OBJECTIVES

When you finish this section, you should be able to:

1. Determine whether the parameter used in a vector function is arc length (p. 775)
2. Find the curvature of a curve (p. 776)
3. Find the curvature of a space curve (p. 778)
4. Find the curvature of a plane curve given by \({y=f( x)}\) (p. 779)
5. Find an osculating circle (p. 780)

THEOREM Criterion for Arc Length as Parameter

Suppose a smooth curve \(C\) is traced out by the vector function \begin{equation*} \mathbf{r}=\mathbf{r}(t)\quad a\leq t\leq b \end{equation*}

The parameter \(t\) is the arc length along \(C\) if and only if \begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED] {\left\Vert \mathbf{r}^{\prime} (t)\right\Vert\;=\;1\quad \hbox{for all}\quad t} \end{equation*}

Proof

We use equation (1). If \(\left\Vert \mathbf{r}^{\prime} (t)\right\Vert\;=\;1\) for all \(t,\) then the arc length \(s\) is \begin{equation*} s(t)=\int_{a}^{t}du=t-a \end{equation*}

That is, the parameter \(t\) is the arc length \(s\) measured along \(C\) from \(t = a\) to \(t= b\). Notice that when \(t=a,\) the length \(s\) is zero, as expected.

Conversely, if the parameter \(t\) is arc length \(s\), then, by differentiating (1), we obtain \[ \frac{ds}{dt}=1=\;\left\Vert \mathbf{r}^{\prime }(t)\right\Vert\qquad \hbox{for all}\quad t \]

Determining Whether the Parameter Used in a Vector Function Is Arc Length

Determine whether the parameter used in each vector function is arc length.

(a) \(C_{1}\): \(\ \mathbf{r}(t)=2\sin \dfrac{t}{2}\mathbf{i}+2\cos \dfrac{t}{2}\mathbf{j}\quad 0\leq t\leq 2\pi \)

(b) \(C_{2}\): \(\ \mathbf{r}(t)=\cos t\mathbf{i}+\sin t\mathbf{j}+t\mathbf{k}\quad t\geq 0\)

Solution (a) We begin by finding \(\mathbf{r}^{\prime} ( t)\) and \(\left\Vert \mathbf{r^{\prime} }( t) \right\Vert .\) \begin{equation*} \begin{array}{rrr} \mathbf{r}^{\prime} (t)=\cos \dfrac{t}{2}\mathbf{i}-\sin \dfrac{t}{2}\mathbf{j} \qquad \left\Vert \mathbf{r}^{\prime} (t)\right\Vert\;=\;\sqrt{ \cos ^{2}\dfrac{t}{2}+\sin ^{2}\dfrac{t}{2}}=1\quad \hbox{for all}~t \end{array} \end{equation*}

Since \(\left\Vert \mathbf{r}^{\prime} (t)\right\Vert\;=\;1\) for all \(t,\) the parameter \(t\) is arc length as measured along \(C_{1}\).

(b) We begin by finding \(\mathbf{r}^{\prime} ( t)\) and \(\left\Vert \mathbf{r^{\prime} }( t) \right\Vert .\) \begin{equation*} \begin{array}{rrr} \mathbf{r}^{\prime} (t)=-\sin t\mathbf{i}+\cos t\mathbf{j+k}\qquad \left\Vert \mathbf{r}^{\prime} (t)\right\Vert\;=\;\sqrt{\sin ^{2}t+\cos ^{2}t+1}=\sqrt{2} \end{array} \end{equation*}

Since \(\left\Vert \mathbf{r}^{\prime }(t)\right\Vert\;\neq 1\) for all \(t\), the parameter \(t\) does not measure arc length along \(C_{2}\).

NOW WORK

Problem 13.

Changing the Parameter to Arc Length

Represent the helix in Example 1(b). \begin{equation*} \mathbf{r}(t)=\cos t\mathbf{i}+\sin t\mathbf{j}+t\mathbf{k}\quad t\geq 0 \end{equation*}

so the parameter is arc length.

Solution The arc length function \(s( t)\) along the curve \(\mathbf{r}=\mathbf{r}( t)\), \(a\leq t\leq b\), from \(t=a\) to an arbitrary \(t\) is given by the integral \begin{equation*} s( t)\;=\;\int_{a}^{t}\left\Vert \mathbf{r}^{\prime} (u)\right\Vert du \end{equation*}

Since the graph of the helix starts at \(t=0,\) and \(\left\Vert \mathbf{r}^{\prime}(t)\right\Vert\;=\;\sqrt{2}\) for all \(t\) (Example 1(b)), we have \begin{equation*} s( t)\;=\;\int_{0}^{t}\sqrt{2}\kern.7ptdu=\Big[ \sqrt{2}\kern.7ptu\Big] _{0}^{t}= \sqrt{2}\kern.7pt t \end{equation*}

Then \(t=\dfrac{s}{\sqrt{2}}.\) The helix using arc length \(s\) as the parameter is expressed as \begin{equation*} \mathbf{r}( s)\;=\;\cos \dfrac{s}{\sqrt{2}}\kern.7pt\mathbf{i}+\sin \dfrac{s}{ \sqrt{2}}\kern.7pt\mathbf{j}+\dfrac{s}{\sqrt{2}}\kern.7pt\mathbf{k}\quad s\geq 0 \end{equation*}

NOTE

\(k\) is the Greek letter kappa.

spanDEFINITIONspan Curvature

Suppose a smooth curve \(C\) is traced out by a twice differentiable vector function \(\mathbf{r}=\mathbf{r}(s)\), where the parameter is the arc length \(s\) along \(C\). The curvature \(\kappa\;=\;\kappa (s)\) at a point \(P\) on \(C\) is \begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED] {\kappa\;=\;\left\Vert \dfrac{d\mathbf{T}}{ds}\right\Vert} \end{equation*}

IN WORDS

Curvature is measured by the magnitude of the rate of change of the unit tangent vector \(\mathbf{T}\) with respect to arc length.

Finding the Curvature of a Line

Show that the curvature of a line is \(0\).

Solution For a line, the unit tangent vector \(\mathbf{T}\) is along the constant direction of the line, so \(\mathbf{T}\) is a constant and \(\dfrac{d\mathbf{T}}{ds}=\mathbf{0}\). The curvature \(\kappa\;=\;\left\Vert \dfrac{d\mathbf{T}}{ds}\right\Vert\) of a line is \(0\).

Finding the Curvature of a Circle

Find the curvature of a circle of radius \(R.\)

Solution For a circle of radius \(R,\) \begin{equation*} \mathbf{r}=\mathbf{r}( t)\;=\;R\;\cos t\mathbf{i}+R\;\sin t\mathbf{j}\qquad 0\leq t\leq 2\pi \end{equation*}

Then we find \(\mathbf{r}^{\prime} (t)\) and \(\left\Vert \mathbf{r^{\prime} }(t) \right\Vert.\) \begin{equation*} \begin{array}{rrr} \mathbf{r}^{\prime} ( t)\;=\;-R\;\sin t\mathbf{i}+R\;\cos t\mathbf{j}\qquad \Vert \mathbf{r}^{\prime} (t) \Vert\;=\;\sqrt{R^{2}\sin ^{2}t+R^{2}\cos ^{2}t}=R \end{array} \end{equation*}

NOTE

The larger a circle, the smaller its curvature. As \(R\rightarrow \infty\), then \(\kappa \rightarrow 0\). and the circle will look more and more like a line.

NOW WORK

Problem 21.

THEOREM Curvature of a Space Curve

If a smooth curve \(C\) in space is traced out by a twice differentiable vector function \(\mathbf{r}=\mathbf{r}(t),\) then its curvature \(\kappa\) is given by the formula \begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED] {\kappa\;=\;\dfrac{\left\Vert \mathbf{r}^{\prime} (t)\times \mathbf{r}^{\prime \prime} (t)\right\Vert }{\left\Vert \mathbf{r}^{\prime} (t) \right\Vert ^{3}}} \tag{4} \end{equation*}

NEED TO REVIEW?

The cross product and its properties are discussed in Section 10.5, pp. 724-729.

Proof

Since \(\mathbf{T}(t)\;=\;\dfrac{\mathbf{r}^{\prime} (t)}{ \left\Vert \mathbf{r}^{\prime} (t)\right\Vert }\) and \( \left\Vert \mathbf{r} ^{\prime} (t)\right\Vert\;=\;\dfrac{ds}{dt}\) [from (2) on p. 775], then \begin{equation} \mathbf{r}^{\prime} (t)\;=\;\left\Vert \mathbf{r}^{\prime} (t)\right\Vert\kern1pt\mathbf{T} (t)\;=\frac{ds}{dt}\kern1pt\mathbf{T}(t) \tag{5} \end{equation}

Now we use the scalar multiplication formula to find \(\mathbf{r^{\prime\prime}} (t)\). \begin{equation} \mathbf{r}^{\prime \prime} (t)=\dfrac{d}{dt}\kern1pt\mathbf{r}^{\prime} (t)=\frac{d}{dt} \left( \frac{ds}{dt}\kern1pt\mathbf{T}(t)\right)\;=\;\frac{d^{2}s}{dt^{2}}\kern1pt\mathbf{T} (t)+\frac{ds}{dt}\kern1pt\mathbf{T}^{\prime} (t) \tag{6} \end{equation}

Using (5) and (6), we get \begin{equation*} \mathbf{r}^{\prime} (t) \times \mathbf{r}^{\prime \prime} (t)\;=\;\dfrac{ds}{dt}\kern1pt\mathbf{T}\times \left( \frac{d^{2}s}{dt^{2}}\kern1pt \mathbf{T}+\frac{ds}{dt}\kern1pt\mathbf{T}^{\prime} \right)\;=\;\dfrac{ds}{dt} \frac{d^{2}s}{dt^{2}} \left( \mathbf{T}\times \mathbf{T}\right) +\left( \dfrac{ds}{dt}\right) ^{2}\left( \mathbf{T}\times \mathbf{T}^{\prime} \right) \end{equation*}

Since \(\mathbf{T}\times \mathbf{T}=\mathbf{0}\), this reduces to \begin{equation} \mathbf{r}^{\prime} (t) \times \mathbf{r}^{\prime \prime} (t)\;=\;\left( \dfrac{ds}{dt}\right)^{2}\left( \mathbf{T}\times \mathbf{T}^{\prime} \right) \tag{7} \end{equation}

Since \(\mathbf{T^{\prime} }\) is parallel to \(\mathbf{N}\) and \(\mathbf{T}\) is orthogonal to \(\mathbf{N}\), it follows that \(\mathbf{T}\) and \(\mathbf{T}^{\prime}\) are orthogonal. This means \(\left\Vert \mathbf{T}\times \mathbf{T}^{\prime} \right\Vert\;=\;\left\Vert \mathbf{T}\right\Vert \left\Vert \mathbf{T}^{\prime} \right\Vert \sin \dfrac{\pi }{2}\;=\left\Vert \mathbf{T}^{\prime} \right\Vert\), so \begin{equation*} \left\Vert \mathbf{r}^{\prime} (t) \times \mathbf{r}^{\prime \prime} (t) \right\Vert\;=\;\left( \dfrac{ds}{dt}\right) ^{2}\left\Vert \mathbf{T}^{\prime} \right\Vert \end{equation*}

Now we solve the above equation for \(\left\Vert \mathbf{T}^{\prime} \right\Vert\) and use the fact that \(\dfrac{ds}{dt}=\;\left\Vert \mathbf{r}^{\prime} (t) \right\Vert.\) Then \begin{equation} \left\Vert \mathbf{T}^{\prime} \right\Vert\;=\;\dfrac{\left\Vert \mathbf{r}^{\prime} (t) \times \mathbf{r}^{\prime \prime} (t) \right\Vert }{ \left\Vert \mathbf{r}^{\prime} (t) \right\Vert ^{2}} \tag{8} \end{equation}

From equation (3), the curvature \(\kappa\) is \begin{equation} \kappa \underset{\underset{\color{#0066A7}{\hbox{\((3)\)}}}{\color{#0066A7}{\uparrow}}}{=}\dfrac{\left\Vert \mathbf{T}^{\prime} \right\Vert }{\left\Vert \mathbf{r}^{\prime} (t) \right\Vert } \underset{\underset{\color{#0066A7}{\hbox{\((8)\)}}}{\color{#0066A7}{\uparrow}}}{=}\dfrac{\left\Vert \mathbf{r}^{\prime} (t) \times \mathbf{r}^{\prime \prime} (t) \right\Vert }{\left\Vert \mathbf{r}^{\prime} (t) \right\Vert ^{3}} \notag \end{equation}

Finding the Curvature of a Space Curve

Find the curvature \(\kappa\) of the space curve \(\mathbf{r}(t)=t\mathbf{i}+t^{2}\mathbf{j}+t^{3}\mathbf{k}\).

Solution To use formula (4), we need to find \(\mathbf{r}^{\prime} (t),\) \(\left\Vert \mathbf{r}^{\prime} (t) \right\Vert,\) \(\mathbf{r}^{\prime \prime} (t),\) \(\mathbf{r}^{\prime} (t) \times \mathbf{r}^{\prime \prime} (t),\) and \(\left\Vert \mathbf{r}^{\prime} (t) \times \mathbf{r}^{\prime \prime} (t) \right\Vert.\) \begin{eqnarray*} \mathbf{r}^{\prime} (t) &=&\mathbf{i}+2t\mathbf{j}+3t^{2}\mathbf{k}\qquad \left\Vert \mathbf{r}^{\prime} (t)\right\Vert\;=\;\sqrt{ 1+4t^{2}+9t^{4}}\\ \mathbf{r}^{\prime \prime} (t) &=&2\mathbf{j}+6t\mathbf{k} \end{eqnarray*} \begin{eqnarray*} \mathbf{r}^{\prime} (t)\times \mathbf{r}^{\prime \prime} (t)&\;=\;&\left\vert \begin{array}{c@{\quad}c@{\quad}c} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[3pt] 1 & 2t & 3t^{2}\\ 0 & 2 & 6t \end{array} \right\vert\;=\;6t^{2}\mathbf{i}-6t\mathbf{j}+2\mathbf{k} \\[6pt] \left\Vert \mathbf{r}^{\prime} (t)\times \mathbf{r}^{\prime \prime} (t)\right\Vert &\;=\;&\sqrt{36t^{4}+36t^{2}+4}=2\sqrt{9t^{4}+9t^{2}+1} \end{eqnarray*}

Now we use formula (4). The curvature \(\kappa\) is \begin{equation*} \kappa\;=\;\dfrac{\left\Vert \mathbf{r}^{\prime} (t)\times \mathbf{r}^{\prime \prime} (t)\right\Vert }{\left\Vert \mathbf{r}^{\prime} (t) \right\Vert ^{3}}=\frac{2\sqrt{ 9t^{4}+9t^{2}+1}}{(1+4t^{2}+9t^{4})^{3/2}} \end{equation*}

NOW WORK

Problem 31.

Finding the Curvature of Plane Curves

(a) Find an equation of the tangent line to the graph of the parabola \(f(x)\;=\;\dfrac{1}{4}x^{2}\) at the point \((2,1).\)

(b) What is the curvature of the graph of \(f\) at the point \((2,1)?\)

(c) Find an equation of the tangent line to the graph of \(g(x)\;=\;\dfrac{x^{3}+4}{12}\) at the point \((2,1).\)

(d) What is the curvature of the graph of \(g\) at the point \(( 2,1)?\)

(e) Graph both curves and their tangent lines.

Solution (a) \(f(x)\;=\;\dfrac{1}{4}x^{2};\) \(f^{\prime} (x)\;=\;\dfrac{1}{2}x\). An equation of the tangent line to the graph of \(f\) at the point \(( 2,1)\) is \begin{eqnarray*} y-1 &=&1\,{\cdot}\, (x-2)\qquad \color{#0066A7}{f^{\prime} (2)\;=\;{1}} \\[4pt] y &=&x-1 \end{eqnarray*}

(b) \(f^{\prime \prime} (x)\;=\;\dfrac{1}{2};\quad f^{\prime} (2)\;=\;1;\quad f^{\prime \prime} (2)\;=\;\dfrac{1}{2}\). We use formula (9) and evaluate \(\kappa\) at \(x=2\). \begin{equation*} \kappa\;=\;\dfrac{\left\vert f^{\prime \prime} ( 2) \right\vert }{ \left( 1+\left[ f^{\prime} ( 2) \right] ^{2}\right) ^{3/2}}=\dfrac{ \dfrac{1}{2}}{( 1+1) ^{3/2}}=\dfrac{1}{2\,{\cdot}\, 2\sqrt{2}}=\dfrac{1 }{4\sqrt{2}}\approx 0.177 \end{equation*}

(c) \(g(x)\;=\;\dfrac{x^{3}+4}{12};\quad g^{\prime} (x)\;=\;\dfrac{3x^{2}}{12}=\dfrac{x^{2}}{4}\). An equation of the tangent line to the graph of \(g\) at the point \((2,1)\) is \begin{eqnarray*} y-1 &=&1\,{\cdot}\, ( x-2)\qquad \color{#0066A7}{g^{\prime} (2)\;=\;{1}} \\[4pt] y &=&x-1 \end{eqnarray*}

Figure 20

(d) \(g^{\prime \prime} (x)\;=\;\dfrac{x}{2};\quad g^{\prime} (2)\;=\;1;\quad g^{\prime \prime} (2)\;=\;1\). The curvature \(\kappa\) at the point \((2,1)\) is \begin{equation*} \kappa\;=\;\dfrac{\vert g^{\prime \prime} ( 2) \vert }{ ( 1+[ g^{\prime} ( 2) ] ^{2}) ^{3/2}}=\dfrac{1 }{( 1+1) ^{3/2}}=\dfrac{1}{2\sqrt{2}}\approx 0.354 \end{equation*}

(e) The functions \(f\) and \(g\) are graphed in Figure 20. Both graphs have the same tangent line at \((2,1),\) but the tangent line to the graph of \(y=\dfrac{1}{4}x^{2}\) at \((2,1)\) turns more slowly \((\kappa \approx 0.177)\) than the tangent line to the graph of \(y=\dfrac{x^{3}+4}{12}\) at \((2,1)\) \((\kappa \approx 0.354)\).

NOW WORK

Problem 39.

NOTE

The term osculating means “kissing.”

NOTE

The osculating circle at a point \(P\) is the limiting circle that results from the circle containing three points \(P,Q\), and \(R\) on the curve \(C\) as the points \(Q\) and \(R\) are made to approach point \(P\).

Finding an Osculating Circle for an Ellipse

Find the radius of the osculating circle of the ellipse traced out by the vector function \begin{equation*} \mathbf{r}(t)=3\sin t\mathbf{i}+4\cos t\mathbf{j} \qquad 0\leq t\leq 2\pi \end{equation*}

(a) At \(t=0\)

(b) At \(t=\dfrac{\pi }{2}\)

Solution To find the radius of the osculating circle, we need to find the curvature \(\kappa\). We treat the ellipse as a curve in space with a \(\mathbf{k}\) component of \(0\) and find \(\mathbf{r}^{\prime} (t),\) \(\left\Vert \mathbf{r}^{\prime} (t) \right\Vert, \mathbf{r}^{\prime \prime} (t),\) \(\mathbf{r}^{\prime} (t) \times \mathbf{r}^{\prime \prime} (t),\) and \(\left\Vert \mathbf{r}^{\prime} (t) \times \mathbf{r}^{\prime \prime} (t) \right\Vert\). \begin{eqnarray*} \mathbf{r^{\prime} }(t) &=& 3\cos t\mathbf{i}-4\sin t\mathbf{j}+0\mathbf{k} \qquad \left\Vert \mathbf{r^{\prime} }(t) \right\Vert\;=\; \sqrt{9\cos ^{2}t+16\sin ^{2}t}\\[9pt] \mathbf{r^{\prime \prime} }(t) &=& -3\sin t\mathbf{i}-4\cos t\mathbf{j}+0\mathbf{k} \end{eqnarray*}

\begin{eqnarray*} \mathbf{r^{\prime} }(t)\times \mathbf{r^{\prime \prime} }(t)= \left|\begin{array}{c@{\quad}c@{\quad}c} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[3pt] 3\cos t & -4\sin t & 0 \\[3pt] -3\sin t & -4\cos t & 0 \end{array}\right| &=&( -12\cos ^{2}t-12\sin ^{2}t) \mathbf{k}=-12\mathbf{k}\\[9pt] \left\Vert \mathbf{r}^{\prime} (t) \times \mathbf{r} ^{\prime \prime} (t) \right\Vert &=&12 \end{eqnarray*}

Using formula (4), the curvature \(\kappa\) of \(C\) is \begin{equation*} \kappa\;=\;\dfrac{\left\Vert \mathbf{r^{\prime} }(t) \times \mathbf{ r^{\prime \prime} }(t) \right\Vert }{\left\Vert \mathbf{r^{\prime} } (t) \right\Vert ^{3}}=\dfrac{12}{( 9\cos ^{2}t+16\sin ^{2}t) ^{3/2}} \end{equation*}

The radius \(\rho\) of the osculating circle is \begin{equation*} \rho\;=\;\frac{1}{\kappa }=\frac{(9\cos ^{2}t+16\sin ^{2}t)^{3/2}}{12} \end{equation*}

(a) At \(t=0\), the curvature of \(C\) is \(\kappa\;=\;\dfrac{4}{9}\), so the radius of the osculating circle is \(\rho\;=\;\dfrac{9}{4}\).

(b) At \(t=\dfrac{\pi }{2},\) the curvature of \(C\) is \(\kappa\;=\;\dfrac{3}{16}\), so the radius of the osculating circle is \(\rho\;=\;\dfrac{16}{3}\).

NOW WORK

Problem 57.