Re: making a function of two coordinates

*To*: mathgroup at smc.vnet.net*Subject*: [mg31638] Re: [mg31617] making a function of two coordinates*From*: Anton Antonov <antonov at wolfram.com>*Date*: Mon, 19 Nov 2001 03:11:22 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

Hi Roderik, First for the exponential you can use the Mathematica build-in function Exp[]. Also, equations in Mathematica are defined in with ==. So this is OK: In[1]:= Clear[U, e, a, t, y] Solve[y == U*Exp[a*t], a] Out[1]:={{a -> Log[y/U]/t}} I used Clear[] because y = U*e^(a*t) sets a value to y. (Try to evaluate y = U*e^(a*t) first and then Solve[y == U*e^(a*t), a], and you will see that nothing happens.) If you try In[2]:= Solve[y == U*e^(a*t), a] you will get Out[2]:= {{a -> Log[y/U]/(t*Log[e])}} since Mathematica doesn't know what e is. Best, Anton ============================================================== Anton Antonov Antonov, PhD ***** Wolfram Research Inc. http://www.imm.dtu.dk/~uniaaa *** tel +1 217 398 0700 #782 ============================================================== Give me wings and I will crawl faster! -------------------------------------------------------------- On Sun, 18 Nov 2001, TheSquaredBun wrote: > > I have got two coordinates out of graph. The line through the > coordinates could be described as: > > y = U * e^(a * t) > > (e is Exponetial e (approx. 2.37)) > > t is the x-value, I have give two points (t1,y1) and (t2,y2) > > I have tried several things (Solve and Fit for example), but I can't > get the function. > Does anyone know what should be the input to let Mathematica solve the > a and U in the formula. > > Much thanks in advance, > > Roderik > r.f.emmerink at st.hanze.nl >