# Curve Lengths

• Apr 1st 2008, 03:40 PM
N736RA
Curve Lengths
Don't have any real idea how to go about this one, even thought its probably really easy... thnanks for the help

Find the length of the curve:
x=[3y^(4/3)]-(3/32)y^(2/3)

-27<y<27 (both those are less than or equal to)

thanks
• Apr 1st 2008, 05:13 PM
Soroban
Hello, N736RA!

Quote:

Find the length of the curve: .$\displaystyle x\:=\:3y^{\frac{4}{3}} - \frac{3}{32}y^{\frac{2}{3}} \qquad -27 \:\leq\: y\:\leq \:27$

We will use this formula: . $\displaystyle L \;=\;\int^b_a\sqrt{1 + \left(\frac{dx}{dy}\right)^2}\,dy$

We have: . . . . . $\displaystyle x \;=\;3y^{\frac{4}{3}} - \frac{3}{32}y^{\frac{2}{3}}$

Differentiate: . $\displaystyle \frac{dx}{dy} \;=\;4y^{\frac{1}{3}} - \frac{1}{16}y^{-\frac{1}{3}}$

. . . . . . . . $\displaystyle \left(\frac{dx}{dy}\right)^2 \;=\;16y^{\frac{2}{3}} - \frac{1}{2} + \frac{1}{256}y^{-\frac{2}{3}}$

. . . . . . $\displaystyle 1 + \left(\frac{dx}{dy}\right)^2 \;=\;1 + \left(16y^{\frac{2}{3}} - \frac{1}{2} + \frac{1}{256}y^{-\frac{2}{3}}\right) \;=\;16y^{\frac{2}{3}} + \frac{1}{2} + \frac{1}{256}y^{-\frac{2}{3}} \;=\;\left(4y^{\frac{1}{3}} + \frac{1}{16}y^{-\frac{1}{3}}\right)^2$

. . . . $\displaystyle \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \;=\;4y^{\frac{1}{3}} + \frac{1}{16}y^{-\frac{1}{3}}$

You must integrate and evaluate: . $\displaystyle L \;=\;\int^{27}_{\text{-}27}\left(4y^{\frac{1}{3}} + \frac{1}{16}y^{-\frac{1}{3}}\right)\,dy$

• Apr 1st 2008, 06:08 PM
N736RA
thanks for the reply! Although, when I do that integral I get 0 (which makes sense), yet the program I'm using tells me that 0 isnt the right answer, any ideas? Edit: nevermind, I figured it out, thanks for your help!