Arc Length and Curvature of Smooth Curves

We discuss piecewise smooth vector functions of one variable. In particular we study arc length and curvature. We define the curvature of a vector-function in term so the unit tangent vector. We give several concrete examples of how to determine the curvature and how to find the maximum curvature.

Smooth Curves

The one variable function $f(x)=|x|$ is not differentiable at $x=0$ because it has a “corner” at $x=0.$ In three dimensions we have the concept of a vector function representing a smooth curve; that is, where there are no so called corners. With an extra degree of freedom we require the derivative to not only exist but also to be continuous and nonzero.

Definition. The graph of the vector function defined by ${F}(t)$ is smooth on any interval of $t$ where ${F}\,’$ is continuous and ${F}'(t)\neq {0}.$

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Example. Find the intervals on which the epicycloid $C$ given by $$ {R}(t)=(5\cos t -\cos 5t) {i} + (5\sin t – \sin 5t ){j}, \qquad 0\leq t \leq 2\pi $$ is smooth.

Solution. The derivative of ${R}(t)$ is $$ {R}(t)=(-5\sin t+5\sin 5t){i}+(5\cos t-5\cos 5t){j}. $$ Notice that ${R}’$ is continuous on $[0,2\pi).$ In the interval $[0,2\pi)$, the only values of $t$ for which ${R}'(t)=0{i}+0{j}$ are found by solving the trigonometric equations: $$ -5\sin t+5\sin 5t =0 \qquad 5\cos t-5\cos 5t=0, \quad \text{ on } [0,2\pi). $$ The first equation yields: $$ t=0, \quad t=\frac{7 \pi }{6}, \quad t=\frac{3 \pi }{2}, \quad t=\frac{11 \pi }{6}, \quad t=\frac{\pi }{6}, \quad t=\frac{\pi }{2}, \quad t=\frac{5 \pi }{6} $$ The second equation yields: $$ t=0, \quad t=\pi, \quad t=\frac{4 \pi }{3}, \quad t=\frac{3 \pi }{2}, \quad t=\frac{5 \pi }{3}, \quad t=\frac{\pi }{3}, \quad t=\frac{\pi }{2}, \quad t=\frac{2 \pi }{3} $$ Therefore, we can conclude that $C$ is smooth on the intervals $$ \left(0,\frac{\pi}{2}\right), \qquad \left(\frac{\pi}{2},\pi\right), \qquad \left(\pi, \frac{3\pi}{2}\right), \qquad \left(\frac{3\pi}{2}, 2\pi\right) $$ as shown above.

The graph of a vector function ${F}$ is piecewise smooth on any interval that can be subdivided into a finite number of subintervals on which ${F}$ is smooth. For instances, the graph is piecewise smooth on $[0,2\pi)$ since the interval $[0,2\pi)$ can be subdivided into a finite number of subintervals on which ${F}$ is smooth. Examples of piecewise smooth curves:

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Example. Determine where the graph of the vector function $$ {F}(t)=\left(2t^3+\left(4-\frac{3\pi }{2}\right)t^2-4\pi t+2\pi \right){i}+\left(-\frac{1}{2}\cos 2 t \right){j}+(\sin t){k}. $$ is piecewise smooth.

Solution. The graph of the vector function ${F}$ is smooth over any interval not containing $t=\pi/2$ because \begin{equation} {F}'(t)=\left(6t^2+(8-3\pi )t-4\pi \right){i}+(\sin 2t){j}+(\cos t){k}, \end{equation} ${F}'(t)\neq {0}$ for any $t$ except $t=\frac{\pi }{2}$, and ${F}’$ is continuous everywhere.

Example. Determine where the graph of the vector function \begin{equation} {F}(t) =\left(2t^2-\pi t+2\pi \right){i}+\left(\frac{1}{2}\sin 2t\right){j}+\left(\frac{-1}{4}\cos 4t\right){k}. \end{equation} is piecewise smooth.

Solution. The graph of the vector function ${F}$ is piecewise smooth everywhere because \begin{equation} {F}'(t)=(4t-\pi ){i}+(\cos 2t){j}+(\sin 4t){k}, \end{equation} ${F}'(t)\neq {0}$ except for $t=\frac{\pi }{4}$, and ${F}’$ is continuous everywhere.

Arc Length

In this section, we first generalize the arc length formula from two to three dimensions. Recall that the arc length of the graph of a differentiable function $y=f(x)$ on the interval $[a,b]$ is given by $$ s=\int_a^b\sqrt{1+\left(\frac{d y}{d x}\right)^2}dx. $$ Now if the graph is parametrized by equations $x=x(t)$ and $y=y(t)$ then by the chain rule \begin{equation} \frac{d y}{d x} =\frac{d y}{d t} \frac{d t}{d x} \end{equation} whenever $d x/d t \neq 0.$ After algebraic manipulation we find that \begin{equation} s=\int_{t_1}^{t_2}\sqrt{\frac{ \left( \frac{d x}{d t} \right)^2 + \left(\frac{d y}{d t}\right)^2}{\left(\frac{d x}{d t}\right)^2}}\left(\frac{dx}{dt}\right) dt = \int_{t_1}^{t_2}\sqrt{\left(\frac{d x}{d t}\right)^2+\left(\frac{d y}{d t}\right)^2}\, dt \end{equation} represents the arc length of the curve $y=f(x)$ from the points corresponding to $x=a$ to $x=b.$

Arc Length and Curvature of Smooth Curves

Suppose that $C$ is a plane curve described by the vector function $$ {R}(t) = x(t){i} +y(t){j} $$ instead. Then ${R}'(t)=x'(t){i}+y'(t){j}$ where $y=f(x)$ and $$ |{R}'(t)|=\sqrt{R'(t)\cdot R'(t)}=\sqrt{[x'(t)]^2+[y'(t)]^2} $$ form which we see that $s$ can also be written in the form $$ s=\int_{t_1}^{t_2}\sqrt{ [x'(t)]^2+[y'(t)]^2 }\, dt = \int_{t_1}^{t_2}|R'(t)| \, dt $$ In $\mathbb{R}^3$ we have $$ s=\int {t_1}^{t_2}\sqrt{\displaystyle \left(\frac{d x}{d t}\right)^2 +\left(\frac{d y}{d t}\right)^2+\left(\frac{d z}{d t}\right)^2}\, dt. $$ But in fact the arc length of a graph is independent of its parametrization and thus, \begin{equation} s(t)=\int_{t_0}^{ t} \sqrt{\left(\frac{d x}{d u}\right)^2 +\left(\frac{d y}{d u}\right)^2+\left(\frac{d z}{d u}\right)^2} \, du\end{equation} where $P_0$ is the base point corresponding to $t_0.$ This discussion motivates the following theorem.

Theorem. (Arc Length Function) Let $C$ be a piecewise smooth curve that is the graph of the vector function described parametrically by $$ {R}(t)=x(t){i}+y(t){j}+z(t){k}, \qquad a \leq t \leq b $$ and let $P_0=P\left(a\right)$ be a particular point on $C$ (base point). If $C$ is transversed exactly once as $t$ increases from $a$ to $t$, then the length of $C$ from the base point $P_0$ to the variable $P(t)$ is given by the arc length function \begin{equation} s(t) =\int_{a}^t\sqrt{\left(\frac{d x}{d u}\right)^2 +\left(\frac{d y}{d u}\right)^2+\left(\frac{d z}{d u}\right)^2}\, du =\int_{a}^{t}||{R}'(t)|| \, dt \end{equation}

Example. Find the arc length of the curve $$ {R}(t)=(1-2\cos t){i}+2\sin t {j}+ 0.3t {k} $$ from $t=-5$ to $t=5.$

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Solution. We find the arc length to be $\sqrt{409}$ units because $$ \frac{d x}{d t}=2\sin t, \quad \frac{d y}{d t}=2 \cos t, \quad \frac{d z}{d t}=0.3 $$ and so the arc length is \begin{equation} \int_{-5}^{5} \sqrt{4\sin ^2t+4\cos ^2t+(0.3)^2} \, dt=\int_{-5}^5 \frac{\sqrt{409}}{10} \, dt
=\sqrt{409}. \end{equation} as claimed.

Theorem. (Speed Along an Arc) Suppose an object moves along a smooth curve $C$ that is the graph of the position function $$ {R}(t)=x(t){i}+y(t){j}+z(t) {k}, $$ where ${R}'(t)$ is continuous on the interval $\left[t_1,t_2\right].$ Then the object has speed $d s/d t$ for $t_1\leq t\leq t_2$ where \begin{equation} s(t)=\int _{t_1}^t\sqrt{\left(\frac{d x}{d u}\right)^2+\left(\frac{d y}{d u}\right)^2+\left(\frac{d z}{d u}\right)^2}\, du. \end{equation}

Proof. Given that ${R}(t)=x(t){i}+y(t){j}+z(t){k}$ is the position vector function for an object which moves along the graph of ${R}$ and given that ${R}'(t)$ is continuous on $\left[t_1,t_2\right]$ we can apply the Fundamental Theorem of Calculus to \begin{equation}
s(t)=\int _{t_1}^t\sqrt{\left(\frac{d x}{d u}\right)^2+\left(\frac{d y}{d u}\right)^2+\left(\frac{d z}{d u}\right)^2}\, du \end{equation} and obtain \begin{equation} \frac{d s}{d t}=\sqrt{\left(\frac{d x}{d t}\right)^2+\left(\frac{d y}{d t}\right)^2+\left(\frac{d z}{d t}\right)^2} =\left|\left| {R}'(t)\right|\right| =\left|\left| {V}(t)\right|\right| \end{equation} as needed.

Example. If a moving object has a position vector function of \begin{equation} {R}(t)=\langle e^{3t},\sqrt{18}t-5,-e^{-3t}\rangle \end{equation} then find the speed of the object at time $t$ and the distance traveled by the object between times $t=0$ and $t=1.$

Solution. The speed of the object at time $t$ is $3e^{3t}+3e^{-3t}$ because \begin{equation}
{V}(t)=\langle 3e^{3t},\sqrt{18},3e^{-3t}\rangle, \end{equation} \begin{equation} |{V}(t)|=\sqrt{9e^{6t}+18+9e^{-6t}}=3\sqrt{\left(e^{3t}+e^{-3t}\right)^2}=3\left(e^{3t}+e^{-3t}\right)
\end{equation} and the distance traveled by the object between times $t=0$ and $t=1$ is \begin{equation} s=\int _0^1\left(3e^{3t}+3e^{-3t}\right)\, dt=e^3-\frac{1}{e^3} \end{equation} as desired.

Example. Express the vector function ${R}(t)=\langle e^{-t},1-e^{-t}\rangle $ in terms of arc length measured from the point corresponding to $t=0$, in the direction of increasing $t.$

Solution. We have \begin{equation} s=s(t)=\int _0^t\sqrt{e^{-2u}+e^{-2u}}\, du=\sqrt{2}\int _0^te^{-u}\, du=\sqrt{2}-\sqrt{2}e^{-t}.\end{equation} Solving for $e^{-t}$ we have $e^{-t}=\frac{\sqrt{2}-s}{\sqrt{2}}.$ Thus $$ {R}(s)=\left\langle\frac{\sqrt{2}-s}{\sqrt{2}},\frac{s}{\sqrt{2}}\right\rangle. $$ which is expressed in terms of arc length $s.$

Theorem. (Normal Vectors) If ${R}(t)$ has a piecewise smooth graph and is represented as ${R}(s)$ in terms of the arc length parameter $s$, then the unit tangent vector ${T}$ and the principal unit normal vector ${N}$ satisfies \begin{equation} {T}=\frac{d {R}}{d s} \qquad \text{and} \qquad {N}=\frac{1}{\kappa } \frac{d {T}}{d s}\end{equation} where $\kappa =\left|\left| d {T} / d s\right|\right| $ is a scalar function of $s.$

Geometrically, the curvature $\kappa$ measures how fast the unit tangent vector to the curve rotates. If a curve keeps close to the same direction, the unit tangent vector changes very little and the curvature is small; where the curve undergoes a tight turn, the curvature is large.

Proof. Given a piecewise smooth graph represented by ${R}(t)$ and in terms of arc length by ${R}(s)$, then by the chain rule, \begin{equation}\frac{d{R}}{d s} = \frac{\frac{ d {R} }{dt} }{\frac{d s}{d t}}=\frac{ {R}'(t) }{ \left|\left| {R}'(t) \right|\right| }={T}.\end{equation} Also \begin{equation}{N}=\frac{{T}'(t)}{\left|\left| {T}'(t)\right|\right| }=\frac{ d {T} / d t }{ \left|\left| {T}'(t) \right|\right| } =\frac{d {T}}{d s}\cdot \frac{d s/d t}{\left|\left| {T}'(t)\right|\right|}\end{equation}and since $d s/ d t >0$ and $\left|\left| {T}'(t)\right|\right| >0$, ${N}$ points in the same direction as $d {T}/d s$ and since ${N}$ is a unit vector \begin{equation} {N} =\frac{\frac{d {T}}{d s}}{\left|\left| d s/d t\right|\right| } =\frac{1}{\kappa }\frac{d {T}}{d s} \end{equation}where $\kappa =\left|\left| d s/ d t. \right|\right|$

Curvature

Suppose the smooth curve $C$ is the graph of the vector function ${R}(s)$, parametrized in terms of the arc length $s.$ Then the curvature of $C$ is the function \begin{equation} \kappa (s) =\left|\left| \frac{d {T}}{d s}\right|\right| \end{equation} where ${T}(s)$ is the unit tangent vector.

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Example. Find the curvature of a circle.

Solution. A circle can be parametrized by ${R}(t)=\langle r \cos t,r \sin t\rangle$ where $r$ is the radius. We determine the arc length function as \begin{equation} s=\int _0^t\sqrt{r^2\cos ^2u+r^2\sin ^2u}\, du=r t. \end{equation} Solving for $t$ we find the component functions to be \begin{equation} x(s)=r \cos \left(\frac{s}{r}\right) \qquad \text{and} \qquad y(s) = r \sin \left(\frac{s}{r}\right). \end{equation} Thus the unit tangent and unit normal vectors are \begin{equation} {T}(s) =\left\langle -\sin \left(\frac{s}{r}\right),\cos \left(\frac{s}{r}\right)\right\rangle\end{equation} and \begin{equation} {N}(s) =\left\langle-\frac{1}{r}\cos \left(\frac{s}{r}\right),-\frac{1}{r}\sin \left(\frac{s}{r}\right)\right\rangle\left(\frac{1}{\kappa }\right) \end{equation} where \begin{equation} \kappa =\sqrt{\frac{1}{r^2}\cos ^2\left(\frac{s}{r}\right)+\frac{1}{r^2}\sin ^2\left(\frac{s}{r}\right)}=\frac{1}{r}. \end{equation} Therefore the curvature of a circle is $\frac{1}{r}.$

Example. Let $C$ be the curve given as the graph of the vector function $$ {R}(t)=(t-\sin t){i}+(1-\cos t){j}+\left(4\sin \frac{t}{2}\right){k}. $$ Find the unit tangent vector ${T}(t)$ to $C,$ $\frac{d {T}}{d s},$ and the curvature $\kappa (t).$

Solution. The derivative is \begin{equation} {R}'(t)=(1-\cos t){i}+(\sin t) {j}+\left(2 \cos \frac{t}{2}\right){k} \end{equation} and the magnitude of the derivative is \begin{align} \left|\left| {R}'(t)\right|\right| & =\sqrt{(1-\cos t)^2+\sin ^2t+4 \cos ^2 \frac{t}{2}} \\ & =\sqrt{2-2\cos t+4 \cos ^2\frac{t}{2}} =\sqrt{4\sin ^2\frac{t}{2}+4\cos ^2\frac{t}{2}} =2 \end{align} thus the unit tangent vector is \begin{equation} {T}(t)=\frac{{R}'(t)}{\left|\left| {R}'(t)\right|\right| }=\frac{1}{2}\left[(1-\cos t){i}+(\sin t){j}+\left(2 \cos \frac{t}{2}\right){k}\right].\end{equation} Then, \begin{equation} {T}'(t)=\frac{1}{2}\left[(\sin t){i}+(\cos t){j}-\left(\sin \frac{t}{2}\right){k}\right] \end{equation} because $\frac{d {T}}{d s}=\frac{{T}(t)}{d s / d t}$ we have \begin{equation} \frac{d {T}}{d s}=\frac{1}{4}\left[(\sin t){i}+(\cos t){j}-\left(\sin ^2\frac{t}{2}\right){k}\right]\end{equation} Therefore the curvature is \begin{equation} \kappa (t)=\left|\left| \frac{d {T}}{d s}\right|\right| =\frac{1}{4}\sqrt{\sin ^2t+\cos ^2t+\sin ^2\frac{t}{2}}=\frac{1}{4}\sqrt{1+\sin ^2\frac{t}{2}}. \end{equation} as a function of $t.$

Theorem. (Curvature in Vector Form) Suppose the smooth curve $C$ is the graph of the vector function ${R}(t).$ Then the curvature is given by \begin{equation} \label{curform} \kappa(t) =\frac{\left|\left| {R}'(t) \times {R}”(t)\right|\right| }{\left|\left| {R}'(t)\right|\right| ^3}. \end{equation}

Proof. Given two vector functions ${R}(t)$ and ${R}(s)$ with the same smooth graph $C$ and where ${R}(s)$ is a parametrization vector function in terms of the arc length of $C$, then according to the rule \begin{equation} \frac{d {T}}{d t}=\frac{d {T}}{d s}\frac{d s}{d t}\end{equation} we have \begin{equation} \kappa(t) =\left|\left| \frac{d {T}}{ds}\right|\right| =\frac{\left|\left| \frac{d {T}}{d t} \right|\right| }{\left|\left| {R}'(t)\right|\right| }. \end{equation} Therefore, \begin{align} \kappa (t) &=\frac{ \left|\left| {T}'(t)\right|\right| }{ \left|\left| {R}'(t)\right|\right| } =\frac{ \left|\left| {T}'(t)\right|\right| \left|\left| {R}'(t)\right|\right| ^2}{\left|\left| {R}'(t)\right|\right| ^3} =\frac{\left|\left| {T}'(t) \right|\right| \left(\frac{d s}{d t}\right)^2}{\left|\left|{R}'(t)\right|\right| ^3} \\ & =\frac{\left|\left| {T}'(t)\right|\right| \left|\left| {T} (t)\right|\right| \left(\frac{d s}{d t}\right)^2}{\left|\left| {R}'(t)\right|\right| ^3} =\frac{\left|\left| {T}'(t)\right|\right| \left|\left| {T} (t)\right|\right| \sin \left(\frac{\pi }{2}\right) \left(\frac{d s}{d t}\right)^2}{\left|\left| {R}'(t)\right|\right| ^3} \\ & =\frac{\left|\left| {T} (t)\times {T}’ (t)\right|\right| \\ \left(\frac{d s}{d t}\right)^2}{\left|\left| {R}'(t)\right|\right| ^3}. \end{align} Since \begin{equation} {R}’ (t)=\left|\left| {R}’ (t)\right|\right| {T} (t)=\left(\frac{d s}{d t}\right){T} (t)\end{equation} and \begin{equation} {R}” (t)=\frac{d}{d t}\left(\frac{d s}{d t}\right){T} (t)+\frac{d s}{d t}{T}’ (t)=\frac{d^2s}{d t^2}{T} (t)+\frac{d s}{d t}{T}’ (t)\end{equation} we have, \begin{align} {R}'(t)\times {R}”(t)&=\left(\frac{d s}{d t}\right){T}(t)\times \left(\frac{d^2s}{d t^2}{T}(t)+\frac{d s}{d t}{T}'(t)\right) \\ & =\left(\frac{d s}{d t}\right)\left(\frac{d^2 s}{d t^2}\right)({T}(t)\times {T}(t))+\left(\frac{d s}{d t}\right)^2({T}(t)\times {T}'(t)) \\ & =\left(\frac{d s}{d t}\right)^2({T}(t)\times {T}'(t)). \end{align} Therefore, \begin{equation} \left|\left| {R}'(t)\times {R}”(t)\right|\right| =\left(\frac{d s}{d t}\right)^2\left|\left| {T}(t)\times {T}'(t)\right|\right| . \end{equation} Finally $ \kappa (t)=\frac{\left|\left| {R}’ (t) \times {R}” (t)\right|\right| }{\left|\left| {R}'(t)\right|\right| ^3} $ as desired.

Example. Given the curve defined by $$ {R}(t)=\sin t {i}+\cos t {j}+t {k} $$ find a unit tangent vector ${T}$ at the point on the curve where $t=\pi $, the curvature at $t=\pi $, and find the length of the curve from $t=0$ to $t=\pi .$

Solution. We have $ {R}'(t)=\cos t {i}-\sin t {j}+{k},$ ${R}”(t)=-\sin t {i} -\cos t {j},$ and \begin{align} \left|\left| {R}'(t)\times {R}”(t) \right|\right| & =\left|\left| \begin{array}{ccc} {i} & {j} & {k} \\ \cos t & -\sin t & 1 \\ -\sin t & -\cos t & 0 \end{array} \right|\right| \\ & =\left|\left| (\cos t){i}+(-\text{sint}) {j}+\left(-\cos ^2t-\sin ^2t\right){k}\right|\right| \\ & =\sqrt{2}. \end{align} Therefore the curvature is $\kappa =\frac{\sqrt{2}}{\left(\sqrt{2}\right)^3}=\frac{1}{2},$ and also \begin{equation} {T}=\left(\frac{1}{\sqrt{2}}\cos t\right) {i}+\left(-\frac{1}{\sqrt{2}}\sin t\right){j}+\left(\frac{1}{\sqrt{2}}\right){k} \end{equation} and $ \int _0^{\pi }\left|\left| {R}’\right|\right| \, dt=\int _0^{\pi }\sqrt{2}\, dt=\sqrt{2}\pi . $

Maximum Curvature

Now that we know that the curvature formula in \eqref{curform} is a one variable function we can apply one variable calculus to find maximum curvature. We will derive two more curvature formulas, one for planar functional form \eqref{pfc} and another for parametric form \eqref{pafc}. There is a fourth formula (for polar form) which we leave for the reader to explore.
For example, consider the cardioid given by equation $r=2(1+\cos \theta)$ with derivative $dr/d\theta=-2\sin\theta.$ We can use this information to determine where the curvature is maximum. From the graph can you tell at which points will the curvature be maximum?

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Theorem. (Curvature in Planar Form) The graph $C$ of the function $y=f(x)$ has curvature \begin{equation} \label{pfc} \kappa(x) =\frac{|f”(x)|}{\left(1+[f'(x)]^2\right)^{3/2}} \end{equation} where $f(x),$ $f'(x),$ and $f”(x)$ all exist.

Proof. Given the vector function $x(t)=t$ and $y(t)=f(t)$ defined by ${y=f(x)}$ we have ${R}(t)=t {i}+f(t) {j}.$ Using the formula \begin{equation} \kappa (t) =\frac{ \left|\left| {R}'(t) \times {R}”(t) \right|\right| }{ \left|\left| {R}'(t) \right|\right|^3 }, \end{equation} we have \begin{equation} \kappa (t) =\frac{\left|\left| \, \left|\begin{array}{ccc} {i} & {j} &{k} \\ 1 & f'(t) & 0 \\ 0 & f”(t) & 0 \end{array} \right|\, \right|\right| }{\left|\left| {R}'(t) \right|\right| ^3} = \frac{|f”(t)|}{\left(\sqrt{1+[f”(t)]^2}\right)^3}. \end{equation} as desired.

Example. Find the maximum curvature for the graph of $y=\ln x.$

Solution. We have $y’=\frac{1}{x},$ $y”=\frac{-1}{x^2},$ and \begin{equation} \kappa (x) =\frac{\frac{1}{x^2}}{\left(1+\frac{1}{x^2}\right)^{3/2}}=\frac{1}{\left(1+\frac{1}{x^2}\right)^{3/2} x^2}\end{equation} To maximize the curvature we find the first derivative \begin{equation}
\frac{d \kappa }{d x}=\frac{3}{\left(1+\frac{1}{x^2}\right)^{5/2} x^5}-\frac{2}{\left(1+\frac{1}{x^2}\right)^{3/2} x^3}=\frac{1-2 x^2}{\sqrt{1+\frac{1}{x^2}} x \left(x^2+1\right)^2} \end{equation} Applying the first derivative test we have the maximum curvature at $x=\frac{1}{\sqrt{2}}$ with curvature of $\kappa \left(\frac{1}{\sqrt{2}}\right)=\frac{2}{3 \sqrt{3}}.$

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Example. Find the maximum curvature for the graph of $y=e^{2 x}.$

Solution. We have $y’=2e^{2x},$ $y\text{”}=4e^{2x},$ and \begin{equation} \kappa =\frac{|y”|}{\left[1+(y’)^2\right]^{3/2}}=\frac{4e^{2x}}{\left(1+4e^{4x}\right)^{3/2}}\end{equation} To maximize the curvature we find the first derivative \begin{equation} \frac{d \kappa }{d x}=-\frac{96 e^{6 x}}{\left(1+4 e^{4 x}\right)^{5/2}}+\frac{8 e^{2 x}}{\left(1+4 e^{4 x}\right)^{3/2}}=-\frac{8 e^{2 x} \left(-1+8 e^{4 x}\right)}{\left(1+4 e^{4 x}\right)^{5/2}} \end{equation} Applying the first derivative test we have the maximum curvature at $x=\frac{-3}{4}\ln2$ with curvature of \begin{equation}
\kappa \left(\frac{-3}{4}\ln2\right)=\frac{4\sqrt{3}}{9}. \end{equation}

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Theorem. (Curvature in Parametric Form) If $C$ is a smooth curve in $\mathbb{R}^2$ described by the parametric equations $x(t)$ and $y(t),$ then the curvature is given by \begin{equation} \label{pafc} \kappa =\frac{|x’y”-y’x”|}{\left[(x’)^2+(y’)^2\right]^{3/2}}. \end{equation}

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Example. Find the point(s) where the ellipse $9x^2+4y^2=36$ has maximum curvature.

Solution. Write the ellipse as $$ \frac{x^2}{2^2}+\frac{y^2}{3^2}=1 $$ and so we parametrize using $x(t)=2 \cos t$ and $y(t)=3 \sin t.$ Then we find $$ x'(t)=-2 \sin t, \quad y'(t)=3 \cos t, \quad x”(t)=-2 \cos t , \quad y”(t)=-3 \sin t. $$ Now using Curvature in Parametric Form, we find \begin{align} \kappa & =\frac{|x’y”-y’x”|}{\left[(x’)^2+(y’)^2\right]^{3/2}} \\ & =\frac{|(-2 \sin t )(-3 \sin t)-(3 \cos t)(-2 \cos t)|}{\left[(-2 \sin t)^2+(3 \cos t)^2\right]^{3/2}} \\ & =\frac{6}{\left(4 \sin ^2t+9\cos ^2t\right)^{3/2}} \\ & = 6\left(4+5\cos ^2t\right)^{-3/2} \end{align}

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Then we apply the first derivative test, using $$ \frac{d\kappa }{dt}=-9\left(4+5 \cos ^2t\right)^{-5/2}(-10\cos t \sin t)=0 $$ when $t=0, \pi/2 ,\pi , 3\pi /2, 2\pi.$ The point(s) where the ellipse $9x^2+4y^2=36$ has maximum curvature are $(0,\pm 3)$ from when $t=\pi/2$ and $t=3\pi/2$ with curvature $\kappa =3/4.$

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Exercises on Arc Length and Curvature

Exercise. Find the length of the given curve over the given interval.

$(1) \quad {R}(t)=t {i}+3 t {j}$ over the interval $[0,4]$

$(2) \quad {R}(t)=t {i}+2t {j}+3 t {k}$ over the interval $[0,2]$

$(3) \quad {R}(t)=\cos ^3t {i}+\cos ^2t {k}$ over the interval $\left[0,\frac{\pi }{2}\right]$

Exercise. Express the following vector functions in terms of the arc length parameter $s$ measured from the point where $t=0$ in the direction of increasing $t.$

$(1) \quad {R}(t)=\langle \sin t,\cos t \rangle$ at the point where $t=0$

$(2) \quad {R}(t)=\langle 3 \cos t+3 t \sin t ,2t^2,3 \sin t-3 t \cos t\rangle$ at the point where $t=0$

$(3) \quad {R}(t)=[\ln (\sin t)]{i}+[\ln (\cos t)]{j}$ at the point where $t=\pi /3$

Exercise. Find the length of the given curve described by the vector function ${R}(t)=3 t {i}+(3 \cos t){j} +(3 \sin t){k}$ over the interval $\left[0,\frac{\pi }{2}\right].$

Exercise. Determine $a$ so that the length of the given curve described by the vector function ${R}(t)=a \cos t {i}+(a \sin t){j} +(5 t){k}$ has length $\sqrt{41}\pi $ over the interval $[0,\pi ].$

Exercise. Express the vector function ${R}(t)=\langle 2+3t,1-t,-4t-9\rangle$ in terms of the arc length parameter $s$ measured from the point where $t=0$ in the direction of increasing $t.$

Exercise. Determine $a$ so that the vector function ${R}(t)=\langle 2-3t,1+t,-4t\rangle$ written in terms of the arc length parameter $s$ measured from the point where $t=0$ in the direction of increasing $t$ is \begin{equation} {R}(s)=\left\langle 2-\frac{3s}{a},1+\frac{s}{a},\frac{-4s}{a}\right\rangle. \end{equation}

Exercise. Find the curvature.${R}(t)=(2t+3){i}+(5-t^2)j$

$(1) \quad {R}(t)=t{i} +\cos t{j}$

$(2) \quad {R}(t)=t{i} + e^t{j}$

$(3) \quad {R}(t)=3\cos t {i} + 2\sin t{j}$

$(4) \quad {R}(t)=(2t-1){i}+(t-3){j}+(2-3t){k}$, at $(1,-2,-1)$

$(5) \quad {R}(t)=t{i}+\ln t{j}+0.5 t{k}$, at $(2,\ln 2,1)$

$(6) \quad {R}(t)=\sin t{i}+\cos t{j}+2t{k}$, at $(1,0,\pi)$

$(7) \quad {R}(t)=(\cos t+t\sin t){i}+(\sin t-t\cos t){j}+3{k}$

$(8) \quad {R}(t)=(e^t\sin t){i}+(e^t \cos t){j}+2{k}$

$(9) \quad {R}(t)=6 \sin 2t {i}+6 \cos 2t {j}+5t{k}$

Exercise. Find the curvature of the plane curve at the given location.

$(1) \quad y=x-\frac{1}{9}x^2$, $x=3$

$(2) \quad y=\sin x$, $x=\frac{\pi }{2}$

$(3) \quad y=\ln x$, $x=1$

$(4) \quad y=1/\ln x$, $x=2$

$(5) \quad y=x^3-3x^2+2$, $x=2$

$(6) \quad y=e^{x^2}$, $x=0$

$(7) \quad y=\sec x$, $\pi/3$

$(8) \quad 1/x^2$, $x=2$

$(9) \quad y=\cos 2x$, $0$

$(10) \quad y=x^4-x^2+3$, $x=-1$

Exercise. Find the curvature of the plane curve at the given location.

$(1) \quad y=\ln (\cos x)$, $-\pi/2 < x <\pi /2$

$(2) \quad x=t-1, y=\sqrt{t}$, $(x,y)=(3,2)$

$(3) \quad x=t-t^2, y=1-t^3$, $(x,y)=(0,1)$

$(4) \quad x=1-\sin t, y=2+\cos t $, $(x,y)=(1,3)$

$(5) \quad x=\cos^3 t, y=\sin^3 t$, $(x,y)=(\sqrt{2}/4,\sqrt{2}/4)$

Exercise. Given the vector function \begin{equation} {R}(t)=\sin t {i}+ \cos t {j}+t {k} \end{equation} and $t_0=\pi ,$ find the unit tangent vector and the curvature at $t_0.$

Exercise. Find all points on the curve at which the curvature is zero.

$(1) \quad y=\cot x$

$(2) \quad y=e^{-x^2}$

$(3) \quad y=x^4-12x^2$

$(4) \quad {R}(t)=t^2{i}-t^2{j}$

$(5) \quad {R}(t)=e^x{i}+\ln x{j}$

$(6) \quad {R}(t)=e^{-x}{i}+e^x{j}$

Exercise. Find the point(s) where the ellipse $36x^2+9y^2=121$ has maximum curvature.

Exercise. Find the points on the curve in the $xy$-plane described by $y=3x-x^3$ at which the curvature is maximum.

Exercise. Show that the curvature of the parabola $y=x^2+3$ approaches 0 as $x\to \infty .$

Exercise. Show that the curvature at every point on a line is 0.

Exercise. Show that the maximum curvature of a parabola occurs at the vertex.

Exercise. Show that for an ellipse, the maximum and minim curvature occurs at ends of the major and minor axes, respectively.

Exercise. Show that for a hyperbola, the maximum curvature occurs at the ends of the traverse axis.

Exercise. Show that the curvature of the helix \begin{equation} {R}(t)= a\cos t {i}+a \sin t {j}+b t \end{equation} is $\kappa=a/(a^2+b^2).$ What is the largest value that $\kappa$ can have for a given value of $b$?

Exercise. Suppose a curve $C$ in the $xy$-plane described by a polar equation $r=f(\theta).$ Show that the curvature of $C$ is \begin{equation} \kappa(\theta) = \frac{| \, \, [f(\theta)]^2+2[f'(\theta)]^2-f(\theta)f”(\theta)\, \, |}{\, {[f(\theta)]^2+[f'(\theta)]^2\, }^{3/2}}. \end{equation}

Exercise. Find the curvature of the polar curve at $P(r,\theta).$

$(1) \quad $ the cardoid $r=a(1-\cos \theta)$, where $0< \theta<2\pi.$

$(2) \quad $ the four-leafed rose $r=\sin 2\theta$, where $0<\theta < \pi/2$

$(3) \quad $ the spiral $r=e^{a\theta}$

$(4) \quad r=1-\cos \theta$, $\theta=3\pi/2$

$(5) \quad r=1-\sin\theta -\cos\theta$, $\theta=2\pi$

Exercise. Show that lines and circles are the only planes curves that have constant curvature. Is this true for space curves?

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