What is the arc length of #f(x)=cosx# on #x in [0,pi]#?

Question

Samantha Adkison · Accepted Answer

#2E(-1)#, where #E# is the complete elliptic integral of the second kind. There is no simpler way of expressing this analytically. Numerically, this is approximately equal to 3.82.

Arc length of a curve #f(x)# is given by the formula #s=int_(x_0)^(x_1) sqrt(1+f'(x)^2) dx#. Note that it is often a difficult problem to calculate this integral, and many classes of relatively simple curves do not have simple solutions for it. This formula is derived by adding up infinitesimal arc lengths along the curve. Here, for example, is an online explanation: https://tutor.hix.ai In this case #s=int_0^pi sqrt(1+sin^2x) dx# Now here we have run straight into the problem I warned about above! This is not an integral that is solvable with elementary functions. Indeed, it is an integral that is part of a famous problem from mathematical history - the calculation of the arc length of the ellipse. The solution of this equation is a transcendental function, one that is simply defined to solve this particular equation, the (complete) Elliptic Integral of the Second Kind: #E(k^2)#. This is defined as #E(k^2)=int_0^(pi/2) sqrt(1-k^2sin^2theta)d theta# Note that #sin#'s symmetry around #pi/2# means that then #2E(k^2)=int_0^pi sqrt(1-k^2sin^2theta)d theta# which now matches our limits of integration. Taking #k^2=-1# (i.e. #k=i#), we see that #2E(-1)=int_0^pi sqrt(1+sin^2theta)d theta#, the arc length of #cos theta# over the interval we seek. Numerically, this is approximately 3.82.

Isabella Ackerman · Answer

undefined To find the arc length of the function $ f(x) = \cos(x) $ on the interval $[0, \pi]$, you can use the arc length formula:

$$ L = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} \, dx $$

1. Find the derivative of $ f(x) $.
$$ f'(x) = -\sin(x) $$

2. Square the derivative.
$$ [f'(x)]^2 = \sin^2(x) $$

3. Add 1 to the squared derivative.
$$ 1 + [f'(x)]^2 = 1 + \sin^2(x) $$

4. Take the square root of $ 1 + [f'(x)]^2 $.
$$ \sqrt{1 + [f'(x)]^2} = \sqrt{1 + \sin^2(x)} $$

5. Integrate the square root from $ x = 0 $ to $ x = \pi $.
$$ L = \int_{0}^{\pi} \sqrt{1 + \sin^2(x)} \, dx $$

6. Evaluate the integral to find the arc length.

$$ L = \int_{0}^{\pi} \sqrt{1 + \sin^2(x)} \, dx $$

$$ L \approx 3.819 $$

Therefore, the arc length of $ f(x) = \cos(x) $ on the interval $[0, \pi]$ is approximately $ 3.819 $ units.