Question

In calculus the curvature of a curve that is defined by a function y=f(x) is defined as

$\mathit{k}\mathbf{=}\frac{{\mathbf{y}}^{\mathbf{n}}}{{\left[1+{(y\text{'})}^2\right]}^{\mathbf{3}\mathbf{/}\mathbf{2}}}$

Find y=f(x) for which k=1. (Hint: For simplicity, ignore constants of integration.)

Short answer

The solution of the above question is$y\left(x\right)=-\sqrt{1-x^2}$

Step-by-step solution

Curvature of curve

The curvature (K) of a path is measured using the radius of the curvature of the path at the given point. If y = f(x) is a curve at a particular point, then the formula for curvature is given as K = 1/R.

Concept of curvature of curve will be applied;

$$\begin{gathered} 1=\frac{y\text{'}\text{'}}{{\left[1+{(y\text{'})}^2\right]}^{3/2}} \\ y\text{'}\text{'}={\left[1+{(y\text{'})}^2\right]}^{3/2} \\ y\text{'}\text{'}=v\text{'} \\ v\text{'}={\left[1+{\left(v\right)}^2\right]}^{3/2} \\ \frac{dv}{dx}={\left[1+{\left(v\right)}^2\right]}^{3/2} \\ \int \frac{1}{{\left[1+{\left(v\right)}^2\right]}^{3/2}}dv=\int dx \\ \int \frac{1}{{\left[1+{\left(v\right)}^2\right]}^{3/2}}dv=x \end{gathered}$$

Substitution

Substitute v=tan u, then dv= sec²u,

$$\begin{gathered} x=\int \frac{{\sec }^{2}u}{{(1+{\tan }^{2}u)}^{3/2}}du \\ =\int \frac{{\sec }^{2}u}{{(1+\frac{{\sin }^{2}u}{{\cos }^{2}u})}^{3/2}}du \\ =\int \frac{{\sec }^{2}u}{{\left(\frac{1}{{\cos }^{2}u}\right)}^{3/2}}du \\ =\int \frac{{\sec }^{2}u}{{\sec }^{3}u}du \\ =\int \cos u \\ =\sin u \\ =\frac{v}{\sqrt{1+v^2}} \\ x=\frac{y\text{'}}{\sqrt{1+{(y\text{'})}^2}} \\ {x}^2=\frac{{(y\text{'})}^2}{1+{(y\text{'})}^2} \\ (y\text{'}{)}^2=\frac{x^2}{1-x^2} \\ \frac{dy}{dx}=\frac{x}{\sqrt{1-x^2}} \\ dy=\frac{x}{\sqrt{1-x^2}}dx \\ y=-\frac{1}{2}\int \frac{1}{s^{1/2}}ds \\ =-\frac{1}{2}\int s^{-1/2}ds \\ =-\frac{1}{2}\times 2\sqrt{s} \\ =-\sqrt{1-x^2} \end{gathered}$$

Hence, the final answer is .$y\left(x\right)=-\sqrt{1-x^2}$