Question

Find the length of the curve.

Short answer

Length of the given curve is\(10\sqrt 5 - 2\).

Step-by-step solution

Definition of curve

A curve is a mathematical entity that is similar to a line but does not have to be straight.

Find the limit L

Consider the given equation and simplify,

We know that the length \(L\)of a curve

\(\begin{aligned}{c}\frac{{dr}}{{d\theta }} = 3 \cdot \frac{1}{3}{\sin ^2}\left( {\frac{\theta }{3}} \right)\cos \left( {\frac{\theta }{3}} \right)\\ = {\sin ^2}\left( {\frac{\theta }{3}} \right)\cos \left( {\frac{\theta }{3}} \right){r^2} + {\left( {\frac{{dr}}{{d\theta }}} \right)^2}\\ = {\sin ^6}\left( {\frac{\theta }{3}} \right) + {\sin ^4}\left( {\frac{\theta }{3}} \right){\cos ^2}\left( {\frac{\theta }{3}} \right)\\ = {\sin ^4}\left( {\frac{\theta }{3}} \right)\left( {{{\sin }^2}\left( {\frac{\theta }{3}} \right) + {{\cos }^2}\left( {\frac{\theta }{3}} \right)} \right)\\ = {\sin ^4}\left( {\frac{\theta }{3}} \right)\end{aligned}\)

Hence, L =\(\int_0^2 6 t\sqrt {1 + {t^2}} dt\).

 Step 3: Change the limit into integration

Let\(u = 1 + {t^2} \to du = 2t\)

Change the limit of integration

If t=0, then u=1

If t=5, then u=26

\(\begin{aligned}{c}\int_0^5 6 t\sqrt {1 + {t^2}} dt = \int\limits_1^{26} {3\sqrt u } du\\ = 3\left( {\frac{2}{3}{u^{3/2}}} \right)_1^{25}\\ = 2(26\sqrt {26} - 1)\end{aligned}\)

Hence, the required length of the curve is L\( = 2(26\sqrt {26} - 1)\).