function \( \vec{r}(t) = \langle 2\cos(t), 2\sin(t) \rangle \). Give your answer in exact, completely factored form.

Solution

2074 video

video by MIP4U

Final Answer

\( K(t) = 1/2 \)

close solution

Log in to rate this practice problem and to see it's current rating.

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

Problem Statement

Find the curvature at \(t=1\) for \( \vec{r}(t) = \langle t, t^2/2, t^3/3 \rangle \).

Final Answer

\( K(1) = \sqrt{2}/3 \)

Problem Statement

Find the curvature at \(t=1\) for \( \vec{r}(t) = \langle t, t^2/2, t^3/3 \rangle \).

Solution

2075 video

video by MIP4U

Final Answer

\( K(1) = \sqrt{2}/3 \)

close solution

Log in to rate this practice problem and to see it's current rating.

\( \vec{r}(t) = \langle -5t, 2t^3, 3t^4 \rangle \), \( t = 2 \).

Problem Statement

Find the curvature of \( \vec{r}(t) = \langle -5t, 2t^3, 3t^4 \rangle \) at \( t = 2 \).

Final Answer

\( K(t) = 24\sqrt{3229}/(9817)^{3/2} \)

Problem Statement

Find the curvature of \( \vec{r}(t) = \langle -5t, 2t^3, 3t^4 \rangle \) at \( t = 2 \).

Solution

2076 video

video by MIP4U

Final Answer

\( K(t) = 24\sqrt{3229}/(9817)^{3/2} \)

close solution

Log in to rate this practice problem and to see it's current rating.

\( \vec{r}(t) = \langle 2\cos(3t),2\sin(3t), \) \(4t \rangle \), \( t = 0 \).

Problem Statement

Find the curvature of \( \vec{r}(t) = \langle 2\cos(3t),2\sin(3t), \) \(4t \rangle \) at \( t = 0 \).

Final Answer

9/26

Problem Statement

Find the curvature of \( \vec{r}(t) = \langle 2\cos(3t),2\sin(3t), \) \(4t \rangle \) at \( t = 0 \).

Solution

This problem is solved in each of these videos. The first video uses the cross product version of the curvature formula, the second video does not use the cross product.

2077 video

video by MIP4U

2077 video

video by MIP4U

Final Answer

9/26

close solution

Log in to rate this practice problem and to see it's current rating.

vector functions curvature 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

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

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

To bookmark this page and practice problems, log in to your account or set up a free account.

Search Practice Problems

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.

calculus motivation - music and learning

The 17Calculus and 17Precalculus iOS and Android apps are no longer available for download. If you are still using a previously downloaded app, your app will be available until the end of 2020, after which the information may no longer be available. However, do not despair. All the information (and more) is now available on 17calculus.com for free.

How to Ace the Rest of Calculus: The Streetwise Guide, Including MultiVariable Calculus

Save Up To 50% Off SwissGear Backpacks Plus Free Shipping Over $49 at eBags.com!

Shop Amazon - Textbook Trade-In - Get Up to 80% Back

As an Amazon Associate I earn from qualifying purchases.

How to Develop a Brilliant Memory Week by Week: 50 Proven Ways to Enhance Your Memory Skills

Save Up To 50% Off SwissGear Backpacks Plus Free Shipping Over $49 at eBags.com!

Shop Amazon - New Textbooks - Save up to 40%

As an Amazon Associate I earn from qualifying purchases.

Practice Instructions

Unless otherwise instructed, calculate the curvature of these vector functions. If a point is given, calculate the curvature at that point also. Give your answers in exact, completely factored form.

Do NOT follow this link or you will be banned from the site!

When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.

DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. In short, use this site wisely by questioning and verifying everything. If you see something that is incorrect, contact us right away so that we can correct it.

Links and banners on this page are affiliate links. We carefully choose only the affiliates that we think will help you learn. Clicking on them and making purchases help you support 17Calculus at no extra charge to you. However, only you can decide what will actually help you learn. So think carefully about what you need and purchase only what you think will help you.

We use cookies on this site to enhance your learning experience.

17calculus

Copyright © 2010-2020 17Calculus, All Rights Reserved     [Privacy Policy]     [Support]     [About]

mathjax.org
Real Time Web Analytics
17Calculus
We use cookies to ensure that we give you the best experience on our website. By using this site, you agree to our Website Privacy Policy.