Question

Graph the curve and find its length..

\(x = {e^t}\cos t,y = {e^t}\sin t,0 \le t \le \pi \)

Short answer

Length of the Curve is \( = \sqrt 2 ({e^\pi } - 1)\)..

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,

\(\frac{{dx}}{{dt}} = {e^t}(\cos t - \sin t)\)and \(\frac{{dy}}{{dt}} = {e^t}(\cos t + \sin t)\)

Step-3(Calculation of length of curve)

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

\({e^t}\sqrt {2({{\cos }^2}t + {{\sin }^2}t)} \)

\( = {e^t}\sqrt 2 \)

By Using the formula,

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

\( = \int\limits_0^\pi {{e^t}\sqrt 2 } dt\)\(\)

\( = \sqrt 2 ({e^\pi } - {e^0})\)

\( = \sqrt 2 ({e^\pi } - 1)\)

\(\)Hence The Length of Given Parametric Curve is \( = \sqrt 2 ({e^\pi } - 1)\) unit.