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!!
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
1 answer