Correct Answer - Option 3 : 4.71
Concept:
The length of the curve in three-dimensional space given in the parametric form is:
\(L = \mathop \smallint \limits_a^b \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2} + \;{{\left( {\frac{{dz}}{{dt}}} \right)}^2}} \;.dt\)
\(a \le t \le b\)
And the length of the curve in two- dimensional space given in the parametric form is:
\(L = \mathop \smallint \limits_a^b \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}\;} \;.dt\)
\(a \le t \le b\)
Calculation:
Given, x(t) = cos t, y(yt) = sin t, z(t) = 2√2 t
\(0 \le t \le \frac{\pi }{2}\)
\(L = \mathop \smallint \limits_0^{\pi /2} \sqrt {{{\left| {\frac{{d\left( {cost} \right)}}{{dt}}} \right|}^2} + {{\left| {\frac{{d\left( {sint} \right)}}{{dt}}} \right|}^2} + {{\left| {2\sqrt 2\frac{{d\left( t \right)}}{{dt}}} \right|}^2}\;} .dt\;\)
\( = \mathop \smallint \limits_0^{\pi /2} \sqrt {{{\sin }^2}t + co{s^2}t +{8}} \;.dt\)
\( = \mathop \smallint \limits_0^{\pi /2} \sqrt {1 +{8}} ~dt\)
\( = 3\mathop \smallint \limits_0^{\pi /2} dt = 3 \times \frac{\pi }{2}\)
= 4.71
