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!
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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
>