Question

I have the problem
x = e^(-t)cos(2012t) y = e^(-t)(2012t)
0≤ t ≥pi calculate its exact length.

so I know I have to use the therom

Answer 1

\[\int\limits_{a}^{b}\sqrt{(dx/dt)^{2} + (dx/dt)^{2}}dx\]

Answer 2

I have so far that

dx/dt = -e^(-t)cos(2012t) - e^(-t)sin(2012t)2012
and
dy/dt = -e^(-t)sin(2012t)+e^(t)cos(2012t)2012

Answer 3

the integral I'm getting is very very ugly I simplified it to

\[\int\limits_{0}^{pi}\sqrt{\sin^{2}(2012t)2012^{2} + \cos^{2}(2012t) -\sin^{2}(2012t) + \cos^{2}(2012t)2012^{2}}dx/e^{t}\]

Answer 4

still very ugly :l should I have used quotient rule instead?

Answer 5

I simplified it down to
\[\int\limits_{0}^{pi} \sqrt{2012^{2} - 1 + 2\cos^{2}(2012t)}dx/e^{t}\]

Answer 6

shouldn't the dx be dt

Answer 7

or

\[\int\limits_{0}^{pi} \sqrt{4048143 - 2\cos{2}(2012t)}dt/e^{t}\]

good call dumbcow :)

Answer 8

It says it should be easy after a simplification :\ i must have did something wrong or he lied

Answer 9

ok i think everything checks out until last step
what happened to sin^2 ?

Answer 10

cos^2(2012t)−sin^2(2012t) = 2cos^(2)(2012t)

Answer 11

by the trig rule sin^(2)(x) + cos^(2)(x) = 1

Answer 12

maybe that simplification made it less simplified :l

Answer 13

oh right i didn't see the "-1" then
i was thinking of identity:
cos(2x) = cos^2 - sin^2

Answer 14

its funny because my prof always said he was evil now I know he wasn't lying

Answer 15

such a long and tedious integral it seems although I may just be dumb

Answer 16

*might

Answer 17

wait, tell me how you factored the e^t out of sqrt
sorry im doing this in my head right now

Answer 18

yes (e^(-2t))^(1/2) = (1/2)^(1)/(e^(2t))^(1/2) = 1/2^(t)

Answer 19

sorry that should be

1/e^(t)

Answer 20

no isn't attached to a sin*cos when you square dx/dt and dy/dt

Answer 21

wow nevermind

Answer 22

?

Answer 23

I could try u substitution for this but it will still be really dirty

Answer 24

due to the e^(t)

Answer 25

maybe integration by parts but even then that will probably even be grosser

Answer 26

no i wouldn't recommend integration by parts

Answer 27

lets see, its in the form
\[\int\limits_{ }^{} \frac{\sqrt{u^{2}+\cos(2u)}}{e^{t}} dt\]
where u = 2012
right

Answer 28

that actually looks a lot better lol

Answer 29

i have no idea what to do...haha
lets see what wolfram can do
http://www.wolframalpha.com/input/?i=integrate+sqrt%281%2Bcos%282t%29%29%2Fe^t+dt

Answer 30

then we have

\[\int\limits_{0}^{pi} \sqrt{u^{2} + \cos^{2}(u) - \sin^{2}(u)-1}du/e^{t}\]

Answer 31

you forgot to include -1

Answer 32

or + 1 rather

Answer 33

no it was -1

Answer 34

hmm
cos(2x) = cos^2 - sin^2
right

Answer 35

right

Answer 36

man this integral is frustrating i need to eat something brb

Answer 37

from first sqrt
\[\rightarrow \sqrt{u^{2}(\sin^{2} + \cos^{2}) + \cos^{2} -\sin^{2}} = \sqrt{u^{2} + \cos(2ut)}\]
the "-1" only comes into play if you use other form
cos^2 - sin^2 = 2cos^2 - 1

Answer 38

it can't be integrated

http://www.wolframalpha.com/input/?i=integrate+sqrt%28a^2%2Bcos%282at%29%29%2Fe^t+dt

Answer 39

so i made a mistake?

Answer 40

It has to be integrable

Answer 41

here is definite integral
...sorry i meant it has no known anti-derivative

http://www.wolframalpha.com/input/?i=integrate+sqrt%282012^2%2Bcos%282%282012%29t%29%29%2Fe^t+dt+from+0+to+pi

Answer 42

you must have made a mistake

Answer 43

why do you have x's in there ?

Answer 44

https://www.wolframalpha.com/input/?i=integrate+%28%28-e^%28-t%29cos%282012t%29+-+e^%28-t%29sin%282012t%292012%29^%282%29+%2B+%28-e^%28-t%29sin%282012t%29%2Be^%28-t%29cos%282012t%292012%29^%282%29%29^%281%2F2%29

Answer 45

it can be solved

Answer 46

ahh i went back with paper and checked everything, you got a sign mixed up somewhere
notice on wolfram answer...all those (sin^2 +cos^2) can be replaced with 1

integral can be reduced to
\[\sqrt{2012^{2} +1}\int\limits_{0}^{\pi} e^{-t} dt\]

Answer 47

the cos(2012)^2 - sin(2012)^2 should be cos(2012)^2 + sin(2012)^2

Answer 48

when I was expanding?

Answer 49

oh nvm I think I see thanks :)

Answer 50

whats funny is you get the same answer of 1925.05 either way

interesting problem your professor came up with

Answer 51

a=2012
\[x(t) =e^{-t} \cos (a\,t)\\
y(t)= e^{-t} \sin (a\, t)\\

\]
Is this our curve?

Answer 52

\[
x'(t)^2+y'(t)^2 =\left(a e^{-t} \cos (a
t)-e^{-t} \sin (a
t)\right)^2+\left(-a e^{-t}
\sin (a t)-e^{-t} \cos (a
t)\right)^2=\\
\left(a^2+1\right) e^{-2 t}\\

\sqrt{ x'(t)^2+y'(t)^2 }= \sqrt{ 1 + a^2} e^{-t}
\]

Answer 53

\[
\sqrt{1 + a^2} \int_0^\pi e^{-t} dt=\sqrt{1 + a^2}(1 - e^{-\pi})\approx 1925.05
\]