Question

Find the curvature of the curve with parametric equations . Exercise 57 in Section 10.

Short answer

Therefore, the value derived \(\kappa = \pi |t|\).

Step-by-step solution

Definition of the parametric equation

A parametric equation is one in which the regression analysis are defined as continuous functions of the independent variable (often represented by t), and the factors considered are not dependent on any other variables. When necessary, more than one parameter is used.

Evaluating the value of parametric equation

Given that,

Therefore, by part 2 of fundamental theorem of calculus,

\(x' = \sin \left( {\frac{1}{2}\pi {t^2}} \right)\;\;\;y' = \cos \left( {\frac{1}{2}\pi {t^2}} \right)\)

Again differentiate, \(x'' = \pi t\cos \left( {\frac{1}{2}\pi {t^2}} \right)\;\;\;y'' = - \pi t\sin \left( {\frac{1}{2}\pi {t^2}} \right)\)

 Substitude the value

Given that, \(\kappa = \frac{{\left| {x'y'' - x''y'} \right|}}{{{{\left( {{{\left( {x'} \right)}^2} + {{\left( {y'} \right)}^2}} \right)}^{3/2}}}}\)

Substitude the value to get,

\(\begin{aligned}{*{20}{c}}{\kappa = \frac{{ - \pi t{{\sin }^2}\left( {\frac{1}{2}\pi {t^2}} \right) - \pi t{{\cos }^2}\left( {\frac{1}{2}\pi {t^2}} \right)}}{{{{\left( {{{\sin }^2}\left( {\frac{1}{2}\pi {t^2}} \right) + {{\cos }^2}\left( {\frac{1}{2}\pi {t^2}} \right)} \right)}^{3/2}}}}}\\\begin{aligned}{l}\kappa = \frac{{| - \pi t|}}{{{{(1)}^{3/2}}}}\\ = \pi |t|\end{aligned}\end{aligned}\)

Therefore, the value derived \(\kappa = \pi |t|\).