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Re: DSolving(?) for a given tangent
*To*: mathgroup at smc.vnet.net
*Subject*: [mg81398] Re: DSolving(?) for a given tangent
*From*: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
*Date*: Sat, 22 Sep 2007 03:22:08 -0400 (EDT)
*Organization*: The Open University, Milton Keynes, UK
*References*: <fct9c4$r6e$1@smc.vnet.net> <fcvrjh$7ur$1@smc.vnet.net>
Jean-Marc Gulliet wrote:
> 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; ie where is this curve's balance point before it
>> really starts taking off?
>
> So what you are looking for is the value of x for which f'[x] == Pi/4
Sorry about this non-sense! Obviously, answering too many threads
simultaneously burnt some of my neurons and nothing surprise me when I
wrote this "answer". The right thing we are looking for is f'[x] == 1.
> (i.e. the slope of the tangent at x is 45 degrees but it must be
> expressed in radians rather than in degrees). The solution can be found
> by solving the equation f'[x] == Pi/4 for x; to do so one can use Solve
> or Reduce for an analytic solution (which implies exact coefficients
> such as 22/100 rather than 0.22) or NSolve or FindRoot for an numerical
> solution. For instance,
>
> In[1]:=
> f[x_] := E^(0.22 x);
> Plot[f[x], {x, 6, 36}]
> sol = NSolve[f'[x] == Pi/4, x]
> x /. sol[[1]]
>
> Out[3]= {{x -> 5.78438}}
>
> Out[4]= 5.78438
The following fixed code yields the correct result.
In[1]:= f[x_] := E^(0.22 x);
Plot[f[x], {x, 6, 36}]
sol = NSolve[f'[x] == 1, x]
x /. sol[[1]]
FindRoot[f'[x] == 1, {x, 6}]
Out[3]= {{x -> 6.8824}}
Out[4]= 6.8824
Out[5]= {x -> 6.8824}
Again, sorry for the confusion.
Regards,
--
Jean-Marc
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