<s>[INST] Consider the curve with implicit equation... I have the last question on my assignment which I cannot answer and was wondering if any of you could help me.
The two part question reads
Consider the curve with implicit equation $x^\frac{2}{3} + y^\frac{2}{3} = 4$. By using implicit differentiation show that:
a) $1 + \left(\frac{dy}{dx}\right)^2 = \frac{4}{x^\frac{2}{3}}$ , and
b) Using the result from (a), show that the arc length of the curve represented by the equation $x^\frac{2}{3} + y^\frac{2}{3} = 4$ over the interval x=1 to x=8 is 9 units.
So far I've managed to find the value of $\frac{dy}{dx}$
$\begin{aligned}
\frac{2}{3}x^\frac{-1}{3} + \frac{2}{3}y^\frac{-1}{3}\frac{dy}{dx} = 0\\
\\
\frac{2}{3}y^\frac{-1}{3}\frac{dy}{dx} = -\frac{2}{3}x^\frac{-1}{3}\\
\\
\frac{dy}{dx} = -\frac{\frac{2}{3}x^\frac{-1}{3}}{\frac{2}{3}y^\frac{-1}{3}}\\
\\
\frac{dy}{dx} = -\frac{x^\frac{-1}{3}}{y^\frac{-1}{3}}\\
\\
\frac{dy}{dx} = -\frac{y^\frac{1}{3}}{x^\frac{1}{3}}\\
\\
\frac{dy}{dx} = -^3\sqrt\frac{y}{x}\\
\end{aligned}$
I'm not sure how to proceed from this point to prove part (a) and complete part (b). If anyone could shed some light would be most appreciated.
Milo
 [/INST] You are almost done. To complete (a),
\begin{align}
1+\left(\frac{dy}{dx}\right)^2 &= 1 + \frac{y^{\frac{2}{3}}}{x^{\frac{2}{3}}}\\
&= 1 + \frac{4-x^{\frac{2}{3}}}{x^{\frac{2}{3}}}\\
&=\frac{4}{x^{\frac{2}{3}}}.
\end{align}
The arc length of $y=f(x)$ from $(a,f(a))$ to $(b,f(b))$ is given by
$$
L=\int_a^b \sqrt{1+(f'(x))^2} dx.
$$
Therefore, you can get $L$ by substituting $1+(f'(x))^2=4x^{-\frac{2}{3}}$ from (a), and $a=1$ and $b=8$ from initial conditions.
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