Question

Sketch the curve and find the arc length of the spiral r=e^(-theta) for theta >= 0

Answer 1

the length of a curve is determined by adding up al the small little hypotenuses of right triangles along its length

Answer 2

im sure there is a way to define in it polar tho but id have to lok for it :)

Answer 3

i have the formula i kinda know how to do it here goes

Answer 4

pg 580 has a nice little formula

Answer 5

\[\int \sqrt{r^2+(r')^2}dt\]

Answer 6

that latex is a little squished looking to me

Answer 7

\[L=\int\limits_{a}^{b}\sqrt{r^2+r'^2}\]

Answer 8

and i just plug it in i guess but i know dont know how to find the limits

Answer 9

t >= 0 is al you get?

Answer 10

my r is = \[e^-\theta\]

Answer 11

you might have to do it in the same manner as an improper integral

Answer 12

use t for theta, its just simpler :)

Answer 13

yea thats all i get.. usually im given from 0<t<pi or somting

Answer 14

e^{-2t} + {-e^t}^2

e^2t + e^2t = 2e^2t

sqrt(2e^2t) = sqrt(2) e^t

i think

Answer 15

yeah; sqrt(e^2t) = e^t

Answer 16

i got sqrt(2) (e^-t)

Answer 17

yeah, that negative thing eluded me :)

Answer 18

so lets try to integrate this thing from 0 to something, then see if we can take the limit of it as that something approaches infinity

Answer 19

so my whole thing is \[\sqrt{2}[-e^-t]\]

Answer 20

and evaluate next but i dont have the limits

Answer 21

-sqrt(2) e^-t

right

-sqrt(2) e^n+sqrt(2) e^0 = -0+sqrt(2)

Answer 22

the limits are from 0 to as this thing approaches infinity

Answer 23

e^-inf = 0

Answer 24

http://www.wolframalpha.com/input/?i=integrate+sqrt%282%29e%5E%28-t%29+from+0+to+inf

Answer 25

so will that be my answer? \[\sqrt{2}\]

Answer 26

if thats all they give you is t >= 0 then yes

Answer 27

ok cool thats all i see in the problems hahah

Answer 28

\[\int_{0}^{inf}\]

Answer 29

can i have a negative \[\sqrt{2}\]

Answer 30

some how i got a -sqrt2