1. exact length of curve

First, here's the problem:

A curve is given by $\displaystyle y=(9-x^\frac{2}{3})^\frac{3}{2} ; 1\leq y\leq 8$. Find the exact length of the curve analytically by antidifferentiation.

First I found the derivative of $\displaystyle y=(9-x^\frac{2}{3})^\frac{3}{2}$, then I tried to integrate it from 1 to 8. The answer I got is 10.6712.

2. Originally Posted by linlinrocks
First, here's the problem:

A curve is given by $\displaystyle y=(9-x^\frac{2}{3})^\frac{3}{2} ; 1\leq y\leq 8$. Find the exact length of the curve analytically by antidifferentiation.

First I found the derivative of $\displaystyle y=(9-x^\frac{2}{3})^\frac{3}{2}$, then I tried to integrate it from 1 to 8. The answer I got is 10.6712.

You need to find the derivative, square it, and evaluate $\displaystyle L=\int_a^b\sqrt{1+\left(\frac{\,dy}{\,dx}\right)^2 }\,dx$

Since $\displaystyle y=\left(9-x^\frac{2}{3}\right)^{\frac{3}{2}}$, it follows that $\displaystyle y^{\prime}=\frac{3}{2}\left(9-x^{\frac{2}{3}}\right)^{\frac{1}{2}}\cdot-\tfrac{2}{3}x^{-\frac{1}{3}}$

Squaring the derivative, we have $\displaystyle \left(y^{\prime}\right)^2=x^{-\tfrac{2}{3}}\left(9-x^{\frac{2}{3}}\right)=9x^{-\frac{2}{3}}-1$

Now, we see that $\displaystyle L=\int_1^8\sqrt{1+\left(9x^{-\frac{2}{3}}-1\right)}\,dx=\int_1^8\sqrt{9x^{-\frac{2}{3}}}\,dx=\int_1^8 3x^{-\frac{1}{3}}\,dx$

We see that this becomes $\displaystyle \tfrac{9}{2}\left.\left[x^{\frac{2}{3}}\right]\right|_1^8=\tfrac{9}{2}\left(8^{\frac{2}{3}}-1^{\frac{2}{3}}\right)=\tfrac{9}{2}\left(4-1\right)=\tfrac{27}{2}$

So the arclength should be $\displaystyle \boxed{L=\tfrac{27}{2}}$.

Does this make sense?

3. Originally Posted by linlinrocks
First, here's the problem:

A curve is given by $\displaystyle y=(9-x^\frac{2}{3})^\frac{3}{2} ; 1\leq y\leq 8$. Find the exact length of the curve analytically by antidifferentiation.

First I found the derivative of $\displaystyle y=(9-x^\frac{2}{3})^\frac{3}{2}$, then I tried to integrate it from 1 to 8. The answer I got is 10.6712.