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

17calculus > vector functions > curvature

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

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

answer

solution

Practice A02

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

answer

solution

Practice A03

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

answer

solution

Practice A04

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

answer

solution

Practice A05

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

answer

solution

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