What is the arc length of #f(x)=sqrt(sinx) # in the interval #[0,pi]#?

Question

Charles Anctil · Accepted Answer

The formula for arc length on interval #[a, b]# is

#A = int_a^b sqrt(1 + (f'x)^2) dx#

The derivative of #f(x)# will be obtained using the chain rule.

#f'(x) = cosx * 1/(2sqrt(sinx))#

#f'(x) = cosx/(2sqrt(sinx)#

Using the given formula:

#A = int_0^pi sqrt(1 + (cosx/(2sqrt(sinx)))^2) dx#

#A = int_0^pi sqrt(1 + cos^2x/(4sinx)) dx#

#A = int_0^pi sqrt(1 + 1/4cotxcosx) dx#

Which according to the integral calculator has no solution through elementary antiderivatives. A numerical approximation for arc length gives

#A= 4.04 # units

to three significant figures.

Hopefully this helps!

Isabella Ackerman · Answer

undefined To find the arc length of the function $ f(x) = \sqrt{\sin(x)} $ over the interval $[0, \pi]$, we use the formula for arc length:

$$ L = \int_a^b \sqrt{1 + (f'(x))^2} \, dx $$

Where $ f'(x) $ is the derivative of $ f(x) $.

First, we find $ f'(x) $:

$$ f(x) = \sqrt{\sin(x)} $$
$$ f'(x) = \frac{1}{2\sqrt{\sin(x)}} \cdot \cos(x) $$

Now, we can substitute $ f'(x) $ into the arc length formula and integrate:

$$ L = \int_0^\pi \sqrt{1 + \left(\frac{\cos(x)}{2\sqrt{\sin(x)}}\right)^2} \, dx $$

$$ L = \int_0^\pi \sqrt{1 + \frac{\cos^2(x)}{4\sin(x)}} \, dx $$

$$ L = \int_0^\pi \sqrt{\frac{4\sin(x) + \cos^2(x)}{4\sin(x)}} \, dx $$

$$ L = \int_0^\pi \sqrt{\frac{4\sin(x) + 1 - \sin^2(x)}{4\sin(x)}} \, dx $$

$$ L = \int_0^\pi \sqrt{\frac{1 + 4\sin(x)}{4\sin(x)}} \, dx $$

This integral is quite complex and cannot be expressed in terms of elementary functions. Therefore, it would typically be approximated numerically.