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Re: DSolving(?) for a given tangent

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  • Subject: [mg81373] Re: DSolving(?) for a given tangent
  • From: Szabolcs Horvát <szhorvat at>
  • Date: Fri, 21 Sep 2007 03:04:35 -0400 (EDT)
  • References: <fct9c4$r6e$>

AngleWyrm wrote:
> Don't know for sure if this is the right function, so here's the scenario:
> f[x_] := E^(0.22 x);
> Plot[f[x], {x, 6, 36}]
> Which plots a nice escalating curve.
> What I would like to know is: Where is the point {x,f[x]} that has a 
> 45-degree tangent line;

f[x_] := Exp[a x]

Solve[f'[x] == 1/2, x]

> ie where is this curve's balance point before it 
> really starts taking off?

Be careful: The exponential function does not have "a point where it 
really starts taking off".  All points are equivalent on an exponential, 
in the sense that shifting the curve along the x axis is equivalent to 
multiplying it by a number (or: it is equivalent to changing the unit on 
the y axis).

But it does have a characteristic "distance" (if 'x' represents a 
distance), 1/a, in the sense that Exp[a (x - (a few times)*(1/a))] is 
negligible compared to Exp[a x] (if a > 0).

People familiar with electronics (I'm not, so don't flame me if I say 
something "less than precise"!) like to say that the voltage drop across 
a Si diode is 0.6-0.7 volts.  But they also say that the diode has an 
exponential characteristic.  So why does the diode's characteristic 
"take off" at 0.6 volts, if an exponential has no special point?  That's 
because the diode is usually operated with currents in the milliampere 
range (much higher currents would destroy it).  The exponential 
(logarithmic) function transforms a current range of several orders of 
magnitude into a tiny voltage range.  If the microampere range were 
preferred (instead of milliamperes), the voltage drop would be smaller 
(as long as the characteristic is still well approximated by an 
exponential in this range).


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