Re: Natural Eq'm of Cardioid
- From: Noone <x@xxxxxxxxx>
- Date: Wed, 02 May 2007 07:53:44 -0400
On Tue, 01 May 2007 22:30:27 -0400, Noone <x@xxxxxxxxx> wrote:
On 1 May 2007 16:57:17 -0700, Rodo <mattrodo@xxxxxxxxx> wrote:
I'm having a difficult time with what is probably an easy problem.
The problem states:<br> "Consider the cardioid
r = 1 - cos(theta), 0 less than or equal to (theta) less than or
equal
to 2pi. Let s(theta) be the arc length from the point (2, pi) on the
cardioid to the point (r, theta) and let "rho" (theta) = 1 / K be the
radius of the curvature at the point (r, theta). Show that s and
"rho" (theta) are related by the equation s^2 + 9("rho")^2 = 16.
I haven't even the slighest clue where to begin with this. Any
insight is greatly appreciated!
Well, given an equation of a curve in polar coordinates, I suppose you
need to know how to find the arc length and the radius of curvature.
In your problem, r = r(t) = 1 - cos(t)
As a check on your work, I got
s = 2 sqrt(2) sqrt[ 1 + cos(t) ]
If I had been more alert last night, I'd have used a half-angle
formula to write s as
s = 4 cos(t/2)
R (radius of curvature) = 4/3 sin(t/2)
And, after a bit of algebra & trig,
s^2 + 9 R^2 = 16
So now we have
s = 4 cos(t/2)
R = 4/3 sin(t/2)
and it now follows immediately that
s^2 + 9 R^2 = 16
.
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