- #1
hungryhippo
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Homework Statement
Find the arc length of the function f(x) = x(sqrt(x/2-x) from 0 to x
Integral forms involving a+bu
--> integral [sqrt(a+bu)]/u^2du
Homework Equations
Arc Length = integral sqrt(1+(f'(x))^2)dx
The Attempt at a Solution
First, I took the derrivate
f'(x) = sqrt(x/2-x) + x/[(2-x)^2(sqrt(x/2-x))]
= [x(2-x)+x]/[(2-x)^2(sqrt(x/2-x))]
=[x(3-x)]/[(2-x)^2(sqrt(x/2-x)]
Then, the square of f'(x)
= [x^2(3-x)^2]/[x(2-x)^3]
=[x(3-x)^2]/[(2-x)^3]
Then I added 1 to the square of the derivative, which simplified to
=[8-3x]/[2-x]^3
Then, the square root of the above
=sqrt([8-3x]/[2-x]^3)
I substituted 2-x = t
=sqrt[(3t+2)/(t^3)] <--multiplied top and bottom by sqrt[t]
=sqrt(3t^2+2t)/t^2
I'm stuck here..I'm allowed to use formulas from a table of integrals (i.e. the one specified above), but I can't seem to get it into that form ..