Respuesta :
we are given
equation of curve is
[tex] x=-2y^{\frac{3}{2}} [/tex]
Bound is
from y=0 to y=9
now, we can use arc length formula
[tex] L=\int\limits^a_b {\sqrt{1+(\frac{dx}{dy})^2}} \, dy [/tex]
Firstly , we will find dx/dy
that is derivative
[tex] x=-2y^{\frac{3}{2}} [/tex]
[tex] \frac{dx}{dy} =-2*\frac{3}{2} *y^{\frac{1}{2}} [/tex]
now, we can simplify it
[tex] \frac{dx}{dy} =-3y^{\frac{1}{2}} [/tex]
Arc length:
now, we can plug it in arc length formula
and we get
[tex] L=\int\limits^0_9 {\sqrt{1+(-3y^{\frac{1}{2}})^2}} \, dy [/tex]
now, we can solve it
Firstly , we will solve integral
[tex] \int \sqrt{1+\left(-3\sqrt{y}\right)^2}dy [/tex]
[tex] =\int \sqrt{9y+1}dy [/tex]
[tex] =\frac{2}{27}\left(9y+1\right)^{\frac{3}{2}} [/tex]
now, we can plug bounds
and we get
[tex] =\frac{164\sqrt{82}}{27}-\frac{2}{27} [/tex]
[tex] L=54.92901 [/tex]...........Answer
