Question

How do you find the arc length of `y=ln(cos(x))` on the interval `[pi/6,pi/4]`?

Answer 1

You can find the Arc Length of a function by first finding its derivative and plugging into the known formula:

`L = int_a^bsqrt(1 + (dy/dx)^2)dx`

Process:

With our function of `ln(cos(x))`, we must first find its derivative. The shortcut for `ln(u)` -type functions is to just take the derivative of the inside and place it over the original of what's inside. Since the derivative of `cos(x)` is (`-sinx`), we end up with:

`dy/dx = -sinx/cosx`,

which is equal to:

`-tanx`.

Plugging into our Arc Length formula, we have:

`L = int_a^b sqrt(1 + (-tanx)^2)dx`.

If we square the `-tanx` term, we get:

`L = int_a^b sqrt(1 + tan^2(x))dx`

Since `1+tan^2(x) = sec^2(x)` is one of our known trig identities, we can change our equation into:

`L = int_a^b sqrt(sec^2(x))dx`, which simplifies to `L = int_a^b secx dx`

Now we must remember that `int secx` = `ln(secx + tanx) + C`, so for our equation we must solve:

`ln(secx + tanx)` from `pi/6` to `pi/4`, giving us:

`L = ln(2/sqrt2 + 1) - ln(2/sqrt3 + 1/sqrt3)`

`2/sqrt2` can be simplified to `sqrt2`, and the right term has a common denominator, which lets you add them together to become `3/sqrt3`, which is simplified`sqrt3`, giving us:

`L = ln(sqrt2 + 1) - ln(sqrt3)`

If you remember that `ln(a) - ln(b) = ln(a/b)`, we can simplify our answer to get:

`L = ln((sqrt2 + 1)/sqrt3)`

We can evaluate this for a decimal answer:

`L ~~ 0.332067...`