Question

Graph the curve and find its length..

\(x = \cos t + \ln (\tan \frac{1}{2}t),y = \sin t,\frac{\pi }{4} \le t \le \frac{{3\pi }}{4}\)

Short answer

Length of the Curve is \( = \ln 2\)..

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, and represent lower limit and uppe\(\alpha \)r limit of t respectively…

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

\(\frac{{dx}}{{dt}} = - \sin t + \frac{1}{{\sin t}}\)and \(\frac{{dy}}{{dt}} = \cos t\)

Step-3(Calculation of length of curve)

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

\(1 - 2 + \frac{1}{{{{\sin }^2}t}}\)

\( = \cos e{c^2}t - 1 = {\cot ^2}t\)

By Using the formula,

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

\( = \int\limits_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {|\cot t|.} dt = 2\int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\cot t.dt} \)\(\)

\( = 2(\ln (\sin \frac{\pi }{2}) - \ln (\sin \frac{\pi }{4}))\)

\(\)

\(2(\ln 1 - \ln (\frac{1}{{\sqrt 2 }}))\)

\( - 2(\frac{{ - 1}}{2})\ln 2\)\(\)\(\)

\( = \ln 2\)

\(\)Hence The Length of Given Parametric Curve is \( = \ln 2\) unit.