Length of a curve in $\Bbb R^2$

Question

How to compute the length of a curve given by the formula $$ f: (0, \frac{\pi}{2}) \ni t \rightarrow ( \cos^3t,\sin^3t) \in \Bbb R^2 $$ I know that the length of a curve in with image in $\Bbb R $ is given by $$ \int_{a}^{b} (1+f_{xx})^{\frac{1}{2}}dx, $$ but don't know what will happen in $\Bbb R^2$.

Answer 1

Well, if $f$ is continuously differentiable, then $$ s[f]=\int_0^{\pi/2}\|\dot{f}(t)\|\ dt, $$ where $\|\cdot\|$ is the usual 2-norm in $\mathbb{R}^2$, and the overdot is differentiation.

Answer 2

Hint.

You have $f(t)=(x(t), y(t))$. The lenght of the curve is given by $$L=\int_0^{\pi/2}\sqrt{(x^\prime(t))^2+(y^\prime(t))^2} dt$$

Answer 3

Hint:

The leght of a plane curve is given by: $$ \int_a^b \sqrt{dx^2+dy^2} $$ as a simple application of the Pythagorean distance between points.

Use $dx=-3\cos^2 t \sin t\;dt$ and $dy=3 \sin^2t \cos t \; dt$ and fix for $a$ and $b$ the values that you want.