Graph the curve and find its exact length.?

Question

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

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Calculus

Answer 1

An infinitesmal arc length, ds, by virtue of pythagoras, can be expressed as:

ds^2 = dy^2 + dx^2

by virtue of calculus, dy = (dy/dx)dx, so

ds = sqrt((dy/dx)^2 + 1)dx.

Integrate this to find exact length.

Answer 2

Take the integral of the speed.

f(x) = (x,y)
v(x) = f'(x) = (1,y')
s(x) = sqrt(1*1 + y'*y')
length = integral(1/2..1)(s(x) dx)

I hope you've talked about these things, otherwise you may not know what's going on.

y' = (x^2)/3 -1/(2x^2)
(y')^2 = (x^4)/9 - 1/3 + 1/(4x^4)
length = integral(1/2..1)(sqrt(1 + (y')^2) dx)

That integral is not very nice to evaluate. You need to do a change of variables.

Answer 3

y= (x^3)/(6) + (1/(2x))
y' = x^/2 -x^-2/2
length = l = integral x=.5 to 1 (1+(x^2/2 -x^-2/2)^2)^.5 dx