Question

Okay I cant get it..
Find the displacement and the distance traveled.
r= e^t i + e^-t j +(2^1/2)t k
over 0<=t<=ln3

Answer 1

displacement is vector formed be integrating the i, j, and k components separately and the distance traveled is the integral of the magnitude.

Answer 2

so displacement s should = <e^t, -e^-t, t^2/sqrt2>

Answer 3

or would you just evaluate r from 0 to ln3 for displacement then the integral of the v would be distance

Answer 4

o wait sorry i thought r was velocity for a second
r itself is displacement and the ittegral of the magnitude of v is distance traveled

Answer 5

where v =(dr1/dt, dr2/dt, dr3/dt)

Answer 6

yeah thats what I was thinking so it just needs to be evaluated. and take the derivitive of r to get v and take the magnitude of that for the distance correct

Answer 7

yes

Answer 8

but you have to find the integral of the magnitude of v not just the magnitude itself

Answer 9

and i assume you know this but jsut in case magnitude is the positive square root of the sum of each component squared

Answer 10

ok so i took the sqrt of each component squared after taking dy/dx of r. I then end up with sqrt((e^t)^2+(-e^(-t))^2+sqrt(2)^2) as the magnitude. So from there I must take the integral of this. It seems like the magnitude must not be correct because its a difficult integration

Answer 11

you might be required to use numerical integration