Question

Finding arc length by calculating definite integral of
(1+[f'(x)]^2)^(1/2)dx for x=-16 to x=16
if f(x) = 19.997/2(e^(x/19.997)+e^(-x/19.997))-19.997
showing steps

Answer 1

Hey Student! Welcome to OpenStudy!
Hmm your function is a little difficult to read in text,
is this what it should look like?\[\large\rm f(x)=\frac{19.997}{2}\left(e^{x/19.997}+e^{-x/19.997}\right)-19.997\]

Answer 2

yes..sorry

Answer 3

Do you understand how to find your derivative?\[\large\rm f'(x)=?\]

Answer 4

Yes ...I think in its simplest form it is f'(x) = \[(1/2)(e^(x/19.997)-e^(-x/19.997))\]
I understand differentiation and integration and when I integrate using this f'(x) in the integral function on my calculator I get the answer which should be close to 36. But i need to show it algebraically and cant get it...,

Answer 5

Just a note for using the equation tool/ LaTeX:
Use curly braces if you want to put more than one character into an exponent.
Example: e^{4x}=\(e^{4x}\)

Answer 6

Algebraically, we can apply an identity,\[\large\rm \sinh(x)=\frac{1}{2}\left(e^x-e^{-x}\right)\]

Answer 7

\[\large\rm f'=\frac12\left(e^{x/19.997}-e^{-x/19.997}\right)\]Do you see how we can apply this identity to our derivative? :)

Answer 8

i see the connection but I haven't learnt that identity so I don't know how to work with it....

Answer 9

Hmm I'm not sure if this will work out nicely otherwise >.<
Thinking...

Answer 10

Ya I'm not really seeing another way to approach this.
\[\large\rm f'=\sinh\left(\frac{x}{19.997}\right)\]If this approach is too confusing, new, we can keep trying stuff maybe...
Lemme know what you're thinking XD

Answer 11

can I just check that if I were to integrate it I would do ((1+(f'(x)^2)^(3/2) and then divide by [(3/2) x the derivative of what was inside the root] ? just want to make sure I am following the correct rule for integration

Answer 12

The thing is is that I am doing a school task that other students have to do and none of us have learnt that identity so I am just assuming there must be a way to do it without it? sorry to be difficult... thanks for helping

Answer 13

No that wouldn't quite work.
Take a simple example like:\[\large\rm \int\limits \sqrt{1+x^2}~dx\quad\ne\quad \frac{2}{3(2x)}(1+x^2)^{3/2}\]It's not even close to that actually.

You would need to apply a trig sub,\[\large\rm \int\limits \sqrt{1+(f')^2}~dx\]\[\large\rm f'=\tan \theta\]\[\large\rm \int\limits \sqrt{1+\tan^2\theta}~dx\]The problem with that approach is...
replacing our differential dx is going to be rather hard.
Maybe there is a clever trick to it that I'm just not thinking of.

Answer 14

\[\large\rm f'=\frac12\left(e^{x/19.997}-e^{-x/19.997}\right)\]Squaring,\[\large\rm (f')^2=\frac14\left(e^{x/19.997}-e^{-x/19.997}\right)^2\]\[\large\rm (f')^2=\frac14\left(e^{2x/19.997}-2+e^{-2x/19.997}\right)\]Adding 1 and square rooting,\[\large\rm \sqrt{1+(f')^2}=\sqrt{1+\frac14\left(e^{2x/19.997}-2+e^{-2x/19.997}\right)}\]Maybe common denominator,\[\large\rm \sqrt{1+(f')^2}=\sqrt{\frac44+\frac14\left(e^{2x/19.997}-2+e^{-2x/19.997}\right)}\]Which we can write as,\[\large\rm \sqrt{1+(f')^2}=\frac{1}{2}\sqrt{4+\left(e^{2x/19.997}-2+e^{-2x/19.997}\right)}\]OH OH OH yes I'm seeing it!

Answer 15

I'll stop there a sec, I don't wanna ruin any of the fun if you see it.
Read that a sec and lemme know if any of it's confusing.

Answer 16

sorry so I understand what you have done but I don't know what to do next....

Answer 17

Not jumping out at you?
That's ok! It's a tricky little trick they're pulling here.

So let's combine the 4 and the -2,\[\large\rm =\frac{1}{2}\sqrt{e^{2x/19.997}+2+e^{-2x/19.997}}\]

Answer 18

yep I get that then can I try and get it all squared so I can remove the root...will that work?

Answer 19

Good good good.
When we had the -2 in the middle, it was a perfect square.

So the +2 in the middle probably gives us some other perfect square!

Answer 20

oh ok so its just the perfect square but with a plus in the middle and then the root cancels with the square and then its just 1/2 (e^2x/19... +e^-2x/19..) and then just integrate that?

Answer 21

\[\large\rm e^{2x/19.997}+2+e^{-2x/19.997}\quad=\quad \left(e^{x/19.997}+e^{-x/19.997}\right)^2\]Mmm yes very good.
Then it's as simple as integrating a couple of exponentials, ya?

Answer 22

yes got 35.525 as length!
thank-you!!!

Answer 23

also I'm new to this - am I supposed to like reward you with something...? I just dont really know this site works

Answer 24

You can click the "Best Response" button by my name if you found my responses helpful. It doesn't really matter though :)

Ooo you got it? Noiceeee.

Answer 25

ok thanks again!

Answer 26

np