Question

The graph of the equation is one of a family of curves called astroids because of their starlike appearance​ (as shown in the accompanying​ figure). Find the length of this particular astroid by finding the length of half the​ first-quadrant portion, , and multiplying by 8.

The length of the given astroid is

Answer: ______.

​(Simplify your answer. Type an exact​ answer, using radicals as​ needed.)

2/3 2/3 = 32/8 y(32/3 -2/3)3/2, JV

Answer 1

We know that length of a curve y f(x),a < x < b is given by 1 + l dx. 1) L 3 Z <x< 3 and multiply it by 8. (32/3 - x2/3)/, We have to find the length of the asteroid y Differentiating y = (32/3-x2/3)3/2 we get dy dr nx 1) dy da Squaring on both sides we get (2) 1 Substituting (2) in (1), and integrating between x< 3, we get L 3(cos e)3dx = (3)3(cos 0)(-sin 0)de Put x 3(cos ) = (cos 0)3 Changing the limits of integration, we get x (12 cos 0= 1 1 Taking the inverse on both sides we get 0 cos (Since, cos = 33(cos 0)3 = 3= (cos e)31 cos 0 And x 1 Taking the inve rse on both sides we get 0 cos 1 = (Since, cos7 = 1) (3(cos 0))3(9(cos 0)2 (3(cos 0) - sin (1 - (cos 0)2)3(cos 0)-2 (9(cos 0)2 sin ((cos 0)(sin 0) -9,/1 + (sin 0)2(cos 0) 9/1+ (tan e) ((cos 0)(sin 0 (sin 0)2/(cos e) = (tan 0)2 (Since (sin 0)(cos e) -9(sec 0)2 ( (cos 0)2 (sin e))de (Since, 1(tan 0)2 = (sec0)

= Si-9sec 0 ((cos 0)2(sin 0))d0 9cos sin ed0 (Since sec 0cos 1) 0sin Od0 (Dividing and multiplying with 2) 4 Esin 20d0 (Since 2sin 0 cos 0 = sin 20) COS 20 (Since sinx dx= - cos x) -I- s2C) G)-1-1 Sof(x)dx= F(b) -F(a)) (Since, +cos 2 - cos () COS Therefore, 36 units Multiplying the value in (3), by '8', we get total length of the Asteroid L8 x