Find the exact length of the curve x = e^y + 1/4e^(-y)

Here we do an example of the arc length of a sideways curve.

The function is x = e^y + 1/4e^(-y) for y in [0,1].

The keys to this particular example are realizing there is a perfect square under the radical once we do the differentiation and add 1 to (x')^2.

Thanks for watching!

[kriptos]

Thank you very much, very clear explanation!

[carultch]

Alternatively, rewrite it as an equivalent linear combination of coshes and sinches, whose arc length integral is extremely simple to do.

Given:
x = e^y + 1/4e^(-y)

Recall definitions:
cosh(y) = (e^y + e^(-y))/2
sinh(y) = (e^y - e^(-y))/2

x = A*cosh(y) + B*sinh(y)
x = A*(e^y + e^(-y))/2 + B*(e^y - e^(-y))/2 = e^y + 1/4e^(-y)

Equate coefficients of like terms:
A/2 + B/2 = 1
A/2 - B/2 = 1/4

Solve for A & B:
A = 5/4; B = 3/4

Thus:
x = 5/4*cosh(y) + 3/4*sinh(y)

Indefinite arc length integral for arc length of cosh, turns out to be sinch. And vice-versa. This coincidentally, is also their respective area integrals.

Proof:
L_cosh(x) = integral sqrt(1 + sinh(x)^2) dx = integral sqrt(cosh(x)^2) dx = integral cosh(x) dx = sinh(x)
L_sinh(x) = integral sqrt(1 + coshh(x)^2) dx = integral sqrt(sinh(x)^2) dx = integral sinh(x) dx = cosh(x)

Thus, the indefinite result for arc length is:
L_x = 5/4*sinh(y) + 3/4*cosh(y)

evaluate from y = 0 to 1:
L_x = [5/4*sinh(1) + 3/4*cosh(1)] - [5/4*sinh(0) + 3/4*cosh(0)]
L_x = -3/4 + 5/4*sinh(1) + 3/4*cosh(1)

Which simplifies to:
L_x = -3/4 - 1/(4*e) + e