How do you find the lengths of the curve $(3y-1)^2=x^3$ for $0<=x<=2$ ?

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Answer 1 · 2018-06-28T05:49:23

Isolate $y$ then apply the arclength formula.

$(3y-1)^2=x^3$ , $x in [0,2]$

Isolate $y$ :

$y=1/3(x^(3/2)+1)$

Take the derivative with respect to $x$ :

$y'=1/2sqrtx$

Arc length is given by:

$L=int_0^2sqrt(1+x/4)dx$

Integrate directly:

$L=8/3[(1+x/4)^(3/2)]_0^2$

Insert the limits of integration:

$L=8/3((3/2)^(3/2)-1)$

How do you find the lengths of the curve (3y-1)^2=x^3(3y-1)^2=x^3 for 0<=x<=20<=x<=2?