Spiral Equation

  • #1
I am looking for a family of curves where if we consider one curve of them and get the tangent of that curve at any arbitrary point on the curve, then you will always find a point in the other curves where the tangent of this point is perpendicular to the tangent of the first point.

My guess for the solution of this, is that i am looking for a family of spirals that is sink in or out of some point.
but this is just a guess without any rigorous prove,

I tried this:
let ##y_n## be the family of curves, consider two adjacent curves ##y_1 (x)## and ##y_2 (x)##, and that first derivatives (slopes of the tangents) are ##y^\prime_1## and ##y^\prime_2##

for those two tangents to be perpendicular we must have ##y^\prime_1 y^\prime_2 = -1##

lets consider ##S## an equal distance around the curves, where ##S_1 = S_2 = S##, then we have :

##S_1 = \int_{x_1}^{x_2} \sqrt{1+{y^\prime_1}^2} \, dx##
##S_2 = \int_{x_3}^{x_4} \sqrt{1+{y^\prime_2}^2} \, dx##

Assuming we know all the integration boundaries ##x_1,x_2,x_3,x_4##
so we can write

##\int_{x_1}^{x_2} \sqrt{1+{y^\prime_1}^2} \, dx = \int_{x_3}^{x_4} \sqrt{1+\frac{1}{{y^\prime_1}^2}} \, dx##

but i don't know what to do next?
 

Answers and Replies

  • #2
Nidum
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  • #4
haruspex
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you will always find a point in the other curves where the tangent of this point is perpendicular to the tangent of the first point.
A sufficient condition would be that for any given slope and any given curve there is a point on the curve at which the tangent has that slope. The family of circles of a given radius and passing through a common point would do.
 

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