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And Another Question About Partial Derivatives

  1. Mar 14, 2016 #1
    1. The problem statement, all variables and given/known data
    [tex]
    \frac{d}{dt}\left(\frac{\dot{q}}{\sqrt{1+\left(\dot{q}\right)^{2}}}\right)=0\Rightarrow\frac{\dot{q}}{\sqrt{1+\left(\dot{q}\right)^{2}}}=const\Rightarrow\dot{q}=A\Rightarrow q=At+B
    [/tex]

    2. Relevant equations
    Why it ok to say that:
    [tex]
    \frac{\dot{q}}{\sqrt{1+\left(\dot{q}\right)^{2}}}=const\Rightarrow\dot{q}=A
    [/tex]

    3. The attempt at a solution
    [tex]
    \left(\frac{\dot{q}}{const}\right)^{2}-\dot{q}-1=0\Rightarrow?
    [/tex]
     
  2. jcsd
  3. Mar 14, 2016 #2

    Samy_A

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    How did you get the last equation?

    Anyway, if you solve ##\frac{\dot{q}}{\sqrt{1+\left(\dot{q}\right)^{2}}}=C## for ##\dot{q}##, what do you get as result?
     
  4. Mar 14, 2016 #3
    If you mean
    [tex]
    \left(\frac{\dot{q}}{const}\right)^{2}-\dot{q}-1=0\Rightarrow?
    [/tex]
    It's by dividing two sides by const and multiplying by
    [tex]
    \sqrt{1+\left(\dot{q}\right)^{2}}
    [/tex]
    and than squaring both sides.
    If I'll continue I'll need to solve quadratic equation which it's results will depend on the value on C.
    It can be either one, two or non complex or real numbers..
     
  5. Mar 14, 2016 #4

    Samy_A

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    I don't see how this operation gives you the term ##\dot q##.
    Maybe. Why would that be a problem? You need to prove that ##\dot q## is a constant, ie doesn't depend on ##t##.
     
  6. Mar 14, 2016 #5

    epenguin

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    Gold Member

    Do that again and do it right! :oldsmile:
     
  7. Mar 14, 2016 #6
    Yes, you're right. It dosen't! my professor accidently wrote it as:
    [tex]
    \frac{\dot{q}}{\sqrt{1+\left(\dot{q}\right)}}
    [/tex]

    and that's how I tried to solve it...

    thank you!
     
  8. Mar 14, 2016 #7

    Ray Vickson

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    This just says that if ##\dot{q}/\sqrt{\dot{q}^2+1}## is constant, the quantity ##\dot{q}## must also be a constant. That is obvious, but if you still cannot see it, imagine plotting a graph of ##y = \dot{q}/\sqrt{\dot{q}^2+1}## in a ##(\dot{q},y)## plot. You can see that if ##\dot{q}/\sqrt{\dot{q}^2+1}## is changed, ##\dot{q}## must be changed as well.
     
  9. Mar 14, 2016 #8
    Hi,
    I understand now what my problem was.
    I can also 'see' why it's true through the algebra, but I think the fact the independent variable is a derivatie of time, makes it a bit un-intuitiable for me.
    I tried plotting it but it came as something ##\delta## like, so it didn't helped so much :sorry:.
    but thank you for the help :)

    and if dont mind I had another question about partial derivatives I'll really appreciate if you'll take a short glimpse at it :)

    and even if you dont - thank you again!
     
  10. Mar 14, 2016 #9

    Ray Vickson

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    Which message are you responding to? You need to use the "Quote" button, in order to keep straight what messages are being addressed.
     
  11. Mar 14, 2016 #10
    Hi, It's: A Question About Partial Derivatives,
    thanks a lot!
     
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