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How do I find the arch length of

$$x^{\frac{2}{3}}+y^{\frac23}=1$$

The hint given was "4x the arc in first quadrant"

I think I am supposed to use the formula:

$$L=\int^b_a \sqrt{1+(f'(x))^2} dx$$


I tried plotting the equation in a graphing utility like https://www.desmos.com/calculator, which results in an error.

So I tried expanding the term :

$$y=\sqrt{(1-x^{\frac{2}{3}})^3} = \sqrt{1-2x^{\frac{2}{3}}-x^2}$$

Then how do I proceed? Complete the square? Doesn't look likes its in an appropriate form?

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3 Answers 3

up vote 5 down vote accepted

Note that the $x^{2/3}+y^{2/3}=1$ can be parametrized by $x(t)=\cos^3 t$ and $y(t)=\sin^3 t$, where $0\leq t\leq 2\pi$. Now you can apply the formula (see here) $$s=\int_0^{2\pi}\sqrt{\left(\frac{dx(t)}{dt}\right)^2+\left(\frac{dy(t)}{dt}\right)^2}dt$$ to find the arclength.

Note added: This is how you can finish the problem: since $x(t)=\cos^3 t$ and $y(t)=\sin^3 t$, we have $\displaystyle\frac{dx(t)}{dt}=-3\cos^2t \sin t$ and $\displaystyle\frac{dy(t)}{dt}=3\sin^2t \cos t$. Substitute it into above formula, we have $$s=\int_0^{2\pi}\sqrt{9\cos^2 t\sin^2 t}dt=4\int_0^{\pi/2}3\cos t\sin tdt=6\int_0^{\pi/2}\sin (2t)dt=-3\cos 2t\Big|_0^{\pi/2}=6.$$

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Ok, I get something like $$y=\sqrt{(1-\cos^2{\theta})^3}=\sqrt{\sin^{\frac{5}{2}}\theta}$$, I must have misunderstood you somewhere? –  Jiew Meng Apr 11 '12 at 14:32
    
You have to use the parametrization for both $x$ and $y$, that is, $x=\cos^3 t$ and $y=\sin^3 t$, and then use the arc length formula I have mentioned instead of your formula $\int^b_a \sqrt{1+(f'(x))^2} dx$. See my edited answer. –  Paul Apr 11 '12 at 14:42
    
$(\frac{dy(t)}{dt})^2 = (3\sin^2t \cos t)^2=9\sin^4{t}\cos^2{t}$. Therefore, $(\frac{dx(t)}{dt})^2+(\frac{dy(t)}{dt})^2=9\cos^4{t}\sin^2{t}+9\sin^4{t}\cos^2{‌​t}=9\cos^2{t}\sin^2{t}$. –  Paul Apr 11 '12 at 15:03
    
How did you get from $$s=\int_0^{2\pi}\sqrt{9\cos^2 t\sin^2 t}dt=4\int_0^{\pi/2}3\cos t\sin tdt$$? How did the 4 appear? I am getting $$\frac{3}{2}\int{\sin{2\theta}}$$ –  Jiew Meng Apr 11 '12 at 15:11
    
@Jiew: limits of the integral. Left hand side integrates from 0 to $2\pi$. Right hand side integrates from 0 to $\pi/2$.(Which is necessary since $\sqrt{x^2} = |x|$ and so we use the symmetry of the trig functions and take four times the integral over the region where the integrand is positive.) –  Willie Wong Apr 11 '12 at 15:21

$$(y')^2=\frac{y^{2/3}}{x^{2/3}} ~\text {and }~ y^{2/3}=1-x^{2/3}$$

Hence :

$$L=\int\limits_{a}^{b} \sqrt{1+\frac{y^{2/3}}{x^{2/3}}} \,dx$$

$$L=\int\limits_{a}^{b} \sqrt{\frac{1}{x^{2/3}}} \,dx$$

$$L=\int\limits_{a}^{b} x^{-1/3} \,dx$$

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How did you get the 1st equation? $(y')^2=\frac{y^{2/3}}{x^{2/3}}$? –  Jiew Meng Apr 11 '12 at 14:36
    
I did $$y=(1-x^{2/3})^{3/2}$$ $$y'= \frac{3}{2}(1-x^{2/3})(-\frac{2}{3} x^{-1/3})=\frac{x^{2/3}-1}{\sqrt[3]{x}}$$? –  Jiew Meng Apr 11 '12 at 14:45
    
@JiewMeng Implicit Differentiation –  pedja Apr 11 '12 at 15:07

The error you found is probably due to trying to take a negative number to the $2/3$rds power. If we assume that $x^{2/3} := (x^{2})^{1/3}$ then the solution set of $x^{2/3}+y^{2/3}=1$ is symmetric about the $x$ axis (since if $(x,y)$ is an element of the solution set, then so to is $(x,-y)$) and symmetric about the $y$-axis (since if $(x,y)$ is an element of the solution set, then so too is $(-x,y)$.) These symmetries imply the hint.

So if $L$ is the arclength of the entire solution set, then \begin{aligned} L=4\int_{0}^{1} \sqrt{1+f'(x)^{2}} \, dx \end{aligned} From here, it's simply aggravation and annoyance: $$ y^{2/3}=1-x^{2/3} \Longrightarrow y=f(x)=(1-x^{2/3})^{3/2} $$ Then we compute the derivative: $$ f'(x) = \frac{3}{2}(1-x^{2/3})^{1/2}\cdot\left(-\frac{2}{3}x^{-1/3}\right) =-\frac{\sqrt{1-x^{2/3}}}{x^{1/3}} $$ and its square $$ f'(x)^{2} = \frac{1-x^{2/3}}{x^{2/3}}=-1+x^{-2/3} $$ so that our integrand is $$ \sqrt{1+f'(x)^{2}}= x^{-1/3} $$ Then $4\int_{0}^{1}x^{-1/3} \, dx = 6$.

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