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anonymous
 5 years ago
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
anonymous
 5 years ago
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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0displacement is vector formed be integrating the i, j, and k components separately and the distance traveled is the integral of the magnitude.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so displacement s should = <e^t, e^t, t^2/sqrt2>

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or would you just evaluate r from 0 to ln3 for displacement then the integral of the v would be distance

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0o 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0where v =(dr1/dt, dr2/dt, dr3/dt)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but you have to find the integral of the magnitude of v not just the magnitude itself

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and i assume you know this but jsut in case magnitude is the positive square root of the sum of each component squared

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you might be required to use numerical integration
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