Question

Find the exact length of curve..

\(x = 1 + 3{t^2},y = 4 + 2{t^3},0 \le t \le 1\)

Short answer

Exact Length of the Curve is \(2(2\sqrt 2 - 1)\)..

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,

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

Step-2(Calculation of length of curve)

\({(\frac{{dx}}{{dt}})^2} + {(\frac{{dy}}{{dt}})^2} = \)\({(6t)^2} + {(6{t^2})^2}\)

By Using the formula,

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

\( = \int\limits_0^1 {\sqrt {{{(6t)}^2} + {{(6{t^2})}^2}} } .dt\)

\( = \sqrt {36{t^2} + 36{t^4}} \)

\( = \int\limits_0^1 {6t\sqrt {1 + {t^2}} } \)

\( = 3\int\limits_0^1 {\sqrt {1 + {t^2}} } (2t)dt\)

Substitute \(1 + {t^2} = u \Rightarrow 2t.dt = du\)

Limits :\(\begin{array}{l}\{ t = 0,u = 1\} \\\{ t = 1,u = 2\} \end{array}\)

\(\therefore L = 3\int\limits_1^2 {\sqrt u } .du\)

\( = 3{(\frac{{{u^{\frac{3}{2}}}}}{{\frac{3}{2}}})_1}^2\)

\( = 2({2^{\frac{3}{2}}} - {1^{\frac{3}{2}}})\)

\( = 2(2\sqrt 2 - 1)\)..

Hence The Length of Given Parametric Curve is \( = 2(2\sqrt 2 - 1)\) unit.