What is the arc length of f(x)=sqrt(sinx) in the interval [0,pi]?
1 Answer
Jan 8, 2018
The formula for arc length on interval
A = int_a^b sqrt(1 + (f'x)^2) dx
The derivative of
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!