Question

Find the exact length of curve..

\(x = 3\cos t - \cos 3t,y = 3\sin t - \sin 3t,0 \le t \le \pi \)

Short answer

Exact Length of the Curve is \( = 12\)..

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,

\(\frac{{dx}}{{dt}} = - 3\sin t + 3\sin 3t\)and \(\frac{{dy}}{{dt}} = 3\cos t - 3\cos (3t)\)

Step-2(Calculation of length of curve)

\({(\frac{{dx}}{{dt}})^2} + {(\frac{{dy}}{{dt}})^2} = \)\({( - 3\sin t + 3\sin 3t)^2} + {(3\cos t - 3\cos 3t)^2}\)

\(18 - 18(\sin t\sin 3t + \cos t\cos 3t)\)

\( = 18(1 - \cos 2t)\)

\( = 36{\sin ^2}t\)

By Using the formula,

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

\( = \int\limits_0^\pi {\sqrt {36{{\sin }^2}t} } .dt\)\(\)

\( = \int\limits_0^\pi {(6\sin t.dt)_0^1} \)

\( = - 6(\cos t)_0^\pi = 12\)..

\(\)Hence The Length of Given Parametric Curve is \( = 12\) unit.