# Minimizing arc length of $y=\ln(\cos x)$

Suppose you know that the difference between $a$ and $b$ is $2$. How can we find the values of $a$ and $b$ (with $-2\leq a,b\leq 2$) which minimizes the arc length of the curve $y=\ln(\cos x)$ from $x=a$ to $x=b$?

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For $\ln(\cos(x))$ to make sense, in the given context, you would need $a,b \in (-\pi/2,\pi/2)$. – user17762 Feb 6 '13 at 22:21

Hint: do you know the formula for arc length? Can you take $\frac {d}{dx} \ln(\cos(x))$? Have you tried plugging this function into it and integrating?

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We are supposed to be minimizing and maximizing the arc length. But if you integrate from -pi/2 to pi/2 you have one pi period. That is greater than 2. So that's a no go – EhBabay Feb 6 '13 at 22:30
@JohnCarpenter: but you can integrate from $a$ to $a+2$. You will get an answer in terms of $a$. Now differentiate that, set it to zero, and show it is a minimum. I don't think there is a maximum. It will increase without bound as $a \to -\frac \pi 2$ or $a \to \frac \pi 2 -2$ – Ross Millikan Feb 6 '13 at 22:31
To add to Ross's answer, the arc length of a curve $y=f(x)$ between $a$ and $b$ is given by $$\int_a^b \sqrt{1 + \left(\dfrac{dy}{dx} \right)^2} dx$$ Hence, in your case, the arc-length is $$\int_a^b \sqrt{1 + \tan^2(x)} dx = \int_a^b \sec(x) dx = \log\left(\left \vert \dfrac{1 + \sin(b)}{1 - \sin(a)}\right \vert \right)$$ Also, for $\ln(\cos(x))$ to make sense, in the given context, you would need $a,b \in (-\pi/2,\pi/2)$. – user17762 Feb 6 '13 at 22:36
Yup got it! Thanks – EhBabay Feb 6 '13 at 22:55

As has been pointed out, We cannot be working in the interval described, since the function is undefined, for example, from $\pi/2$ to $2$. We replace the interval by $(-\pi/2,\pi/2)$.

We want to maximize, minimize the arclength from $t-1$ to $t+1$. By the usual formula for arclength, the arclength from $t-1$ to $t+1$ is $L(t)$, where $$L(t)=\int_{t-1}^{t+1} \sqrt{1+\tan^2 x}\,dx,$$ For brevity, call the integrand $f(t)$.

Note that $f(t)$ is symmetric about $t=0$, decreasing in the interval $(-\pi/2,0)$, and increasing in $(0,\pi/2)$. It follows that $L(t)$ attains a minimum at $t=0$. There is no maximum. Alternately, we can observe that by the Fundamental Theorem of Calculus, we have $L'(t)=f(t+1)-f(t-1)$. This is $0$ only at $t=0$.

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