Solution.

\(\displaystyle {K = \dfrac {\left |y''\right | }{\Big [ 1 + (y')^2\Big ]^{\dfrac {3}{2}}}}\)

\(\displaystyle {y' = 1 - \frac {1}{2}x\hspace {0.3cm} , \hspace {0.2cm} y'(2) = 0}\)

\(\displaystyle {y'' = -\frac {1}{2}\implies y''(2) = \frac {-1}{2}}\)

\[K = \dfrac {\left |\dfrac {-1}{2}\right | }{\Big [ 1 + 0^2\Big ]^{\dfrac {3}{2}}}= \frac {1}{2}\]

Note that \(K\) is a non-negative constant.

The curvature of a curve \(y = f(x)\) is given by \[K = \dfrac {\left |\dfrac {d^2y}{dx^2}\right | }{\Bigg [ 1 + \Big (\dfrac {dy}{dx}\Big )^2\Bigg ]^{\dfrac {3}{2}}}\]

The reciprocal of the curvature is called the radius of curvature and is denoted by \(\rho \). i.e \(\rho = \dfrac {1}{K}\) \[\rho = \frac {\Bigg [ 1 + \Big (\dfrac {dy}{dx}\Big )^2\Bigg ]^{\dfrac {3}{2}} }{\left |\dfrac {d^2y}{dx^2}\right | }\]

The curvature in terms of intrinsic coordinates in given by \[K =\frac {d\Psi }{ds} = \frac {1}{\dfrac {ds}{d\Psi }} = \frac {1}{\rho }\]

\[\text {i.e}\hspace {0.3cm} \rho = \frac {ds}{d\Psi }\]

Solution.

For the curve \(s = 4(1 - \sin \Psi )\)

\(\displaystyle {\frac {ds}{d\Psi } = - 4\cos \Psi }\hspace {0.3cm}\) at \(\hspace {0.3cm}\Psi = \dfrac {\pi }{4}\)

\[\rho = \left |-4\cos \dfrac {\pi }{4}\right | = 2\sqrt {2}\]

Hence, the curvature at \(\Psi = \dfrac {\pi }{4}\) is \(\displaystyle {K = \frac {1}{\rho } =\frac {1}{2\sqrt {2}}}\).