4.13 Curvature
A curvature \(C\) defined by \(\overline {r}(t)\) is smooth if \(\overline {r}(t)\) is continuous and \(\overline {r}'(t) \neq 0\) on an interval \(I\). If \(C\) is a smooth curve defined by the vector function \(\overline {r}(t)\), then the unit tangent vector \(T'(t)\) is given by \[ T(t) = \frac {\overline {r}(t)}{\left |\overline {r}'(t)\right | }\] and indicates the direction of the curve.
The curvature of \(C\) at a given point is a measure of how quickly the curve changes direction at that point. i.e it is the magnitude of the rate of change of the unit tangent vector with respect to arc length. It is given by \[K = \left |\dfrac {d T}{d S}\right |\] where \(T\) is the unit tangent vector and is the arclength function. \[\frac {d T}{d t} = \frac {d T}{d S}\cdot \frac {d S}{dt}\]
\[\frac {dT}{dS} = \frac {\dfrac {dT}{dt}}{\dfrac {dS}{dt}}\]
but \(\dfrac {dS}{d t} = \left |\overline {r}(t)\right |\) so that \(K(t) = \left |\dfrac {T'(t)}{r'(t)}\right |\).
The curvature of the curve given by the vector \(\overline {r}(t)\) is \[K(t) = \frac {\left |\overline {r}'(t)\times \overline {r}''(t)\right | }{\left |\overline {r}'(t)\right |^3 }\]
Find the curvature of the curve \(\overline {r}(t) = \big \langle t, t^2 , t^3\big \rangle \) at a general point and at \((0,0,0)\).
Solution. \[\overline {r}'(t) =\big \langle 1, 2t, 3t^2\big \rangle \hspace {0.2cm} , \hspace {0.2cm} \overline {r}''(t) = \big \langle 0, 2, 6t\big \rangle \]
\[\overline {r}'(t) \times \overline {r}''(t) = \begin {vmatrix} \textbf {i} & \textbf {j} & \textbf {k}\\ 1 & 2t & 3t^2\\ 0 & 2 & 6t\\ \end {vmatrix}= 6t^2\textbf {i} - 6t\textbf {j} + 2\textbf {k}\]
\[\left |\overline {r}'(t) \times \overline {r}''(t)\right |=\sqrt {36t^4 + 36t^2 + 4}\]
\[\left |\overline {r}'(t)\right |= \sqrt {1 + 4t^2 + 9t^4}\]
\[\implies \hspace {0.5cm} K(t) = \frac {2\sqrt {9t^4 + 9t^2 + 1}}{\Big (9t^4 + 4t^2 + 1\Big )^3}\]
At \((0,0,0)\hspace {0.2cm},\hspace {0.2cm} x = 0 \hspace {0.2cm},\hspace {0.2cm} y = 0 \hspace {0.2cm},\hspace {0.2cm} z = 0\) from \(\overline {r}(t)\) \[x =t \hspace {0.2cm},\hspace {0.2cm} y = t^2 \hspace {0.2cm},\hspace {0.2cm} z = t^3\implies \hspace {0.2cm} t = 0\hspace {0.2cm}\text {at}\hspace {0.2cm} (0,0,0)\] Curvature at \((0,0,0)\) is \(K(0) = 2\).
For the special case in the plane where \(y = f(x)\) we choose \(x\) as the parameter and write \(r(x) = x\textbf {i} + f(x)\textbf {j}\). Then \(\overline {r}'(x) = \textbf {i} + f'(x)\textbf {j}\) and \(\overline {r}''(x) = f''(x)\textbf {j}\), \(\hspace {0.2cm} \left |\overline {r}'(x)\times \overline {r}''(x)\right |= f''(x)\) and \(\left |\overline {r}'(x)\right | = \sqrt {1 + (f'(x))^2}\)
\[K(x) = \frac {\left |f''(x)\right | }{\Big [ 1 + f'(x)\Big ]^{\dfrac {3}{2}}}\hspace {1cm} K(t) = \frac { \left |\overline {r}'\times \overline {r}''\right | }{\left |\overline {r}'\right |^3 }\]
Find the curvature of the parabola \(y = x^2\) at the point \((0,0), (1,1)\) and \((2,4)\).
Solution. \[K(x) = \frac {\left |y''\right | }{\Big [1 + (y')^2\Big ]^{3/2}} = \frac {2}{\Big [1 + 4x\Big ]^{3/2}}\]
\[K(0) = 2 \hspace {0.2cm},\hspace {0.2cm} K(1) = \frac {2}{5^{3/2}} \hspace {0.2cm},\hspace {0.2cm} K(2) = \frac {2}{(17)^{3/2}}\]
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