How do you find the arc length of $y = ln (cos (x))$ $y=ln(cos(x))$ on the interval $[\frac{π}{6}, \frac{π}{4}]$ $[pi/6,pi/4]$ ?

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How do you find the arc length of $y = ln (cos (x))$ $y=ln(cos(x))$ on the interval $[\frac{π}{6}, \frac{π}{4}]$ $[pi/6,pi/4]$ ?

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Answer 1 · 2014-08-21T17:10:01

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

$L = \int_{a}^{b} \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x$ $L = int_a^bsqrt(1 + (dy/dx)^2)dx$

Process:

With our function of $ln (cos (x))$ $ln(cos(x))$ , we must first find its derivative. The shortcut for $ln (u)$ $ln(u)$ -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 $cos (x)$ $cos(x)$ is ( $- sin x$ $-sinx$ ), we end up with:

$\frac{d y}{d x} = - \frac{sin x}{cos x}$ $dy/dx = -sinx/cosx$ ,

which is equal to:

$- tan x$ $-tanx$ .

Plugging into our Arc Length formula, we have:

$L = \int_{a}^{b} \sqrt{1 + {(- tan x)}^{2}} d x$ $L = int_a^b sqrt(1 + (-tanx)^2)dx$ .

If we square the $- tan x$ $-tanx$ term, we get:

$L = \int_{a}^{b} \sqrt{1 + {tan}^{2} (x)} d x$ $L = int_a^b sqrt(1 + tan^2(x))dx$

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

$L = \int_{a}^{b} \sqrt{{sec}^{2} (x)} d x$ $L = int_a^b sqrt(sec^2(x))dx$ , which simplifies to $L = \int_{a}^{b} sec x d x$ $L = int_a^b secx dx$

Now we must remember that $\int sec x$ $int secx$ = $ln (sec x + tan x) + C$ $ln(secx + tanx) + C$ , so for our equation we must solve:

$ln (sec x + tan x)$ $ln(secx + tanx)$ from $\frac{π}{6}$ $pi/6$ to $\frac{π}{4}$ $pi/4$ , giving us:

$L = ln (\frac{2}{\sqrt{2}} + 1) - ln (\frac{2}{\sqrt{3}} + \frac{1}{\sqrt{3}})$ $L = ln(2/sqrt2 + 1) - ln(2/sqrt3 + 1/sqrt3)$

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

$L = ln (\sqrt{2} + 1) - ln (\sqrt{3})$ $L = ln(sqrt2 + 1) - ln(sqrt3)$

If you remember that $ln (a) - ln (b) = ln (\frac{a}{b})$ $ln(a) - ln(b) = ln(a/b)$ , we can simplify our answer to get:

$L = ln (\frac{\sqrt{2} + 1}{\sqrt{3}})$ $L = ln((sqrt2 + 1)/sqrt3)$

We can evaluate this for a decimal answer:

$L ~~ 0.332067...$

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How do you find the arc length of $y = ln (cos (x))$ $y=ln(cos(x))$ on the interval $[\frac{π}{6}, \frac{π}{4}]$ $[pi/6,pi/4]$ ?

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