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re: parabola formula
18 apr 2000
zamboni wrote:
>...the problem was to determine the surface area of a parabolic mirror...
>please if anyone spots a mistake, for this student's sake do point it out...
>...i used a general form of the parabola:
>
> y = ax^b.
say, y = x^2, with a = 1 and b = 2...
>...at a given value y, the cross-section of the parabola is a circle
>with radius x. therefore, we can integrate the circumference of an
>infinte number of differentially small circles' circumferences over
>the range 0 -> h where h is the height of the mirror.
we need to add up the centrally-tilted areas of lots of rings, right?
>this is set up as:
>
> h|
> int | 2*pi*x dy
> 0|
the 2*pi*x is the inner circumference, but we need to multiply that
by the tilted width, ie sqrt(dx^2+dy^2), no?
>surface area = 2*pi * ab/(1+b) * (h/a)^(b+1/b) ***final generalized***
> ********answer*********
> where h is the max height of the parabola
hmmm. surface area = 4*pi/3 = 4.19 for a = h = 1 and b= 2...
how about da = 2*pi*sqrt(y)*sqrt(dy^2+dx^2) = 2*pi*sqrt(ydy^2+ydx^2)
= 2*pi*sqrt*(ydy^2+y(dy^2/(4y)) = 2*pi*sqrt(y+1/4)dy?
h |
then a = 2*pi*int |(1/4+y)^0.5 dy = 4*pi/3*((1/4+1)^1.5-(1/4)^1.5) = 5.33?
0 |
nick
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