Question

Set up the integral that represent the length of the curve. Then use your calculator to find the length correct to four decimal places. \(x = t + {e^{ - t}},y = t - {e^{ - t}},0 \le t \le 2\)

Short answer

The length of the given curve is given as: \(L \approx 3.1416\)

Step-by-step solution

Find \(\frac{{dx}}{{dt}},\frac{{dy}}{{dt}}\)

The equation of curve is:

\(\begin{array}{l}x = t + {e^{ - t}}\\y = t - {e^{ - t}}\\\frac{{dx}}{{dt}} = 1 - {e^{ - t}}\\\frac{{dy}}{{dt}} = 1 + {e^{ - t}}\\{\left( {\frac{{dx}}{{dt}}} \right)^2} + {\left( {\frac{{dy}}{{dt}}} \right)^2} = {\left( {1 - {e^{ - t}}} \right)^2} + {\left( {1 + {e^{ - t}}} \right)^2}\\{\left( {\frac{{dx}}{{dt}}} \right)^2} + {\left( {\frac{{dy}}{{dt}}} \right)^2} = 2 + 2{e^{ - 2t}}\end{array}\)

Find the length

The length of the curve can be obtained by the following formula:

\(L = \int\limits_\alpha ^\beta {\sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} } dt\)

The length can be given as:

\(\begin{array}{l}L = \int\limits_\alpha ^\beta {\sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} } dt\\L = \int\limits_0^2 {\sqrt {2 + 2{e^{ - 2t}}} } dt\\L = 2 - \sqrt 2 {\sinh ^{ - 1}}(1) + \sqrt 2 \left( {{{\sinh }^{ - 1}}({e^2}) - \frac{{\sqrt {1 + {e^4}} }}{{{e^2}}}} \right)\\L \approx 3.1416\end{array}\)

Therefore, the length of the given curve can be given as: \(L \approx 3.1416\)