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

*To*: mathgroup at smc.vnet.net*Subject*: [mg27459] Re: [mg27407] Deriviation d/dt(x(t))*From*: "Mark Harder" <harderm at ucs.orst.edu>*Date*: Mon, 26 Feb 2001 01:29:25 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

Christian, 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: In[325]:= eqn = f[x[t], y[t] ] == r^2 Out[325]= 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): In[333]:= Deqn = D[eqn /. f[x[t], y[t] ] -> x[t]^2 + y[t]^2, t] Out[333]= 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: In[337]:= Flatten[Solve[Deqn, # ] & /@ {x'[t], y'[t]} ] Out[338]= 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 sbox.tu-graz.ac.at> To: mathgroup at smc.vnet.net 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 > > > >