\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \)
\( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \)
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You CAN Ace Calculus

17calculus > vector functions > curvature

ATTENTION INSTRUCTORS: The new 2018 version of 17calculus will include changes to the practice problem numbering system. If you would like advance information to help you prepare for spring semester, send us an email at 2018info at 17calculus.com.

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Curvature Using Vector Functions

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 κ for curvature.

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 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 the curvature a bit better.

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

Here is a longer video explaining the derivation of some of the equations.

Louis Saumier - Curvature: Definition, Derivation [25min-30secs]

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Practice Problems

Instructions - - Unless otherwise instructed, find the curvature of these vector functions. Give your answers in exact, completely factored form.

Level A - Basic

Practice A01

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



Practice A02

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



Practice A03

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



Practice A04

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



Practice A05

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



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