Question

How do you find the length of the curve `x^(2/3)+y^(2/3)=1` for the first quadrant?

Answer 1

`L = 3/2`

Explanation:

We can write the curve in parametric form by posing:

`x^(2/3)=t`

so that:

`y^(2/3) = 1-t`

As in the first quadrant `x` and `y` are positive:

`{(x=t^(3/2)),(y=(1-t)^(3/2)):}`

where `0 <= t <= 1`.

The length of the curve is therefore:

`L= int_0^1 sqrt( (dx/dt)^2+(dy/dt)^2)dt`

`L= int_0^1 sqrt( (3/2t^(1/2))^2+(-3/2(1-t)^(1/2))^2)dt`

`L= int_0^1 3/2sqrt( t+(1-t))dt`

`L= 3/2 int_0^1dt = 3/2`

graph{x^(2/3)+y^(2/3) = 1 [-2.5, 2.5, -1.25, 1.25]}