Question

Find the exact length of curve..

\(x = t\sin t,y = t\cos t,0 \le t \le 1\)

Short answer

Exact Length of the Curve is \( = \frac{{\sqrt 2 + \ln (1 + \sqrt 2 )}}{2}\)..

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,

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

Step-2(Calculation of length of curve)

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

\( = {t^2}{\cos ^2}t + {\sin ^2}t + 2t\sin t\cos t + {\cos ^2}t + {t^3}{\sin ^3}t - 2t\sin t\cos t\)

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

\( = {t^2} + 1\)

By Using the formula,

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

\( = \int\limits_0^1 {\sqrt {{t^2} + 1} } .dt\)

\( = \int\limits_0^1 {(\frac{1}{2}t\sqrt {{t^2} + 1} + \frac{1}{2}\ln (t + \sqrt {{t^2} + 1} ))_0^1} dt\)

\( = \frac{{\sqrt 2 + \ln (1 + \sqrt 2 )}}{2}\)..

\(\)Hence The Length of Given Parametric Curve is \( = \frac{{\sqrt 2 + \ln (1 + \sqrt 2 )}}{2}\) unit.