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

Question

15687 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-10T15:13:47

#L = 3/2#

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]}