Answer:
6
Step-by-step explanation:
An astroid is a hypocycloid with four cusps. It has an equal length in all four quadrants due to its symmetric graph.
The formula for the arc length of a curve is given by:
[tex]\displaystyle L=\int^b_a\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\;dx[/tex]
First, rewrite the equation of the astroid as a function of x by isolating y:
[tex]x^{\frac23}+y^{\frac23}=1\\\\\\y^{\frac23}=1-x^{\frac23}\\\\\\y=\left(1-x^{\frac23}\right)^\frac{3}{2}[/tex]
Now, differentiate with respect to x:
[tex]\dfrac{dy}{dx}=\dfrac{3}{2}\left(1-x^{\frac23}\right)^{\frac{1}{2}}\cdot \dfrac{d}{dx}\left(1-x^{\frac23}\right)\\\\\\\\\dfrac{dy}{dx}=\dfrac{3}{2}\left(1-x^{\frac23}\right)^{\frac{1}{2}}\cdot \left(-\dfrac{2}{3}x^{-\frac13}\right)\\\\\\\\\dfrac{dy}{dx}=-\left(1-x^{\frac23}\right)^{\frac{1}{2}}\cdot \left(x^{-\frac13}\right)\\\\\\\dfrac{dy}{dx}=-\dfrac{\left(1-x^{\frac23}\right)^{\frac{1}{2}}}{x^{\frac13}}[/tex]
Now determine 1 + (dy/dx)²:
[tex]1+\left(\dfrac{dy}{dx}\right)^2=1+\left(-\dfrac{\left(1-x^{\frac23}\right)^{\frac{1}{2}}}{x^{\frac13}}\right)^2[/tex]
[tex]1+\left(\dfrac{dy}{dx}\right)^2=1+\dfrac{\left(1-x^{\frac23}\right)}{x^{\frac23}}[/tex]
[tex]1+\left(\dfrac{dy}{dx}\right)^2=\dfrac{x^{\frac23}}{x^{\frac23}}+\dfrac{\left(1-x^{\frac23}\right)}{x^{\frac23}}[/tex]
[tex]1+\left(\dfrac{dy}{dx}\right)^2=\dfrac{x^{\frac23}+1-x^{\frac23}}{x^{\frac23}}[/tex]
[tex]1+\left(\dfrac{dy}{dx}\right)^2=\dfrac{1}{x^{\frac23}}[/tex]
[tex]1+\left(\dfrac{dy}{dx}\right)^2=x^{-\frac23}[/tex]
Substitute this expression into the arc length formula over the interval [0, 1] to calculate the arc length within quadrant I. To obtain the total arc length, multiply the integral by 4, as the astroid exhibits symmetry with respect to both the x-axis and y-axis.
[tex]\displaystyle L=4\int^1_0\sqrt{x^{-\frac23}}\;dx[/tex]
Integrate:
[tex]\displaystyle L=4\int^1_0\left(x^{-\frac23}\right)^{\frac12}\;dx[/tex]
[tex]\displaystyle L=4\int^1_0x^{-\frac13}\;dx[/tex]
[tex]\displaystyle L=4\left[\dfrac{x^{-\frac{1}{3}+1}}{-\frac{1}{3}+1}\right]^1_0[/tex]
[tex]\displaystyle L=4\left[\dfrac{x^{\frac{2}{3}}}{\frac{2}{3}}\right]^1_0[/tex]
[tex]\displaystyle L=4\left[\dfrac{3x^{\frac{2}{3}}}{2}\right]^1_0[/tex]
[tex]\displaystyle L=4\left[\left(\dfrac{3(1)^{\frac{2}{3}}}{2}\right)-\left(\dfrac{3(0)^{\frac{2}{3}}}{2}\right)\right][/tex]
[tex]\displaystyle L=4\left[\dfrac{3}{2}-0\right][/tex]
[tex]\displaystyle L=\dfrac{12}{2}[/tex]
[tex]\displaystyle L=6[/tex]
Therefore, the arc length of the astroid is 6.
