Question

How do you use the arc length formula to find lengths of spirals?

Answer 1

the parametric arc length formula for polar coordinates is\[ds=\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\]

Answer 2

from the formula for changing from rectangular to polar coordinates is\[x=r\cos\theta\]\[y=r\sin\theta\]which would seem to suggest that\[r(t)=e^{-t}\]and\[\theta=t\]

Answer 3

so the formula becomes\[r=e^{-\theta}\]you can get \(ds\) from that

Answer 4

I understand your first and second posts but can you explain your third one.

Answer 5

compare\[x=e^{-t}\cos t\]\[y=e^{-t}\sin t\]with\[x=r\cos\theta\]\[y=r\sin\theta\]

Answer 6

Oh I see. I understand. Let me try to work it out.

Answer 7

seems to imply that \(r=e^{-t}\) and \(\theta=t\)
I will try to work it out as well :)

Answer 8

well I got an answer, I hope it's the same as yours

Answer 9

I think I messed up. I get e^(-2t) -e^-(2t) under the square root.

Answer 10

that minus between them should go away when you square the derivative

Answer 11

Oh yeah. So it would be the square root of 2e^(-2t)?

Answer 12

\[(\frac{dr}{d\theta})^2=(-e^{-t})^2=e^{-2t}\]yup

Answer 13

Then what do you do? It is from infinity to 0

Answer 14

0 to infinity*

Answer 15

yes

Answer 16

Now do we plug it in and subtract?

Answer 17

well, you can't exactly "plug in" infinity, and you haven't mentioned integrating yet
though basically you are right
what do you have after the integral?

Answer 18

After you integrate aAfter I integrate sqrt(2e^(-2t)), I get -sqrt(2) e^-t

Answer 19

right, now you can evaluate

Answer 20

So do you get sqrt(2) ?

Answer 21

technically the correct way to deal with the infinity in the bounds is to say\[\left.\sqrt2e^{-t}\right|_0^\infty=\lim_{n\to\infty}\left.\sqrt2e^{-t}\right|_0^n\]which yes, gives the elegant solution of \(\sqrt2\)

Answer 22

oops that should be a negative in the formula before evaluating

Answer 23

Wouldn't that still give you a -sqrt(2)?

Answer 24

\[\left.-\sqrt2e^{-t}\right|_0^\infty=\lim_{n\to\infty}\left.-\sqrt2e^{-t}\right|_0^n=-\sqrt2(0-1)=\sqrt2\]so no

Answer 25

Actually I think it should still be positive

Answer 26

arc length should always be positive

Answer 27

Oh that makes a lot of sense.

Answer 28

if you got negative you probably messed up somewhere

Answer 29

Yeah

Answer 30

I've got another question. SHould I just create a new question?

Answer 31

Well I guess that means another mission accomplished :)
Yes, please post each Q separately

Answer 32

Okay. Thank you very much for your help. I will post it shortly.

Answer 33

Welcome!