How do you find the arc length of the curve $f (x) = cosh x$ $f(x)=coshx$ over the interval [0, 1]?

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Answer 1 · 2016-10-28T18:46:51

I used WolframAlpha
$s = \int_{0}^{1} \sqrt{1 + {sinh}^{2} (x)} d x \approx 1.1752$ $s = int_0^1 sqrt(1 + sinh^2(x))dx ≈ 1.1752$

From the reference on Arc Length we write the equation:

$s = \int_{a}^{b} \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x$ $s = int_a^b sqrt(1 + (dy/dx)^2)dx$

Given: $a = 0, b = 1, and y = cosh (x)$ $a = 0, b = 1, and y = cosh(x)$

We know that $\frac{d y}{d x} = sinh (x)$ $(dy)/(dx) = sinh(x)$

The arc length integral is:

$s = \int_{0}^{1} \sqrt{1 + {sinh}^{2} (x)} d x \approx 1.1752$ $s = int_0^1 sqrt(1 + sinh^2(x))dx ≈ 1.1752$

How do you find the arc length of the curve f(x)=coshxf(x)=coshxf(x)=coshx over the interval [0, 1]?