Find the perimeter of an astroid with equation x^(2/3) + y^(2/3) = 64. (Hint: find the arclength of the portion of the curve in the first quadrant and use symmetry).

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lfryerda's profile pic

lfryerda | High School Teacher | (Level 2) Educator

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An astroid is symmetrical in reflections of both the x-axis and the y-axis.  This means that the perimeter of the astroid is 4 times the length of the segment in the first quadrant.

Also, the intercepts of the astroid in the first quadrant are (0, 16) and (16,0).  The formula for the length of a line segment is 

`s=int_a^bsqrt{1+({dy}/{dx})^2}dx`

In this case, we can isolate y to get:

`x^{2/3}+y^{2/3}=64`

`y^{2/3}=64-x^{2/3}`    now differentiate

`2/3 y^{-1/3}{dy}/{dx}=-2/3x^{-1/3}`   divide coefficients

`{dy}/{dx}=-y^{1/3}/x^{1/3}`   square both sides

`({dy}/{dx})^2=y^{2/3}/x^{2/3}`   sub in equation from above

`={64-x^{2/3}}/x^{2/3}`

`=64x^{-2/3}-1`

Now sub into the arclength formula

`s=int_0^16sqrt{1+64x^{-2/3}-1}dx`

`=int_0^16sqrt{64x^{-2/3}}dx`

`=int_0^16 8 x^{-1/3}dx`   now use power law for integrals

`=24/2(x^{2/3})|_0^16`   simplify

`=12(16)^{2/3}`

The perimeter of the astroid is four times this, which is `48(16)^{2/3}` .

mathsworkmusic's profile pic

mathsworkmusic | (Level 2) Educator

Posted on

We have `x^(2/3) + y^(2/3) =64`

Make a change of variables `u = x^(2/3)`  and `v = y^(2/3)`   so that we have

`u + v = 64`

Now, the arclength `s` over `[a,b]` satisfies

`ds^2 = dx^2 + dy^2 = ((dx)/(du))^2du^2 + ((dy)/(dv))^2 dv^2`

`implies`

`(ds^2)/(du^2) = ((dx)/(du))^2 + ((dy)/(dv))^2(dv^2)/(du^2)`

`implies`

`s = int_a^b sqrt(((dy)/(dv))^2(dv^2)/(du^2) + ((dx)/(du))^2) du `

In the first quadrant `u` ranges from 0 to 64

Also, `(dy)/(dv) = 3/2sqrt(v)`  `implies ((dy)/(dv))^2 = 9/4v`   and `(dv)/(du) = -1`  and `(dx)/(du) = 3/2sqrt(u)`

`implies ((dx)/(du))^2 = 9/4u`

`therefore`

`s = int_0^64 (sqrt(9/4(64-u)(-1)^2 + 9/4u)) du`

` = int_0^64 sqrt(144) du = int_0^64 12 du = 12u|_0^64 = 768`

 The total perimeter of the astroid is 4s = 4(768) = 3072

 

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