Arc length problem?

T

twohaha

New member
Joined
Apr 7, 2012
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18
Find the arc length of

\(\displaystyle \ x^5/10 +1/6x^3 \) in [1,2]

First i found the derivative:

\(\displaystyle \ x^4 - 1/2x^4 \)

then, using the integral for arc lengths:
\(\displaystyle \ \int\ sqrt(1+ (x^4 - 1/2x^4)^2) \)

After simplifying, i get:

\(\displaystyle \ \int\ sqrt(x^8 - 1/4x^8) \)

i can't seem to integrate this, and wolfram gives me a ridiculously complicated integral...

Does anyone have any suggestions?

Thank you and Happy New Year!
 
twohaha said:
Find the arc length of

x^5/10 +1/(6x^3) in [1,2] <<<<<<<<< Please correct your post using proper grouping symbols

First i found the derivative:

x^4/2 - 1/2x^4 <<<<<<<<<<<<<<<< Incorrect dy/dx

then, using the integral for arc lengths:
\(\displaystyle \ \int\ sqrt(1+ (x^4 - 1/2x^4)^2) \)

After simplifying, i get:

\(\displaystyle \ \int\ sqrt(x^8 - 1/4x^8) \)

i can't seem to integrate this, and wolfram gives me a ridiculously complicated integral...

Does anyone have any suggestions?

Thank you and Happy New Year!
Click to expand...

.
 
Last edited by a moderator:
[h=2]From Wolfram:
Definite integral:[/h]
img
 
Hello, twohaha!

Your algebra is off . . .

\(\displaystyle \text{Find the arc length of: }\:y \:=\:\frac{1}{10}x^5 +\frac{1}{6x^3}\:\text{ in }[1,2]/\)
Click to expand...

We have: .\(\displaystyle y \:=\:\tfrac{1}{10}{x^5} + \tfrac{1}{6}x^{-3}\)

Then: .\(\displaystyle \frac{dy}{dx}\;=\;\frac{1}{2}x^4 - \frac{1}{2}x^{-4}\)

Then: .\(\displaystyle \left(\frac{dy}{dx}\right)^2 + 1 \;=\;\left(\frac{1}{2}x^4 - \frac{1}{2}x^{-4}\right)^2 + 1 \;=\;\frac{1}{4}x^8 - \frac{1}{2} + \frac{1}{4}x^{-8} + 1\)

. . . . . . . . . . . . . . \(\displaystyle =\;\frac{1}{4}x^8 + \frac{1}{2} + \frac{1}{4}x^{-8} + 1 \;=\;\left(\frac{1}{2}x^4 + \frac{1}{2}x^{-4}\right)^2\)

Hence: .\(\displaystyle \sqrt{\left(\frac{dy}{dx}\right)^2 + 1} \;=\;\sqrt{\left(\frac{1}{2}x^4 + \frac{1}{2}x^{-4}\right)^2} \;=\; \frac{1}{2}x^4 + \frac{1}{2}x^{-4}\)


Therefore: .\(\displaystyle \displaystyle L \;=\;\tfrac{1}{2}\int^2_1\left(x^4 + x^{-4}\right)\,dx\)

Got it?
 
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