Finding dv/dx when given x and v as functions of t

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First use the chain rule and then the rule for the derivative of an inverse function: $$ \frac{dv}{dx}=\frac{dv}{dt}\cdot\frac{dt}{dx}=\frac{dv}{dt}\left(\frac{dx}{dt}\right)^{-1}=\frac{a}{v} $$ This is basically the identity that you wanted to investigate.

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Shinaolord
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Shinaolord

Bachelors in Applied and Computational Mathematics, Loving user of Mathematica, physics nerd.

Updated on August 18, 2020

Comments

  • Shinaolord
    Shinaolord about 3 years

    This isn't a homework question, I tried to word this as a physical problem and it got re tagged as homework by someone

    Say I am given the equation of position in one dimensional kinematics. Say its the usual free fall kinematics, with x being the up down motion, and v= dx/dt. starting from rest. How do we find dv/dx when given x(t)? Would taking the derivative of v(t)/x(t) work?

    I assume I'll need the chain and quotient rules. If you're wondering, I want to see how to use the derived relation a = v dv/dx. I know this holds, but I don't know how to apply it when given with regard to time.

    I'm using parametric equations at the moment

    As per the comments, I realized I mis-wrote seine things. I corrected it.

    As per comments: so there's nothing more to it, I was just overthinking it? What about quotient rule? I'm thinking w the chain rule it would be dx/dt = dx/dv * dv/dt then I would divide by dx/dv to invert it? Which would give me dv/dx * dx/dt = dv/dt Which gives me dv/dt = dv/dt?

    I figured it out. Thanks for the answer though Dv/dt= dv/dx dx/dt => dv/dt * (dx/dt)^-1 = a/v. So to get a, wee multiply by v giving us a= v*dv/dx. I get it meows

    • Danu
      Danu about 9 years
      Like you said, use the chain rule.
    • zephyr
      zephyr about 9 years
      what on earth is y? Have a look at the third equation on here, should explain it: en.wikipedia.org/wiki/Chain_rule
  • Danu
    Danu about 9 years
    Just a heads up: It's generally discouraged here to answer homework questions with a full solution.