# Thread: Length of a curve

1. ## Length of a curve

I calculate the length of a curve. Task but I can not modify (derivate, cube) that I could substitute in the formula

2. Originally Posted by Ahplym
I calculate the length of a curve. Task but I can not modify (derivate, cube) that I could substitute in the formula

$\displaystyle{y^{2/3}+x^{2/3}=a^{2/3}\Longrightarrow y=\sqrt{\left(a^{2/3}-x^{2/3}\right)^3}}$ , and now you can calculate $y'$.

Note: Don't forget to take into account the domain of definition of y !

Tonio

3. Thank you. I got derivate but I can not potentiate.

4. Hello, Ahplym!

tonio told you what has to be done.
But it takes some Olympic-level gymnastics.

$\text{Calculate the length of the curve: }x^{\frac{2}{3}} + y^{\frac{2}{3}} \:=\:a^{\frac{2}{3}}$

$\displaystyle \text{But I cannot modify it to substitute in the formula: }\: \int^b_a\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx$

We have: . $y^{\frac{2}{3}} \;=\;\ a^{\frac{2}{3}} - x^{\frac{2}{3}} \quad\Rightarrow\quad y \;=\;\left(a^{\frac{2}{3}} - x^{\frac{2}{3}}\right)^{\frac{3}{2}}$
Differentiate: . $\dfrac{dy}{dx} \;=\;\frac{3}{2}\left(a^{\frac{2}{3}}-x^{\frac{2}{3}}\right)^{\frac{1}{2}}\left(\text{-}\frac{2}{3}x^{-\frac{1}{3}}\right) \;=\;\dfrac{-\left(a^{\frac{2}{3}} - x^{\frac{2}{3}}\right)^{\frac{1}{2}}} {x^{\frac{1}{3}}}$

Square: . $\left(\dfrac{dy}{dx}\right)^2 \;=\;\dfrac{a^{\frac{2}{3}} - x^{\frac{2}{3}}}{x^{\frac{2}{3}}}$

$\displaystyle \text{Add 1: }\;1 + \left(\dfrac{dy}{dx}\right)^2 \;=\;1 + \dfrac{a^{\frac{2}{3}} - x^{\frac{2}{3}}} {x^{\frac{2}{3}}} \;=\; \frac{a^{\frac{2}{3}}}{x^{\frac{2}{3}}}$

Take the square root: . $\displaystyle \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \;=\;\sqrt{\frac{a^{\frac{2}{3}}}{x^{\frac{2}{3}}} } \;=\;\frac{a^{\frac{1}{3}}}{x^{\frac{1}{3}}}$

We have: . $\displaystyle{L \;=\;\int^b_a \frac{a^{\frac{1}{3}}}{x^{\frac{1}{3}}}\,dx \;=\;a^{\frac{1}{3}}\!\!\int^b_a \!\!x^{\frac{1}{3}}\,dx$

The graph is an astroid (a hypocycloid of four cusps).
It is symmetric to both axes and the Origin.

We can integrate from 0 to $\,a$ and multiply by 4.

. . $\displaystyle L \;=\;4a^{\frac{1}{3}}\!\!\int^a_0\!\! x^{-\frac{1}{3}}\,dx \;=\;4a^{\frac{1}{3}}\cdot\tfrac{3}{2}x^{\frac{2}{ 3}}\,\bigg]^a_0$

. . . . $=\;6a^{\frac{1}{3}}\bigg[a^{\frac{2}{3}} - 0^{\frac{2}{3}}\bigg] \;=\;6a^{\frac{1}{3}}\cdot a^{\frac{2}{3}} \;=\;6a$

5. Thank you so much.
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