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

17Calculus Vector Functions - Curvature

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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

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

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

Krista King Math - 2073 video solution

video by Krista King Math

Final Answer

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

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

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

MIP4U - 2074 video solution

video by MIP4U

Final Answer

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

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

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

MIP4U - 2075 video solution

video by MIP4U

Final Answer

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

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

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

MIP4U - 2076 video solution

video by MIP4U

Final Answer

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

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

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.

MIP4U - 2077 video solution

video by MIP4U

MIP4U - 2077 video solution

video by MIP4U

Final Answer

9/26

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Find the curvature, \(\kappa\), for the curve \( \vec{r}(t) = \langle (2/3)t^3, t, t^2 \rangle \). Simplify your answer as much as possible.

Problem Statement

Find the curvature, \(\kappa\), for the curve \( \vec{r}(t) = \langle (2/3)t^3, t, t^2 \rangle \). Simplify your answer as much as possible.

Solution

Steve Butler - 4328 video solution

video by Steve Butler

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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.

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