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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

1897.0. "Where is Maximum Curvature of y = x^n ?" by TROOA::RITCHE (From the desk of Allen Ritche...) Mon Sep 26 1994 21:24

Consider each graph of the form y = x^n ,    where  n >= 2 , x >= 0

(a) Find the point of greatest curvature on each of these curves and 
    describe the locus of such points.

(b) Find n (integer) for which the locus in (a) has a maximum value, y.

(c) Do the same above for n real.


Allen 
(who has always been fascinated by the simplicity and power of y=x^n)
T.RTitleUserPersonal
Name		DateLines
1897.1TROOA::RITCHEFrom the desk of Allen Ritche...Wed Oct 05 1994 00:276
    > 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
1897.2AUSSIE::GARSONachtentachtig kacheltjesWed Oct 05 1994 02:564
    re .0
    
    Do you want to refresh my RAM and exhibit the appropriate calculus
    expression that gives the radius of curvature?
1897.3RAM refreshed?TROOA::RITCHEFrom the desk of Allen Ritche...Wed Oct 05 1994 11:3229
1897.4IOSG::TEFNUT::carlinDick Carlin IOSG ReadingWed Oct 05 1994 13:0918
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
1897.5TROOA::RITCHEFrom the desk of Allen Ritche...Wed Oct 05 1994 21:3232
>     -> 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