Question

Find the radius of curvature xy=c^2

Answer 1

Answer:

Solution.

Write the derivatives of the quadratic function:

y′=(x2)′=2x;y′′=(2x)′=2.

Then the curvature of the parabola is defined by the following formula:

K=y′′[1+(y′)2]32=2[1+(2x)2]32=2(1+4x2)32.

At the origin (at x=0), the curvature and radius of curvature, respectively, are

K(x=0)=2(1+4⋅02)32=2,R=1K=12.

Answer 2

Answer:

The radius of curvature for curve xy=c² is equal to  $\rho = \frac{(\big x^4+\big c^4)^\frac{3}{2} }{\big 2\big c^2\big x^3}$.

Step-by-step explanation:

The reciprocal of the curvature at any point is called radius of curvature.

The radius of curvature at any point of curve y= f(x) is

$\rho =\frac{(\big 1 \big + \big y_{1}^ 2)^\frac{3}{2} }{\big y_{2}}$                                                                   .................(1)

where, y₁ = dy/dx  and  y₂ = d²y/dx²

We have given the equation of the curve is: xy = c²

Find the value of y from above expression:

$y=\frac{\big c^2}{\big x}$

Now the value of  y₁ will be:

$y_{1}=\frac{\big dy}{\big dx} =\frac{\big d}{\big dx} \big (\frac{ c^2}{x} \big )$

$y_{1}=\frac{-c^2}{x^2}$

Now the value of  y₂ will be:

$y_{2}=\frac{d^2y}{dx^2}=\frac{d}{dx}[\frac{d}{dx}(\frac{c^2}{x} )]=\frac{d}{dx}(\frac{-c^2}{x^2} )$

$y_{2}=\frac{2c^2}{x^3}$

Substitute the value of y₁  and y₂  in eq.(1):

$\rho = \frac{\big (1+\frac{\big c^4}{\big x^4}\big )^\frac{3}{2} }{\frac{\big 2\big c^2}{\big x^3} }$

$\rho = \frac{(\big x^4+\big c^4)^\frac{3}{2} }{\big 2\big c^2\big x^3}$

Hence the radius of curvature for the given curve is obtained.