| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1897.1 | | TROOA::RITCHE | From the desk of Allen Ritche... | Tue Oct 04 1994 21:27 | 6 |
| > Find the point of greatest curvature on each of these curves
The point of interest is where the radius of curvature, R is a minimum.
(Since "curvature" at a point is defined as the reciprocal of R).
Allen
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| 1897.2 | | AUSSIE::GARSON | achtentachtig kacheltjes | Tue Oct 04 1994 23:56 | 4 |
| re .0
Do you want to refresh my RAM and exhibit the appropriate calculus
expression that gives the radius of curvature?
|
| 1897.3 | RAM refreshed? | TROOA::RITCHE | From the desk of Allen Ritche... | Wed Oct 05 1994 08:32 | 29 |
| RE: .2
> Do you want to refresh my RAM and exhibit the appropriate calculus
> expression that gives the radius of curvature?
Spoiler follows...
R = ds/d� :-)
(i.e. rate of change of arc length wrt angle of tangent)
If y = f(x),
[ 1 + y'^2 ] ^3/2
R = -----------------
y''
where y' and y'' are 1st and 2nd derivatives wrt variable x.
Allen
|
| 1897.4 | | IOSG::TEFNUT::carlin | Dick Carlin IOSG Reading | Wed Oct 05 1994 10:09 | 18 |
| Sorry I can't generate one of these spoiler thingies on the pc client.
My completely unrigorous derivation (happily the same result as yours) was
as follows:
y' = tan(phi)
-> y'' = sec^2(phi)*(dphi/dx)
-> y'' = sec^2(phi)*(dphi/ds)*(ds/dx)
-> y'' = sec^3(phi)*(dphi/ds)
-> y'' = (1+tan^2(phi))^(3/2)*(dphi/ds)
-> y'' = (1+y'^2)^(3/2)*(dphi/ds)
-> ds/defy = (1+y'^2)^(3/2)/y'' :-)
I had a quick look and was surprised that y=x^4 didn't have its minimum
radius of curvature at the origin like y = x^2.
Dick
|
| 1897.5 | | TROOA::RITCHE | From the desk of Allen Ritche... | Wed Oct 05 1994 18:32 | 32 |
| > -> y'' = sec^2(phi)*(dphi/ds)*(ds/dx)
> -> y'' = sec^3(phi)*(dphi/ds)
It took me a few moments to make the connection that ds/dx is cos(phi).
The derivation for R that I recall, which is probably equivalent is...
phi==p = arctan y'
dp/dx = 1/(1+y'^2) x y'' (deriv of arctan and chain rule)
ds/dx = sqrt(1+y'^2) (standard form for arc length)
ds/dp = ds/dx / dp/dx
> -> ds/defy = (1+y'^2)^(3/2)/y'' :-)
Same result. Same rigour. :-)
>
>I had a quick look and was surprised that y=x^4 didn't have its minimum
>radius of curvature at the origin like y = x^2.
Yes indeed. It is quite interesting to "see" how this innocuous curve
behaves as n increases. We normally only think of y=x^n getting very large
and steep for x>1 but for 0<x<1 it gets quite flat and therefore has to make
a very sharp turn before making it through (1,1).
Now everyone should have the tools to solve .0
Allen
|