Question

Find the arc length of the curve x=−2y^3/2 from y=0 to y=9.

Answer 1

we are given

equation of curve is

`x=-2y^{(3)/(2)}`

Bound is

from y=0 to y=9

now, we can use arc length formula

`L=\int\limits^a_b {\sqrt{1+((dx)/(dy))^2}} \, dy`

Firstly , we will find dx/dy

that is derivative

`x=-2y^{(3)/(2)}`

`(dx)/(dy) =-2*(3)/(2) *y^{(1)/(2)}`

now, we can simplify it

`(dx)/(dy) =-3y^{(1)/(2)}`

Arc length:

now, we can plug it in arc length formula

and we get

`L=\int\limits^0_9 {\sqrt{1+(-3y^{(1)/(2)})^2}} \, dy`

now, we can solve it

Firstly , we will solve integral

`\int \sqrt{1+\left(-3√(y)\right)^2}dy`

`=\int √(9y+1)dy`

`=(2)/(27)\left(9y+1\right)^{(3)/(2)}`

now, we can plug bounds

and we get

`=(164√(82))/(27)-(2)/(27)`

`L=54.92901`...........Answer