Question

Find the exact length of the curve.

\({\rm{y = 3 + }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{cosh2x,0}} \le {\rm{x}} \le {\rm{1}}\)

Short answer

The length of the curve is\(\frac{{\rm{1}}}{{\rm{2}}}{\rm{sinh(2)}}\).

Step-by-step solution

To find\(\frac{{dy}}{{dx}}\).

Given: \({\rm{y = 3 + }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{cosh2x}}\)

\(\frac{{{\rm{dy}}}}{{{\rm{dx}}}}{\rm{ = sinh2x}}\)

To find the length of the curve.

The formula for arc length: \({\rm{L = }}\int_{\rm{a}}^{\rm{b}} {\sqrt {{\rm{1 + }}{{\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)}^{\rm{2}}}} } {\rm{dx}}\)

\({\rm{L = }}\int_{\rm{0}}^{\rm{1}} {\sqrt {{\rm{1 + sin}}{{\rm{h}}^{\rm{2}}}{\rm{x}}} } {\rm{dx}}\)

Known value\({\rm{cos}}{{\rm{h}}^{\rm{2}}}{\rm{x = 1 + sin}}{{\rm{h}}^{\rm{2}}}{\rm{x}}\)

\(\begin{aligned}{}{\rm{L = }}\int_{\rm{0}}^{\rm{1}} {\sqrt {{\rm{cos}}{{\rm{h}}^{\rm{2}}}{\rm{2x}}} } {\rm{dx}}\\{\rm{ = }}\int_{\rm{0}}^{\rm{1}} {{\rm{cos}}{{\rm{h}}^{\rm{2}}}} {\rm{2xdx}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{|sinh2x|}}_{\rm{0}}^{\rm{1}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{sinh(2) - }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{sinh(0)}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{sinh(2)}}\end{aligned}\)

Therefore, the length of the curve is\(\frac{{\rm{1}}}{{\rm{2}}}{\rm{sinh(2)}}\).