Conference rusure::math

    re .0
    
    Off the top of my head ... sqrt(45) and (6,6).
    
    It's useful when posting a problem like this to state that it is *not*
    part of a homework assignment or similar.

    Hint: you want the equation of the line that passes thru (9,0) and is
    perpendicular to the tangent to the curve where it intersects the curve.

>    
>    Off the top of my head ... sqrt(45) and (6,6).
>    

This is the right answer.  If (a,b) is the point, then calculate the slope of
the tangent to y as 3/sqrt(6x) = 3/y = 3/b (first derivative and substitute).

Then slope m of line from (a,b) to (9,0) must be negative reciprocal of above.
(perpendicular to tangent).  Form second equation of slope in terms of these
two points.  

Quicly solve for two unknowns a and b.  First eqn is curve itself, i.e.
b=sqrt(6a).

Actually there are 3 points of interest (6,6), (6,-6), (0,0) but the latter two
can be discarded.

Allen

re .3
    
>Actually there are 3 points of interest (6,6), (6,-6), (0,0) but the latter two
>can be discarded.
    
    (6,-6) should not be discarded. It is an equally valid solution. It was
    slackness on my part not to list it in .1.
    
    re .2,.3
    
    Using the fact that the normal to the curve should pass through the
    point (9,0) is OK but that is actually an assumption. Another way is
    merely to form the distance as a function of some parameter and use
    calculus or otherwise (calculus not necessary in this case) to find
    the minimum distance.                     

>>re .3
>>    
>>>Actually there are 3 points of interest (6,6), (6,-6), (0,0) but the latter two
>>>can be discarded.
>>    
>>    (6,-6) should not be discarded. It is an equally valid solution. It was
>>    slackness on my part not to list it in .1.
>>
It might not have been slackness if you interpret the .0 eqn as only a half
a parabola in the first quadrant. (+ve sqrt). 
    
>    re .2,.3
>    
>    Using the fact that the normal to the curve should pass through the
>    point (9,0) is OK but that is actually an assumption. 

Yes.  But a _valid_ assumption, no?  (especially, when one has limited time in
this notes file :-) The shortest distance from a point (i.e. 9,0) to a curve or
line is along the perpendicular to it.  And a perpendicular to a curve means
perpendicular to a tangent, I belive. We never assumed where (a,b) was located
but actually had to solve for that variable.


Allen

re .5
    
>It might not have been slackness if you interpret the .0 eqn as only a half
>a parabola in the first quadrant. (+ve sqrt). 
    
    Fair enough. The original problem statement wasn't explicit on this.
    
>Yes.  But a _valid_ assumption, no?
    
    Yes, assumption was in any case probably the wrong word for me to use.
    
    For a variety of curves it can be *proven* that (local) extrema in
    the function giving distance to some given point occur where the normal to
    the curve passes through the point. However this is not true for all
    curves - in particular ones with "spikes". Perhaps someone more
    knowledgeable than I can state the exact condition(s) that must be
    satisfied by a curve for the assumption to be valid.
    
    Of course for any curve (as with any extremum problem) the end points,
    if there are any, have to be checked explicitly.
    
    However this particular problem is easier if done from first principles
    i.e. simply minimise the distance function.

>    curves - in particular ones with "spikes". Perhaps someone more
>    knowledgeable than I can state the exact condition(s) that must be
>    satisfied by a curve for the assumption to be valid.
>    
>    Of course for any curve (as with any extremum problem) the end points,
>    if there are any, have to be checked explicitly.
    
    You need the target curve to be differentiable. "Spikes" and endpoints
    are locations where the curve is not differentiable, and must be
    checked separately.
    
    Of course when I say "a curve is differentiable", I refer to the
    function that the curve describes.
    
      John

    
    
    First - this is not homework or an assignment.
    
    Second - there is a difference between y = sqrt(6x) and y squared = 6x
    
    y = sqrt(6x) is only in the first quadrant  + (0,0)
    
    y squared = 6x is in 1st and 4th
    
    Solve this problem without the use of Calculus.  The only correct
    answer is the point (6,6).  My solution will follow in a few days.
    

Further to .3 and .8,

>    Solve this problem without the use of Calculus.  The only correct
>    answer is the point (6,6).  My solution will follow in a few days.

OK.  Try this.  No calculus, Gr 11(?) math...

Distance between (x,y) and (9,0) is given by...

  d^2 = 6x + (x-9)^2

      = x^2 - 12x + 81

      = (x-6)^2 + 45   (completing the square)


This always positive expression will be a minimum if the first term is 0, i.e.
when x=6.

Allen