Question

Verify that the curvature at any point \((x, y)\) on the graph of \(y=\cosh x\) is \(1 / y^{2}\)

Short answer

The curvature at any point \( (x, y) \) on the graph of \( y=\cosh x \) is actually equal to \( 1 / (2y^{2}) \), not \( 1 / y^{2} \).

Step-by-step solution

Expression for Curvature and Derivatives

First determine the expression for curvature of a curve, which is given by \(\kappa = |y''| / (1 + (y')^{2})^{3/2}\). Next, identify the given function: \(y=\cosh x\) and find its first and second derivatives. The first derivative \(y'=\sinh x\) and the second derivative \(y''=\cosh x\).

Substitution into Curvature Expression

Next substitute the derivatives into the curvature expression: \(\kappa = |\cosh x| / (1 + (\sinh x)^{2})^{3/2}\). We can simplify further by recognizing the Pythagorean Identity of hyperbolic functions which is \(\cosh^{2} x - \sinh^{2} x = 1\). Hence, \(\kappa = |\cosh x| / (2\cosh^{2} x)^{3/2} = 1/(2\cosh^{2}x)\).

Verifying Equality

Finally, we need to verify that our expression for curvature is equal to the given expression \(1/y^{2}\). Since \(\cosh x = y\), substituting \(\cosh x\) by \(y\) in the curvature expression obtained in the last step will yield \(1/(2y^{2})\), which is half the given curvature. Hence, the given expression for curvature should be \(1/(2y^{2})\) rather than \(1/y^{2}\).