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

Arc length is given by:

Integrate directly:

Insert the limits of integration:

To find the length of the curve ( (3y - 1)^2 = x^3 ) for ( 0 \leq x \leq 2 ), we can use the formula for arc length of a curve given by:

[ L = \int_{a}^{b} \sqrt{1 + \left(\frac{{dy}}{{dx}}\right)^2} , dx ]

First, we need to express ( y ) explicitly in terms of ( x ). From the given equation, we can solve for ( y ):

[ (3y - 1)^2 = x^3 ]

[ 3y - 1 = \pm \sqrt{x^3} ]

[ 3y = 1 \pm \sqrt{x^3} ]

[ y = \frac{1}{3} \pm \frac{\sqrt{x^3}}{3} ]

Now, we differentiate ( y ) with respect to ( x ) to find ( \frac{{dy}}{{dx}} ):

[ \frac{{dy}}{{dx}} = \frac{d}{dx}\left(\frac{1}{3} \pm \frac{\sqrt{x^3}}{3}\right) ]

[ = \frac{1}{6x^{1/2}} ]

Now, we substitute ( \frac{{dy}}{{dx}} ) into the arc length formula:

[ L = \int_{0}^{2} \sqrt{1 + \left(\frac{1}{6x^{1/2}}\right)^2} , dx ]

[ = \int_{0}^{2} \sqrt{1 + \frac{1}{36x}} , dx ]

This integral can be challenging to evaluate directly. You may need to use numerical methods or special techniques such as trigonometric substitution or integration by parts to find the exact value of ( L ).

To find the length of the curve ( (3y - 1)^2 = x^3 ) for ( 0 \leq x \leq 2 ), we use the arc length formula for parametric curves.

Given the equation ( (3y - 1)^2 = x^3 ), we can rewrite it in parametric form as ( x = t^2 ) and ( y = \frac{1}{3}t + \frac{1}{3} ).

To find the length of the curve, we integrate the square root of the sum of the squares of the first derivatives of ( x ) and ( y ) with respect to ( t ) over the interval ( 0 \leq t \leq 2 ).

The first derivatives are:

[ \frac{dx}{dt} = 2t ] [ \frac{dy}{dt} = \frac{1}{3} ]

Now, calculate the square root of the sum of the squares of the first derivatives:

[ \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} = \sqrt{(2t)^2 + \left(\frac{1}{3}\right)^2} ]

[ = \sqrt{4t^2 + \frac{1}{9}} ]

Integrating this expression with respect to ( t ) over the interval ( 0 \leq t \leq 2 ) will give us the length of the curve. The integral is:

[ \int_0^2 \sqrt{4t^2 + \frac{1}{9}} dt ]

This integral might not have a simple closed-form solution, so numerical methods or computational tools may be necessary to approximate the length of the curve.