calc: arc length

Posted by COFFEE on Monday, June 11, 2007 at 11:48pm.

find the exact length of this curve:

y = ( x^3/6 ) + ( 1/2x )

1/2 <or= x <or= 1

im looking over my notes, but i'm getting stuck.

here's my work so far:

A ( 1 , 2/3 )
B ( 1/2 , 49/48 )

y' = [1/6 (3x^2)] + [1/2 (-1x^-2)]

y' = ( x^2 / 2 ) - ( x^-2 / 2 )

(y')^2 = [( x^2 / 2 ) - ( x^-2 / 2 )]^2

y = (x^4 / 4) - (1/2) + (x^-4 / 4)

Integral: [from 1 to 1/2]
( 1 )
∫ sqrt[1 + (x^4 / 4) - (1/2) + (x^-4 / 4)] dx
( 1/2 )

( 1 )
∫ sqrt[(x^4 / 4) + (1/2) + (x^-4 / 4)] dx
( 1/2 )

..now i'm stuck. hopefully i computed it correctly.. how do i finish this to get a numerical answer? please help! thanks!!

For Further Reading

* calc: arc length - Count Iblis, Tuesday, June 12, 2007 at 9:37am

Write the integral as:

Integral of
sqrt[(x^2/2 - x^(-2)/2)^2 + 1] dx

from x = 1/2 to x = 1

substitute x = Exp(t)

Then

x^2/2 - x^(-2)/2 =

[Exp(2t) - Exp(-2t)]/2 = Sinh(2t)

And thus:

[(x^2/2 - x^(-2)/2)^2 + 1 =

Sinh(2t)^(2) + 1 = Cosh(2t)^(2)

And we have:

sqrt[(x^2/2 - x^(-2)/2)^2 + 1] =

Cosh(2t)

Finally, using that:

dx = Exp(t) dt

we can write the integral as:


Integral of Cosh(2t) Exp(t) dt

from t = -Log(2) to 0 =

Integral of 1/2[Exp(3t) + Exp(-t)] dt

from t = -Log(2) to 0 =

95/144

... this answer was wrong. did Count Iblis do it correctly or is there another way to end this problem? thanks!

I don't see any mistakes here. Are you sure that 1/2x means 1/(2x)? this is not the standard convention...

Anyway, just try to find the error for a few more minutes and then just move on to other problems. Sometimes there is an error in the textbook, your Prof. can make errors.

You don't want to waste hours of your time just to find out that that you made a trivial error somewhere or someone else did. You don't learn anything from that.

What matters is that you understand calculus, in this case how to compute arc length and to do that you must master integration.

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