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anonymous

  • one year ago

solve dv/dt=32-.8v

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  1. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \frac{dv}{dt} =32-0.8v }\) ?

  2. anonymous
    • one year ago
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    yup

  3. anonymous
    • one year ago
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    we gotta seperate

  4. anonymous
    • one year ago
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    dv/(32-.8v)=dt

  5. anonymous
    • one year ago
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    then integrate

  6. SolomonZelman
    • one year ago
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    yes, that is fine.

  7. SolomonZelman
    • one year ago
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    What I would tend to do is: \(\large\color{black}{ \displaystyle \frac{dv}{dt} =32-0.8v }\) \(\large\color{black}{ \displaystyle \frac{1}{32-0.8v}\frac{dv}{dt} =1 }\) \(\large\color{black}{ \displaystyle \color{red}{\int}\frac{1}{32-0.8v}\frac{dv}{dt}\color{red}{dt} =\color{red}{\int}1\color{red}{dt} }\) and then cancel the differentials on the left side.

  8. SolomonZelman
    • one year ago
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    So that you aren't just adding an integral sign (which is technically meaningless), but integrating with respect to a variable. But, what you did is fine too.

  9. anonymous
    • one year ago
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    how do i do it my way

  10. anonymous
    • one year ago
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    like integrating dv/(32-.8v)

  11. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \int\frac{1}{32-0.8v}dv =\int dt}\) Do a u substitution on the left side, and the right side should be obvious obvious at this point.

  12. anonymous
    • one year ago
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    int dt = t but idk wat int dv/32-.8v is

  13. anonymous
    • one year ago
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    u usb

  14. anonymous
    • one year ago
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    oh ok

  15. anonymous
    • one year ago
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    u=32-.8v?

  16. SolomonZelman
    • one year ago
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    ok, go ahead and solve the integral on the left side, and tell me what you get:)

  17. anonymous
    • one year ago
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    should u = 32-.8v

  18. SolomonZelman
    • one year ago
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    yes, that is the correct u-sub.

  19. anonymous
    • one year ago
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    ln u

  20. anonymous
    • one year ago
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    is tat correct

  21. SolomonZelman
    • one year ago
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    you had du=?

  22. anonymous
    • one year ago
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    du= v?

  23. anonymous
    • one year ago
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    no dv

  24. SolomonZelman
    • one year ago
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    what is the derivative of: 32-0.8v (with respect to v?

  25. anonymous
    • one year ago
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    -.8

  26. SolomonZelman
    • one year ago
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    Yes, so you have: du/dv=-0.8 and then, du=-0.8 dv ---> du=-(8/10) dv then, (-10/8) du=dv

  27. anonymous
    • one year ago
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    -10/8(lnU)

  28. anonymous
    • one year ago
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    =t

  29. SolomonZelman
    • one year ago
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    |dw:1443636158125:dw|

  30. SolomonZelman
    • one year ago
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    yes, *CORRECT* !

  31. SolomonZelman
    • one year ago
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    -10/8 ln(u) = t Now, solve for u.

  32. anonymous
    • one year ago
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    so -10/8 ln (32-.8v)=t

  33. SolomonZelman
    • one year ago
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    yes, right, now solve for v

  34. SolomonZelman
    • one year ago
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    (you can solve for u and then subsitute the v back, or as you want to do - to sub back the v first and then to solve for v)

  35. anonymous
    • one year ago
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    v= e^(8t/-10)-32 all over -.8

  36. anonymous
    • one year ago
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    ? right

  37. SolomonZelman
    • one year ago
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    Yes

  38. SolomonZelman
    • one year ago
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    hold on...

  39. SolomonZelman
    • one year ago
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    -10/8 ln (32-.8v)=t ln (32-.8v)=(-8/10)t e^ln (32-.8v)=e^(-8/10)t 32-.8v=e^(-8/10)t -0.8v=e^(-8/10)t-32 0.8v=32-e^(-8/10)t v=[ 32-e^(-8/10)t ] 0.8

  40. SolomonZelman
    • one year ago
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    yes, yes, you are right...

  41. anonymous
    • one year ago
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    thank u

  42. SolomonZelman
    • one year ago
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    the result is simplifiable, you can make it into: v = 40 - (5/4)e\(\large ^{-8t/10}\)

  43. SolomonZelman
    • one year ago
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    Good luck

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