Question

find the arc length of the graph of the function over the indicated interval. y= ln(cosx) [0, pi/3]

Answer 1

So if I remember correctly, the formula for arclength is:\[\large \int\limits ds\]It's just a bunch of little pieces of arc summed up.
Which we can write as,\[\large \int\limits \sqrt{1+\left(\frac{dy}{dx}\right)^2}dx\]

So we need the derivative as part of this process.
Do you know the derivative of ln(cos x)? :)

Answer 2

yes the formula is correct and no thats one of the parts I'm having trouble on is finding the derivative.

Answer 3

You remember the derivative of \(\large \ln x\) correct? \(\large \dfrac{1}{x}\)


Our derivative here will work out using the same rule.
We put in the denominator the entire contents of the log.

So for our derivative,
\[\large \left[\ln(\cos x)\right]' \qquad = \qquad \frac{1}{\cos x}\color{royalblue}{\left(\cos x\right)'}\]

Understand what I did so far?

Just as ln x produces 1/x,
ln(cos x) will produce 1/cos x.

The blue term is showing up due to the chain rule.
We have to multiply by the derivative of the inner function.
The little prime is to show that we still need to take it's derivative.

Answer 4

Do you remember your Trig derivatives? :)
What's the derivative of the blue term?

Answer 5

yes i undersatnd, and i believe its sinx?

Answer 6

It's actually -sin x,
\[\large \frac{1}{\cos x}\color{royalblue}{\left(\cos x\right)'} \qquad =\qquad \frac{1}{\cos x}\color{orangered}{\left(-\sin x\right)} \qquad = \qquad -\left(\frac{\sin x}{\cos x}\right)\]

Ok good, so we've got our derivative.
Hmm, sin x divided by cosine x.. that looks like another trig identity.
Does that simplify down to something?

Answer 7

ummm i dont remember

Answer 8

Ok I'll just tell you the identity I was thinking of.
Maybe it will refresh your memory :)

\[\large \frac{\sin x}{\cos x} \qquad = \qquad \tan x\]

So for our derivative we found this out,\[\large \left[\ln(\cos x)\right]' \qquad = \qquad -\tan x\]

Answer 9

\[\large y=\ln(\cos x)\]\[\large \color{green}{\frac{dy}{dx}=-\tan x}\]

\[\large \int\limits_0^{\pi/3}\sqrt{1+\left(\color{green}{\frac{dy}{dx}}\right)^2}dx\]

Ok understand where we're going with this?

Answer 10

yes, you plug in the -tanx where the dy/dx is in the formula?

Answer 11

yah that sounds good.
So the negative will go away when we square it.

\[\large \int\limits\limits_0^{\pi/3}\sqrt{1+\tan^2x}dx\]We should be left with something like this.
We need to remember another trig identity.

\(\large \color{orangered}{1+\tan^2x=?}\)

Do you remember? :)

Answer 12

is it the Pythagorean identity?

Answer 13

Yes, which gives us what? c:

Answer 14

sec^2x? :D

Answer 15

\[\large \int\limits\limits\limits_0^{\pi/3}\sqrt{\sec^2x}\;dx\]Ok good.

Answer 16

\[\large \int\limits\limits_0^{\pi/3}\sqrt{\sec^2x}\;dx \qquad =\qquad \int\limits\limits_0^{\pi/3}\sec x\;dx\]

Answer 17

This is an integral that you'll simply want to remember.
It's not one that you could easily work out.

\[\large \int\limits \sec x \;dx\qquad =\qquad \ln\left(\sec x+\tan x\right)\]

Answer 18

\[\large \int\limits\limits_0^{\pi/3}\sec x\;dx \qquad = \qquad \ln\left(\sec x+\tan x\right)\;|_0^{\pi/3}\]

Do you understand how to solve it from here? :)

Answer 19

ummm yea i just plug in the pi over 3 and zero?

Answer 20

ya sounds good!
plug in the pi/3's, then subtract the 0's

Answer 21

umm idk what the sec of pi/3 is...