Question

Find the length of the loop of the curve

\(x = 3t - {t^3},y = 3{t^2}\)

Short answer

Length of the Curve is \( = 12\sqrt 3 \)..

Step-by-step solution

Step-1(Graph)

Step-2(Calculation of limits)

Loop start fromand end at \(x = 0\)

\((3t - {t^3}) = 0\)

\(t(3 - {t^2}) = 0\)

\(t = 0, \pm \sqrt 3 \)

\( - \sqrt 3 \le t \le \sqrt 3 \)

Step-3(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,

\(\frac{{dx}}{{dt}} = 3 - 3{t^2}\)and \(\frac{{dy}}{{dt}} = 6t\)

Step-4(Calculation of length of curve)

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

\(9{(1 - {t^2})^2} + 36{t^2}\)

\( = (9 + 9{t^4} - 18{t^2}) + 36{t^2}\)

\( = 9{(1 + {t^2})^2}\)

By Using the formula,

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

\( = \int\limits_{ - \sqrt 3 }^{\sqrt 3 } {\sqrt {9{{(1 + {t^2})}^2}} } dt = 3\int\limits_{ - \sqrt 3 }^{\sqrt 3 } {(1 + {t^2}).dt} \)\(\)

\( = 3(t + \frac{{{t^3}}}{3})_{ - \sqrt 3 }^{\sqrt 3 }\)

\(\)

\(3(\sqrt 3 + \frac{{3\sqrt 3 }}{3} + \sqrt 3 + \frac{{3\sqrt 3 }}{3})\)

\( = 3(4\sqrt {3)} \)\(\)\(\)

\( = 12\sqrt 3 \)

\(\)Hence The Length of Given Parametric Curve is \( = 12\sqrt 3 \) unit.