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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

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  1. anonymous
    • 5 years ago
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    displacement is vector formed be integrating the i, j, and k components separately and the distance traveled is the integral of the magnitude.

  2. anonymous
    • 5 years ago
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    so displacement s should = <e^t, -e^-t, t^2/sqrt2>

  3. anonymous
    • 5 years ago
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    or would you just evaluate r from 0 to ln3 for displacement then the integral of the v would be distance

  4. anonymous
    • 5 years ago
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    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

  5. anonymous
    • 5 years ago
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    where v =(dr1/dt, dr2/dt, dr3/dt)

  6. anonymous
    • 5 years ago
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    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

  7. anonymous
    • 5 years ago
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    yes

  8. anonymous
    • 5 years ago
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    but you have to find the integral of the magnitude of v not just the magnitude itself

  9. anonymous
    • 5 years ago
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    and i assume you know this but jsut in case magnitude is the positive square root of the sum of each component squared

  10. anonymous
    • 5 years ago
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    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

  11. anonymous
    • 5 years ago
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    you might be required to use numerical integration

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