Re: Circle Laying On Curves
- From: "Ioannis" <morpheus@xxxxxxxxxxxx>
- Date: Wed, 28 Feb 2007 20:27:17 +0200
"Narek Saribekyan" <narek.saribekyan@xxxxxxxxx> wrote in message
news:1172681916.585957.261320@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Consider curves e^(-x) and ln(x) and a tangent unit circle. How can I
fins the center coordinates?
The set of points at distance 1 above the curve exp(-x), is exp(-x) + 1.
Similarly, the set of points at distance 1 below the curve ln(x), is ln(x) -
1.
Therefore the center of the circle (x,y) will lie at the intersection of the
curves:
exp(-x) + 1 and ln(x) - 1.
x then must satisfy the equation:
exp(-x) + 1 = ln(x) - 1 (1)
(1) cannot be solved exactly, so you have to use numerical methods. Here I use
a Hyper-Lambert HW function to solve this equation numerically as:
ln(x) - exp(-x) = 2 =>
x*(ln(x)/x - exp(-x)/x) = 2 =>
x*exp(ln(ln(x)/x - exp(-x)/x)) = 2 =>
x = HW(ln(ln(x)/x - exp(-x)/x);2) =>
x =~ 7.393603058
Therefore the circle's origin will be at: (x, ln(x) - 1)
--
I.N. Galidakis
http://ioannis.virtualcomposer2000.com/
.
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