Question

Find the curvature and radius of curvature of the plane curve at the given value of \(x\). $$ y=2 x^{2}+3, \quad x=-1 $$

Short answer

The curvature is \(k = \frac{4}{\sqrt{4913}}\) and radius of curvature is \(R = \frac{\sqrt{4913}}{4}\).

Step-by-step solution

Find the first differential

Start by finding the first differential of \(y\), which is given by \(y' = \frac{dy}{dx}\). So, if \(y=2x^2+3\), then \(y'=4x\).

Find the second differential

Next, find the second differential of \(y\), which is given by \(y'' = \frac{d^2y}{dx^2}\). The derivative of \(y'=4x\) is \(y''=4\).

Substitute in the curvature formula

Substitute the values of \(y''\) and \(y'\) into the formula for curvature at \(x=-1\). So, the curvature \(k = \frac{|4|}{(1 + (4*-1)^2)^\frac{3}{2}}\).

Simplification and calculation of curvature

Simplify the above expression to the form \(k = \frac{4}{(1 + 16)^\frac{3}{2}}\). Therefore, \(k = \frac{4}{\sqrt{4913}}\).

Find the radius of curvature

Finally, find the radius of curvature using \(R = \frac{1}{|k|}\). So, \(R = \frac{\sqrt{4913}}{4}\).