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MathGroup Archive 2001

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Re: Deriviation d/dt(x(t))

  • To: mathgroup at
  • Subject: [mg27459] Re: [mg27407] Deriviation d/dt(x(t))
  • From: "Mark Harder" <harderm at>
  • Date: Mon, 26 Feb 2001 01:29:25 -0500 (EST)
  • Sender: owner-wri-mathgroup at

    If all you want is the *relationship* that defines the velocity vector,
you can do the following:

Set up the equation you want to differentiate:
eqn = f[x[t], y[t] ] == r^2
f[x[t], y[t]] == r^2

Find the total derivative of both sides of eqn with respect to t, giving
replacement rule for f
  (this wouldn't be necessary if I had defined f more explicitly above):
Deqn = D[eqn /. f[x[t], y[t] ] -> x[t]^2 + y[t]^2, t]
2 x[t] x'[t] + 2 y[t] y'[t] == 0

    If you want to know the velocity vector, {x'[t],y'[t]}, the best you can
do without adding extra
information about the circle is the following:

Flatten[Solve[Deqn, # ] & /@ {x'[t], y'[t]}  ]

               y[t] y'[t]                  x[t] x'[t]
{x'[t] -> -(----------), y'[t] -> -(----------)}
                 x[t]                        y[t]
I hope that helps.
-mark harder

-----Original Message-----
From: Christian Maier <maier1 at>
To: mathgroup at
Subject: [mg27459] [mg27407] Deriviation d/dt(x(t))

>I have a problem, having a definition of a circle: x^2+y^2=r^2
>Where x and y are functions of the time:  x(t), y(t) and r ist a constant.
>Now I want to deriviate the equation to get the velocity vector.
>It should be : 2 x dx/dt +2 y dy/dt = 0
>Can anyone tell me how to do?
>Thans in advance!
>Christian Maier

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