Question

Find the curvature and radius of curvature of the plane curve at the given value of \(x\). $$ y=\frac{3}{4} \sqrt{16-x^{2}}, \quad x=0 $$

Short answer

The curvature at \(x=0\) is \(3\) and the radius of curvature is \(\frac{1}{3}\).

Step-by-step solution

Differentiation of Function

Given \(y=\frac{3}{4} \sqrt{16-x^{2}}\), let's start by finding the first derivatives of \(y\). Using the chain rule, we get \(y'=\frac{-3x}{4\sqrt{16-x^{2}}}\). Then proceed to find the second derivative. Again applying chain rule, we end up with \(y''=\frac{-3(16-x^{2})^{\frac{1}{2}}+3x^{2}(16-x^{2})^{-\frac{1}{2}}}{16}\)

Plugging into Curvature Formula

The curvature formula is given by \(\kappa = \frac{|y''|}{(1+(y')^2)^{\frac{3}{2}}}\). At, \(x=0\), the functions \(y'\) and \(y''\) will simplify to \(y'(0)=0\) and \(y''(0)=\frac{-3}{2}\sqrt{16}= -3\). Plugging these values into the formula will give a curvature \( \kappa= |-3|= 3\)

Finding Radius of Curvature

The radius of curvature is given by \(R = \frac{1}{|\kappa|}\). Substituting for \(\kappa\), we get \(R = \frac{1}{3}\)