Arc Length of an Astroid

Question

I want to find the arc length of the equation: $ x^{2/3} + y^{2/3} = 4 $

My steps follow as:

$ y= f(x) = (4-x^{2/3})^{3/2} $

$ f'(x)= -x^{-1/3}(4-x^{2/3})^{1/2} $

$ [f'(x)]^2 = 4x^{-2/3}-1 $

$ [f'(x)]^2 + 1 = 4x^{-2/3} $

$ L = \int_\sqrt8^4 \sqrt{f'(x)^2 + 1} dx = 0.0955 $

Bounds of the upper integral are found by:

$ y= (4-x^{2/3})^{3/2} = x $

$ x= 2 $ , $ x = \sqrt8 $

The integral gives us the asteroid's 1/8th of length. We multiply by 8 and finally get the result 0.7646.

I solved this problem by imitating the steps exactly of a similar problem with $ x^{2/3} + y^{2/3} = 1 $ instead of this problem's $ x^{2/3} + y^{2/3} = 4 $.

Change of the number from 1 to 4 should not change the method of solving this question right? According to my professor's answer, I should be getting 8 as a result. I am wondering if I did something wrong or I wrote my professor's answer as wrong.

PS: This is my first stackexchange post, feel free to point out if I could have done something better in my post.

There is a careless mistake at differentiation step. You forgot to times the original exponent. — Karn Watcharasupat, Nov 16 '17 at 08:42
$$a^{2/3}=4\to a=4^{3/2}=8;;L=48$$ https://en.wikipedia.org/wiki/Astroid#Equations — Raffaele, Nov 16 '17 at 08:44
You should get $6$ as a result, then multiply by 8 and get $L=48$ — Raffaele, Nov 16 '17 at 08:49
@KarnWatcharasupat Instead of multiplying it I divided apparently. Thanks for pointing it out. Now I get 6.88 which is much closer to the professor's answer. — utaysi, Nov 16 '17 at 08:55
@Raffaele But $ L = \int_\sqrt8^4 \sqrt{4x^{-2/3}} dx $ gives me 0.86, which I then multiply by 8 to get 6.88. How did I set the integral wrong? — utaysi, Nov 16 '17 at 09:03
@utaysi write it as $$\int_0^8\left( 2x^{-1/3}\right),dx$$ and multiply the result by $4$ — Raffaele, Nov 16 '17 at 09:11
Possible duplicate of Arc Length of $x^{\frac{2}{3}}+y^{\frac23}=1$ — nmasanta, Jul 02 '19 at 08:35

score 4 · Answer 1 · answered Nov 16 '17 at 19:13

An alternative method is to use the parametric form of the astroid, i.e.,

$$ x=8\cos^3t\\ y=8\sin^3t $$

Employing symmetry we can say that

$$ \begin{align} s &=4\int_0^{\pi/2}\sqrt{\dot x^2+\dot y^2}~dt\\ &=4\cdot8\cdot3\int_0^{\pi/2}\sqrt{\cos^4t\sin^2t+\sin^4t\cos^2t}~dt\\ &=4\cdot8\cdot3\int_0^{\pi/2}\cos t\sin t\sqrt{\cos^2t+\sin^2t}~dt\\ &=4\cdot8\cdot3\int_0^{\pi/2}\cos t\sin t~dt\\ &=4\cdot8\cdot3\cdot\frac{1}{2}\\ &=48 \end{align} $$

score 1 · Answer 2 · answered Jul 02 '19 at 08:51

The function is indeed $f(x)=(4-x^{2/3})^{3/2}$; let's go slowly with the derivative: $$ f'(x)=\frac{3}{2}(4-x^{2/3})^{1/2}\left(-\frac{2}{3}x^{-1/3}\right)=-\frac{(4-x^{2/3})^{1/2}}{x^{1/3}} $$ Hence $$ 1+(f'(x))^2=1+\frac{4-x^{2/3}}{x^{2/3}}=4x^{-2/3} $$ Now you want to compute the bounds. You get a quarter of the length with $$ \int_0^8\sqrt{1+(f'(x))^2}\,dx=\int_0^8 2x^{-1/3}\,dx=2\Bigl[\frac{3}{2}x^{2/3}\Bigr]_0^8=12 $$ Your bounds are wrong. If you want an eighth, intersect the astroid with $y=x$, which yields $2x^{2/3}=4$, so $x=2^{3/2}=\sqrt{8}$, but the upper bound should be when $y=0$, that is, $x=8$.

Arc Length of an Astroid

2 Answers2