Question

Find the exact length of curve..

\(x = {e^t} + {e^{ - t}},y = 5 - 2t,0 \le t \le 3\)

Short answer

Exact Length of the Curve is \( = {e^3} - {e^{ - 3}} \approx 20.036\)..

Step-by-step solution

Step-1(Using Integration to represent length of Curve)

The formula for length of curve is

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

Where, \(\alpha \)and \(\beta \) represent lower limit and upper limit of t respectively…

On Differentiating x and y w.r.t \(t\).We get,

\(\frac{{dx}}{{dt}} = {e^t} - {e^{ - t}}\)and \(\frac{{dy}}{{dt}} = - 2\)

Step-2(Calculation of length of curve)

\({(\frac{{dx}}{{dt}})^2} + {(\frac{{dy}}{{dt}})^2} = \)\({({e^t} - {e^{ - t}})^2} + {( - 2)^2}\)

By Using the formula,

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

\( = \int\limits_0^3 {\sqrt {{{({e^t} - {e^{ - t}})}^2} + {{( - 2)}^2}} } .dt\)

\( = \int\limits_0^3 {\sqrt {{e^{2t}} - 2{e^t}.{e^t} + {e^{ - 2t}} + 4} } .dt\)

\( = \int\limits_0^3 {\sqrt {{e^{2t}} + 2 + {e^{ - 2t}}} } .dt\)

\( = \int\limits_0^3 {\sqrt {({e^t} + {e^{ - t}}} {)^2}} dt\)

\(\therefore L = \int\limits_0^3 {({e^t} + {e^{ - t}}} ).dt\)

\(\therefore L = {\int\limits_0^3 {({e^t} - {e^{ - t}})} ^3}_0\)

\( = ({e^3} - {e^{ - 3}} - (1 - 1))\)

\( = {e^3} - {e^{ - 3}} \approx 20.036\)..

Hence The Length of Given Parametric Curve is \( = {e^3} - {e^{ - 3}} \approx 20.036\) unit.