Find the arc length

Question

Find the arc length of $y=\left(\frac x2\right)^{2/3}$ with $x\in(0,2)$.

I tried solving it from x but I couldn't find it, problem says I must pass it into terms of Y in order to solve it.

Answer 1

arc length $= \int_0^2 (1+(\frac {dy}{dx})^2)^\frac 12 dx$

$\frac {dy}{dx} = \frac 13 (\frac {x}{2})^{-\frac 13}\\ (\frac {dy}{dx})^2 = k x^{-\frac 23}\\ k = \frac {2^\frac 23}{9}$

$= \int_0^2 (1+ kx^{-\frac 23})^\frac 12\ dx$

hmmm that looks ugly.

lets integrate with respect to y.

$= \int_0^1 (1+(\frac {dx}{dy})^2)^\frac 12 dx$

$x = 2y^\frac 32\\ \frac {dx}{dy} = 3 y^\frac 12\\ (\frac {dx}{dy})^2 = 9 y\\ \int_0^1 (1+9y)^\frac 12 dy\\ \frac {2}{3}\frac {1}{9}(1+9y)^\frac 32|_0^1\\ (\frac {2}{27})((10)^\frac 32 - 1)$

Answer 2

The formula to find arc length is:

$$\int_{a}^{b} \sqrt{1 + \left(\dfrac{dy}{dx}\right)^{2}} dx$$

Here we have $y = \left(\frac{x}{2}\right)^{\frac{2}{3}}$, so that $\dfrac{dy}{dx} = \frac{1}{3}x^{-\frac{1}{3}}$, and hence we're integrating:

$$\int_{0}^{2} \sqrt{1 + \frac{1}{9}x^{-\frac{2}{3}}}dx$$

Can you integrate this function?

Answer 3

I think in order to avoid the non-integral exponents, its better to find the arc length by parametrizing the curve, so I present a solution using parametric equations first, and then the solution you proposed, in Cartesian coordinates in the second section.

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Method 2: Parametric Equations

In order to parametrize the curve, we see that: \begin{align*} y &= \left( \frac{x}{2} \right)^{2/3} \\ y^3 &= \left( \frac{x}{2} \right)^2 \\ 4y^3 &= x^2 \end{align*} So we can parametrize the curve as: \begin{align*} x &= 2t^3 & y &= t^2 \end{align*} We see that going from $x = 0$ to $x = 2$ corresponds to $t = 0$ to $t = 1$. So our arc length integral is: \begin{align*} L &= \int_0^1 \sqrt{\left(\frac{dx}{dt} \right)^2 + \left(\frac{dy}{dt} \right)^2} \, dt \\ &= \int_0^1 \sqrt{\left(6t^2 \right)^2 + \left(2t \right)^2} \, dt \\ &= \int_0^1 \sqrt{36t^4 + 4t^2} \, dt \\ &= \int_0^1 2t\sqrt{9t^2 + 1} \, dt \\ &= \int_0^1 \sqrt{9u + 1} \, du \\ &= \left[\frac{2}{27}(9u+1)^{3/2} \right]_0^1 \\ &= \frac{20\sqrt{10}-2}{27} \end{align*}

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Method 1: Cartesian Coordinates

We recall that the formula for the arclength of $f(x)$ from $x = a$ to $x=b$ is: $$ L = \int_a^b \sqrt{1+ \left(\frac{dy}{dx} \right)^2} \, dx $$ We have our curve as: $$ y = \left( \frac{x}{2} \right)^{2/3} $$ Differentiating, the power rule will give us a power of $x$ with the denominator still 3, but we would a denominator of 2, so that we get an integral power in square root of the arc length formula. So we take another view on the curve: \begin{align*} y &= \left( \frac{x}{2} \right)^{2/3} \\ y^{3/2} &= \frac{x}{2} \\ 2y^{3/2} &= x \\ \end{align*} So now we look at the curve from the $x(y)$ perspective rather than the $y(x)$ perspective, and note that the curve goes from $x = 0$ to $x = 2$, for choice of $y = 0$ to $y = 1$. Therefore, we have: \begin{align*} L &= \int_0^1 \sqrt{1 + \left(\frac{dx}{dy} \right)^2} \, dy \\ &= \int_0^1 \sqrt{1 + \left(3y^{1/2} \right)^2} \, dy \\ &= \int_0^1 \sqrt{1 + 9y} \, dy \\ &= \left[\frac{2}{27}(1+9y)^{3/2} \right]_0^1 \\ &= \frac{20\sqrt{10}-2}{27} \end{align*}

Answer 4

I demonstrate how to carry out the solution in the complex plane, for a different point of view. Consider that

$$z=x+iy=x+i\left(\frac{x}{2}\right)^{2/3}$$

and that arc length is given in general by

$$s=\int|\dot z|du$$

Thus, for this problem we have

$$\dot z=1+i\frac{1}{3}\left(\frac{x}{2}\right)^{-1/3}\\ |\dot z|=\sqrt{1+\frac{1}{9}\left(\frac{x}{2}\right)^{-2/3}}\\ s=\int_0^2\sqrt{1+\frac{1}{9}\left(\frac{x}{2}\right)^{-2/3}}\ dx $$

Or, we could change variables from $x$ to $y$ with

$$s=\int_0^1\sqrt{1+\frac{1}{9y}}\ \frac{dy}{dx}\ dy=\int_0^1\sqrt{1+9y}\ dy, \quad \frac{dy}{dx}=3\sqrt{y} $$

By either means we obtain

$$s=\frac{2}{27}(10^{3/2}-1)$$

This result is clearly in agreement with the other answers.