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

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.
