Question

Graph the curve and find its length..

\(x = {e^t} - t,y = 4{e^{\frac{t}{2}}}, - 8 \le t \le 3\)

Short answer

Length of the Curve is \( = 11 + {e^3} - {e^{ - 8}}\)..

Step-by-step solution

Step-1(Graph)

Step-2(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,

\(\frac{{dx}}{{dt}} = {e^t} - 1\)and \(\frac{{dy}}{{dt}} = 4\frac{1}{2}{e^{\frac{t}{2}}} = 2{e^{\frac{t}{2}}}\)

Step-3(Calculation of length of curve)

\({(\frac{{dx}}{{dt}})^2} + {(\frac{{dy}}{{dt}})^2} = \)\({({e^t} - 1)^2} + {(2{e^{\frac{t}{2}}})^2}\)

\({e^{2t}} + 1 - 2{e^t} + 4{e^t} = {({e^t} + 1)^2}\)

By Using the formula,

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

\( = \int\limits_{ - 8}^3 {\sqrt {{{({e^t} + 1)}^2}} } dt\)\(\)

\( = \int\limits_{ - 8}^3 {({e^t} + 1).dt} \)

\(({e^t} + t)_{ - 8}^3\)

\( = {e^3} + 3 - {e^{ - 8}} + 8\)

\( = 11 + {e^3} - {e^{ - 8}}\)

\(\)Hence The Length of Given Parametric Curve is \( = 11 + {e^3} - {e^{ - 8}}\) unit.