Author Topic: vector (Read 1849 times) Tweet Share

TonyHem · « **on:** July 14, 2009, 06:23:36 pm »

11) an object is moving so that its position r at time t is given by: $r(t)=(e^t+e^{-t})i + (e^t-e^{-t})J\; t\ge 0$
~~A) find the initial position of the object = 2i~~
~~b) find the position at t = log_e(2) 2.5i + 3j/2~~
c) Find the cartesian equation of the graph

I tried $x = e^t+e^{-t}, y = e^t-e^{-t}$
$X(e^t)= e^2t+ 1$
$x(e^t)-e^2t= 1$
$e^t(x-e^t) = 1$
$t(x-e^t) = 0$
$x = e^t$
$loge(x) = t$
And tried subbing that into y=e^t-e^{-t} and it was wrong

Done a and b.
Need help with C

thanks

dcc · « **Reply #1 on:** July 14, 2009, 06:29:10 pm »

Have you tried something like $x^2 - y^2$ ? I'm lazy, but that might work out.

TonyHem · « **Reply #2 on:** July 14, 2009, 06:33:23 pm »

Yup I get it now, thanks

edit: btw, j/w, why doesn't my way work?

dcc · « **Reply #3 on:** July 14, 2009, 06:36:55 pm »

Quote from: TonyHem on July 14, 2009, 06:23:36 pm

$r(t)=(e^t+e^{-t})i + (e^t-e^{-t})J\; t\ge 0$

$e^t(x-e^t) = 1$
$t(x-e^t) = 0$
$x = e^t$

First off, you forgot to take the logarithm of the $x - e^t$ term. Secondly, the conclusion that $x = e^t$ when $x = e^t + e^{-t}$ is implying that $e^{-t}$ is zero always, which is not the case. Try to look out for things like this when you are doing problems.

kamil9876 · « **Reply #4 on:** July 14, 2009, 06:37:44 pm »

$x=e^t + e^{-t}$
$y=e^t - e^{-t}$

$x+y=2e^t$

$x-y=2e^{-t}$
$=\frac{2}{0.5(x+y)}$

$(x-y)(x+y)=4$
$x^2-y^2=4$

that's how I did it back in the days, of course now i would just use hyperbolic functions haha.

Jordan23 · « **Reply #5 on:** September 09, 2009, 09:05:43 pm »

cant see latex

TrueTears · « **Reply #6 on:** September 09, 2009, 09:12:48 pm »

Quote from: Jordan23 on September 09, 2009, 09:05:43 pm

cant see latex

learn the commands, or wait for a while.

Mao · « **Reply #7 on:** September 09, 2009, 09:56:11 pm »

I think the latex server is down..

Search the forums now!

We have moved!

Author Topic: vector (Read 1849 times) Tweet Share

TonyHem

vector

dcc

Re: vector

TonyHem

Re: vector

dcc

Re: vector

kamil9876

Re: vector

Jordan23

Re: vector

TrueTears

Re: vector

Mao

Re: vector

Recent Posts

User:
Password: