Descartes' circle method for finding normal

[Rob]

I'm having trouble with this. I need to use Descartes' circle method
to determining the normal and the slope of the tangent to the curv y=
x^(3/2).

I think I understand why is method works. It takes advantage of the
fact that a circle's radius is normal to the point on the circle that
it intersects. So if we can get he curv y and a circle to intersect at
a single point, and if we know that radius of the circle to that
intersection, then we know a normal.

Problem is, I'm not able to work out the answe. Here's what I've
started to do, and any assistance is appreciated!

Take the point (a, a^(3/2)) on the curve y=x^(3/2). Intersect the
curve y at this point with a circle centred at (x_0, 0). So, we knoe
the eq'n of the cirle is (x - x_0)^2 + y^2 = r^2.

i.e. y^2 = (r^2) - (x - x_0) ^2
y = sqrt [ (r^2) - (x - x_0) ^2 ]
so now equate this with curve y=x^(3/2)
x^(3/2) = y = sqrt [ (r^2) - (x - x_0) ^2 ]
x^3 = (r^2) - (x - x_0)^2
(x^3) - (r^2) - (x - x_0)^2 = 0

Now Descarte does some sort somthing where he compares this to a
polynomial and compares coefficients. I'm not sure how to do this, or
where to go from here.

Thanks for any help!
Rob

[Paul Sperry]

In article <9d11b6b.03041...@posting.google.com>, Rob
<rm11...@yahoo.com> wrote:

This was a new one on me.

I'm going to change your notation a little and assume the circle has
center (a, 0) and passes through y = x^(3/2) at (b, b^(3/2)).

The circle's equation is x^2 - 2ax + a^2 + y^2 = r^2 and (b, b^(3/2))
must satisfy it so b^2 -2ab + a^2 + b^3 = r^2. So, we have
x^2 - 2ax + a^2 + x^3 - (b^2 - 2ab + a^2 + b^3) = 0.

Now, x = b is a root so x - b is a factor and a little quick long
division gives

x^2 - 2ax + a^2 + x^3 - (b^2 - 2ab + a^2 + b^3) =
(x - b)*(x^2 + (1 + b)x + (b^2 + b - 2a)

In order to have only one intersection point, b must be a root of
x^2 + (1 + b)x + (b^2 + b - 2a). That gives a = (1/2)*(3b^2 + 2b).
[This step is a little shaky. _If_ x^2 + (1 + b)x + (b^2 + b - 2a) has
a real root, it must be b.]

Finally, to wrap it up, the line from (a, 0) to (b, b^(3/2)) is the
normal, it has slope (-2/3)b^(-1/2) so the tangent has slope
(3/2)b^(1/2).

Lo! A miracle has occurred! That is (by calculus) the correct slope.

[Jonathan Miller]

Paul Sperry wrote:

Why do you call this a miracle? The correct slope is the correct slope.

Jon Miller

[Paul Sperry]

In article <3E963896...@comcast.net>, Jonathan Miller
<jonathan...@comcast.net> wrote:

The miracle is that I got it right.

[Jonathan Miller]

Paul Sperry wrote:

>The miracle is that I got it right.
>

You could be right about that. The reason we use Newton's calculation
rather than Descartes' is that it's really better. More user-friendly.

Jon Miller