फाइंड द पेरिमीटर ऑफ़ द एस्ट्रॉयड
Answers
Step-by-step explanation:
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.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
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} .
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.