Question

Use Theorem 10 to show that the curvature of a plane parametric curve \(x = f(t),y = g(t)\) is

\(\kappa = \frac{{|\dot x\dot y - \dot y\dot x|}}{{{{\left( {{{\dot x}^2} + {{\dot y}^2}} \right)}^{3/2}}}}\)where the dots indicate derivatives with respect to \(t.\)

Short answer

The curvature of a plane parametric curve \(x = f(t),y = s(t)\) is, \(\kappa = \frac{{|\dot x\ddot y - \dot y\ddot x|}}{{{{\left( {{{\dot x}^2} + {{\dot y}^2}} \right)}^{\frac{3}{2}}}}},\) where the dots indicate derivatives with respect to\(t\) .

The required proof is explained inside.

Step-by-step solution

Step 1: To find the curvature of a plane parameter

Using the notation \(\dot x = {f^'}(t),\dot y = {g^'}(t),\ddot x = {f^{''}}(t)\) and \(\ddot y = {g^{''}}(t),\) let us compute \({r^'}(t)\) and \({r^{''}}(t)\)

\(r(t) = < x,y,0 > \)

\({r^'}(t) = < \dot x,\dot y,0 > \)

\({r^{''}}(t) = < \ddot x,\ddot y,0 > \)

Next, we have to compute the cross- product, \({r^'}(t) \times {r^{''}}(t)\)

\({r^'}(t) \times {r^{''}}(t) = \left| {\begin{aligned}{*{20}{c}}i&j&k\\{\dot x}&{\dot y}&0\\{\ddot x}&{\ddot y}&0\end{aligned}} \right| = (\dot x\ddot y - \dot y\ddot x)k\)

\(\dot x \dot y 0\ddot x \ddot y 0 = (\dot x\ddot y - \dot y\ddot x)k\)

Step 2: Final Proof

Therefore, using theorem 10, we can find the curvature

\(\kappa = \frac{{\left| {{r^'}(t) \times {r^{''}}(t)} \right|}}{{{{\left| {{r^'}(t)} \right|}^3}}}.\)

\(\kappa = \frac{{|\dot x\ddot y - \dot y\ddot x|}}{{\sqrt {{{\dot x}^2} + {{\dot y}^2}} }} \)

\(\kappa = \frac{{|\dot x\ddot y - \dot y\ddot x|}}{{{{\left( {{{\dot x}^2} + {{\dot y}^2}} \right)}^{\frac{3}{2}}}}}\)

The curvature of a plane parametric curve \(x = f(t),y = s(t)\) is, \(\kappa = \frac{{|\dot x\ddot y - \dot y\ddot x|}}{{{{\left( {{{\dot x}^2} + {{\dot y}^2}} \right)}^{\frac{3}{2}}}}},\) where the dots indicate derivatives with respect to\(t\) .

The required proof is explained inside.