DEFINITION Curvature

Let \({\bf r}(s)\) be an arc length parametrization and \(\TT\) the unit tangent vector. The curvature at \({\bf r}(s)\) is the quantity (denoted by a lowercase Greek letter “kappa”) \[ \boxed{\bbox[#FAF8ED,5pt]{\kappa(s) = \left\|\frac{\TT}s\right\|}} \tag{1} \]

EXAMPLE 1 A Line Has Zero Curvature

Compute the curvature at each point on the line \({\bf r}(t) = \langle x_0,y_0,z_0\rangle + t{\bf u}\), where \(\|{{\bf u}}\|=1\).

Solution First, we note that because \({\bf u}\) is a unit vector, \({\bf r}(t)\) is an arc length parametrization. Indeed, \({\bf r}'(t)={\bf u}\) and thus \(\|{{\bf r}'(t)}\|= \|{{\bf u}}\|=1\). Thus we have \(\TT(t) = {\bf r}'(t)/\|{\bf r}'(t)\| = {\bf r}'(t)\) and hence \(\TT'(t) = {\bf r}”(t)=\mathbf{0}\) (because \({\bf r}'(t) = {\bf u}\) is constant). As expected, the curvature is zero at all points on a line: \[ \kappa(t) = \|{\frac{\TT}t}\| = \|{{\bf r}\|(t)} = 0 \]

EXAMPLE 2 The Curvature of a Circle of Radius \(R\) Is \(1/R\)

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

Solution Assume the circle is centered at the origin, so that it has parametrization \({\bf r}(\theta)= \langle R\cos\theta, R\sin\theta\rangle\) (Figure 13.23). This is not an arc length parametrization if \(R \ne 1\). To find an arc length parametrization, we compute the arc length function: \[ s(\theta)= \int_0^{\theta}\|{{\bf r}'(u)}\|\,du = \int_0^{\theta}R\,du = R\theta \]

Thus \(s=R\theta\), and the inverse of the arc length function is \(\displaystyle{\theta =g(s) = {s}/R}\). In Section 13.3, we showed that \({\bf r}_1(s) = {\bf r}(g(s))\) is an arc length parametrization. In our case, we obtain \[ {\bf r}_1(s) = {\bf r}(g(s)) = {\bf r}\left(\frac{s}R\right) = \langle R\cos \frac{s}R , R\sin\frac{s}R\rangle \]

The unit tangent vector and its derivative are \[ \begin{align*} \TT(s) &= \frac{{\bf r}_1}{s} = \frac{}{s}\langle R \cos \frac{s}{R}, R \sin \frac{s}{R} \rangle =\langle {-\sin \frac{s}R} ,\cos\frac{s}R\rangle\\ \frac{\TT}s &= -\frac1R\langle\cos\frac{s}R, \sin\frac{s}R\rangle \end{align*} \]

By definition of curvature, \[ \begin{align*} \kappa(s) &= \|{\frac{\TT}{s}}\| = \frac1R\|{\langle\cos\frac{s}R, \sin\frac{s}R\rangle}\| = \frac1R \end{align*} \]

This shows that the curvature is \(1/R\) at all points on the circle.

Reminder

To prove that \(\TT(t)\) and \(\TT'(t)\) are orthogonal, note that \(\TT(t)\) is a unit vector, so \(\TT(t)\cdot\TT(t)=1\). Differentiate using the Product Rule for Dot Products: \[ \frac{d}{dt} \TT(t)\cdot\TT(t) = 2\TT(t)\cdot\TT'(t)=0 \]

This shows that \displaystyle{\TT(t)\cdot\TT'(t)=0}

THEOREM 1 Formula for Curvature

If \({\bf r}(t)\) is a regular parametrization, then the curvature at \({\bf r}(t)\) is \[ \kappa(t) = \frac{\|{{\bf r}'(t)\|\times {\bf r}”(t)}}{\|{{\bf r}'(t)}\|^3}\tag{3} \]

Note

To apply Eq.(3) to plane curves, replace \({\bf r}(t)=\langle x(t), y(t)\rangle\) by \({\bf r}(t)=\langle x(t), y(t),0\rangle\) and compute the cross product.

Proof

Since \(v(t)=\|{{\bf r}'(t)}\|\), we have \({\bf r}'(t) = v(t)\TT(t)\). By the Product Rule, \[ {\bf r}”(t) = v'(t)\TT(t) + v(t)\TT'(t) \]

Now compute the following cross product, using the fact that \(\TT(t)\times\TT(t)=\mathbf{0}\): \[ \begin{align} {\bf r}'(t)\times{\bf r}”(t) &= v(t)\TT(t)\times\big(v'(t)\TT(t)+v(t)\TT'(t)\big)\notag\\ &=v(t)^2\TT(t)\times\TT'(t)\tag{4} \end{align} \]

Because \(\displaystyle{\TT(t)\) and \(\displaystyle{\TT'(t)\) are orthogonal, \[ \|{\TT(t)\times \TT'(t)}\| = \|{\TT(t)}\|\,\, \|{\TT'(t)}\|\sin\frac{\pi}2 = \|{\TT'(t)}\| \]

Eq.(4) yields \(\displaystyle{\|{{\bf r}'(t)\times{\bf r}''(t)\)\| = v(t)^2\|{\TT'(t)}\|}\). Using Eq.(2), we obtain \[ \|{{\bf r}'(t)\times{\bf r}''(t)}\| = v(t)^2\|{\TT'(t)}\| = v(t)^3\kappa(t) = \|{{\bf r}'(t)}\|^3\kappa(t) \]

This yields the desired formula.

EXAMPLE 3 Twisted Cubic Curve

Calculate the curvature \(\kappa(t)\) of the twisted cubic \({\bf r}(t) = \langle t,t^2,t^3\rangle\). Then plot the graph of \(\kappa(t)\) and determine where the curvature is largest.

Solution The derivatives are \[ {\bf r}'(t) = \langle 1,2t,3t^2\rangle ,\qquad {\bf r}”(t) = \langle 0,2,6t\rangle\\ \]

The parametrization is regular because \({\bf r}'(t)\ne{\bf 0}\) for all \(t\), so we may use Eq. (3): \begin{align*} {\bf r}'(t)\times {\bf r}”(t) &= \left| \begin{array}{ccc} {\bf i} & {\bf j} & {\bf k} \\ 1 & 2t & 3t^2 \\ 0 & 2 & 6t \\ \end{array} \right| = 6t^2{\bf i} -6t{\bf j}+2{\bf k} \\ \kappa(t) &= \frac{\|{{\bf r}'(t)\times {\bf r}”(t)}}\|{\|{{\bf r}'(t)}\|^3} = \frac{\sqrt{36t^4+36t^2+4}}{(1+4t^2+9t^4)^{3/2}} \end{align*}

The graph of \(\kappa(t)\) in Figure 13.24 shows that the curvature is largest at \(t = 0\). The curve \({\bf r}(t)\) is illustrated in Figure 13.25. The plot is colored by curvature, with large curvature represented in blue, small curvature in green.

THEOREM 2 Curvature of a Graph in the Plane

The curvature at the point \((x,f(x))\) on the graph of \(y = f(x)\) is equal to \[ \boxed{\bbox[#FAF8ED,5pt]{\kappa(x) = \frac{|f”(x)|}{\big(1+f'(x)^2\big)^{3/2}}}}\tag{5} \]

Proof

The curve \(y=f(x)\) has parametrization \({\bf r}(x) = \langle x,f(x)\rangle\). Therefore, \({\bf r}'(x)=\langle 1,f'(x)\rangle\) and \({\bf r}”(x)=\langle 0,f”(x)\rangle\). To apply Theorem 1, we treat \({\bf r}'(x)\) and \({\bf r}”(x)\) as vectors in \({\bf R}^3\) with \(z\)-component equal to zero. Then \[ {\bf r}'(x)\times {\bf r}”(x) = \left| \begin{array}{ccc} {\bf i} & {\bf j} & {\bf k} \\ 1 & f'(x) & 0 \\ 0 & f”(x) & 0 \\ \end{array} \right| = f”(x) {\bf k} \]

Since \(\|{{\bf r}'(x)}\| = \bigl\|\langle 1, f'(x)\rangle\bigr\| = (1+f'(x)^2)^{1/2}\), Eq. (3) yields \[ \kappa(x) = \frac{\|{{\bf r}'(x)\times {\bf r}”(x)}}\|{\|{{\bf r}'(x)}\|^3} = \frac{|f”(x)|}{\big( 1+f'(x)^2\big)^{3/2}} \]

CONCEPTUAL INSIGHT

Curvature for plane curves has a geometric interpretation in terms of the angle of inclination, defined as the angle \(\theta\) between the tangent vector and the horizonal (Figure 13.26). The angle \(\theta\) changes as the curve bends, and we can show that the curvature \(\kappa\) is the rate of change of \(\theta\) as you walk along the curve at unit speed (see Exercise 61).

EXAMPLE 4

Compute the curvature of \(f(x) = x^3 - 3x^2 + 4\) at \(x = 0, 1, 2, 3\).

Solution We apply Eq. (5): \[ \begin{align*} f'(x)&=3x^2-6x=3x(x-2),\qquad f”(x)=6x-6\\ \kappa(x) &= \frac{|f”(x)|}{\big(1+f'(x)^2\big)^{3/2}} = \frac{|6x-6|}{\big(1+9x^2(x-2)^2\big)^{3/2}} \end{align*} \]

We obtain the following values: \[ \begin{eqnarray*} \kappa(0) &=& \frac{6}{(1+0)^{3/2}} = 6,& \quad\kappa(1) &=& \frac{0}{(1+9 )^{3/2}} = 0 \\ \kappa(2) &=& \frac{6}{(1+0)^{3/2}} = 6,& \quad\kappa(3) &=& \frac{12}{82^{3/2}} \approx 0.016 \end{eqnarray*} \]

Figure 13.27 shows that the graph bends more where the curvature is large.

EXAMPLE 5 Unit Normal to a Helix

Find the unit normal vector at \(t=\frac{\pi}4\) to the helix \({\bf r}(t) = \langle \cos t, \sin t, t\rangle\).

Solution The tangent vector \({\bf r}'(t)= \langle -\sin t,\cos t, 1\rangle\) has length \(\sqrt{2}\), so \[ \begin{align*} \TT(t) &= \frac{{\bf r}'(t)}{\norm{{\bf r}'(t)}} = \frac1{\sqrt{2}}\langle -\sin t,\cos t,1\rangle \\ \TT'(t) &= \frac1{\sqrt{2}}\langle -\cos t, - \sin t,0\rangle \\ \NN(t) &= \frac{\TT'(t)}{\norm{\TT'(t)}} = \langle -\cos t,-\sin t, 0\rangle \end{align*} \]

Hence, \(\displaystyle{\NN\left(\frac{\pi}4\right)=\langle -\frac{\sqrt 2}2, -\frac{\sqrt 2}2, 0\rangle}\) (Figure #).

EXAMPLE 6

Parametrize the osculating circle to \(y=x^2\) at \(x=\frac12\).

Solution Let \(f(x) = x^2\). We use the parametrization \[ {\bf r}(x)=\langle x, f(x)\rangle = \big\langle x, x^2\big\rangle \] and proceed by the following steps.

Figure 13.31: Among all circles tangent to the curve at \(P\), the osculating circle is the ”best fit” to the curve.

Step 1. Find the radius

Apply Eq. (71) to \(f(x)=x^2\) to compute the curvature: \[ \kappa(x) = \frac{|f”(x)|}{\big(1+f'(x)^2\big)^{3/2}}= \frac{2}{\big(1+4x^2\big)^{3/2}},\qquad \kappa\left(\frac12\right)=\frac{2}{2^{3/2}} = \frac1{\sqrt{2}} \]

The osculating circle has radius \(R = 1/\kappa\big(\frac12\big) = \sqrt{2}\).

Step 2. Find \(\NN\) at \(t=\frac12\)

For a plane curve, there is an easy way to find \(\NN\) without computing \(\TT'\). The tangent vector is \({\bf r}'(x)=\langle 1,2x\rangle\), and we know that \(\langle 2x,-1\rangle\) is orthogonal to \({\bf r}'(x)\) (because their dot product is zero). Therefore, \(\NN(x)\) is the unit vector in one of the two directions \(\pm\langle 2x,-1\rangle\). Figure 13.32 shows that the unit normal vector points in the positive \(y\)-direction (the direction of bending). Therefore, \[ \NN(x) = \frac{\langle -2x,1\rangle}{\norm{\langle -2x,1\rangle}} = \frac{\langle -2x,1\rangle}{\sqrt{1+4x^2}},\qquad \NN\left(\frac12\right)=\frac1{\sqrt{2}}\langle -1,1\rangle \]

Figure 13.32: The osculating circle to \(y=x^2\) at \(x=\frac12\) has center \(Q\) and radius \(R = \sqrt 2\).

Step 3. Find the center \(Q\)

Apply Eq. (8) with \(t_0=\frac12\): \[ \vc{OQ} = {\bf r}\left(\frac12\right)+\kappa\left(\frac12\right)^{-1}\NN\left(\frac12\right) = \langle \frac12,\frac14\rangle + \sqrt 2\left(\frac{\langle -1,1\rangle}{\sqrt 2}\right) = \langle -\frac12,\frac54\rangle \]

Step 4. Parametrize the osculating circle

The osculating circle has radius \(R = \sqrt 2\), so it has parametrization \[ \boldc(t) = \underbrace{\langle -\frac12,\frac54\rangle}_{\textrm{Center}} + \sqrt{2}\langle \cos t, \sin t\rangle \]

Note

If a curve \(C\) lies in a plane, then this plane is the osculating plane. For a general curve in three-space, the osculating plane varies from point to point.