Question

Find the length of the curve given by x= t^3, y= 2t^2, in 0≤ t≤2

Answer 1

$\text{length of the curve}=\int_a^b \sqrt{x_t^2+y_t^2}dt\newline =\int_0^2 \sqrt{(3t^2)^2+(4t)^2}dt\newline =\int_0^2 \sqrt{9t^4+16t^2}dt\newline =\int_0^2 t\sqrt{9t^2+16}dt\newline Put\space u=9t^2+16\newline \implies du=18t dt\newline Therefore, \newline \int_0^2 \sqrt{(3t^2)^2+(4t)^2}dt=\frac{1}{18}\int \sqrt{u}du\newline =\frac{2}{27} {u^{3/2}}\newline =\frac{2}{27} [(9t^2+16)^{3/2}]_0^2\newline =11.52$