# exact length of curve

• May 8th 2009, 09:06 PM
linlinrocks
exact length of curve
First, here's the problem:

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

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

• May 8th 2009, 09:29 PM
Chris L T521
Quote:

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

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

First I found the derivative of $y=(9-x^\frac{2}{3})^\frac{3}{2}$, 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 $L=\int_a^b\sqrt{1+\left(\frac{\,dy}{\,dx}\right)^2 }\,dx$

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

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

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

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

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

Does this make sense?
• May 8th 2009, 11:38 PM
matheagle
Quote:

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

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

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