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Calculus Level 2

What is the length of the arc of the curve $x^{\frac{2}{3}} + y^{\frac{2}{3}}=4$ ?

24 48 12 96

2 solutions

Lokeshwar Tabjula
Aug 22, 2014

We use the parametric equations $x = 8\cos^3t$ and $y = 8\sin^3t$ to describe the given curve.

Now, we use the arclength formula for parametric equations: $L = \int_{a}^{b}\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}dt$

We'll find the length of the curve that exists solely in the first quadrant and then multiply this by 4. $\int_{0}^{\frac{\pi}{2}}\sqrt{\left(-24\cos^2t \sin{t} \right)^2 + \left(24\sin^2t \cos{t} \right)^2}dt$ $\int_{0}^{\frac{\pi}{2}}24\sin{t}\cos{t}dt = 12\sin^2t \Big|_{0}^{\frac{\pi}{2}} = 12$

Multiplying this by four gives an area of 48.

Multiply by 4 gives an arc length of 48

Jerry McKenzie - 8 months, 3 weeks ago

You don’t need to do any of this. If you retool the equation into the equation of a circle using some algebra voodoo, you can use s = r(theta) to find the exact arc length as 16pi.

x^(2/3) + y^(2/3) = 4

If you write 4 as 2^(2) and put both the left and right sides to the third power, you’ll see that cuberoot(2^6) = 4, so the equation’s arc length is equivalent to

x^2 + y^2 = 64.

The angle in radians is 2pi, so 8*2pi = 16pi ~ 50.26

Jesse Olsson - 4 months ago

how can you come up with the parametrization of $x$ and $y$ ?

Bostang Palaguna - 2 months, 3 weeks ago

Bostang Palaguna
Mar 20, 2021

Standard length of curve calculation (without using parametrization):

$x^{2/3} + y^{2/3} = 4 \leftrightarrow y = (4-x^{2/3})^{3/2}$

so, $y' = -\sqrt{4-x^{2/3}}x^{-1/3}$

notice that the graph would produce the same curve for 4 quadrants, so:

$L = 4 \times \int_{0}^{8} \sqrt{1 + y'^2}$ $= 4 \times \int_{0}^{8} \sqrt{1 + \frac{4-x^{2/3}}{x^{2/3}}}$ $= 4 \times \int_{0}^{8} 2 \sqrt{x^{-2/3}}$ $= 4 \times \int_{0}^{8} 2 x^{-1/3}$ $= \boxed{48}$

Astroid!

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