I Spiral Equation

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?
 

Nidum

Science Advisor
Gold Member
2,990
844
tangents.jpg
 
thank you for your interest in the question, i was not quite clear in stating the problem, by family of curves i mean a family curves that all intersect at one single point, and they are identical such that you can have them all by rotating one of them around an axis at the point of intersection.

something like this
https://www.google.com/search?q=rotation+of+curve&pws=0&gl=us&biw=1366&bih=667&source=lnms&tbm=isch&sa=X&ved=0ahUKEwi6jaq0hZTOAhXiJcAKHdrvABYQ_AUIBigB#pws=0&gl=us&tbm=isch&q=rotated+curves&imgrc=6pYzj4IXzb-XsM:
 

haruspex

Science Advisor
Homework Helper
Insights Author
Gold Member
2018 Award
31,233
4,557
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.
 

Want to reply to this thread?

"Spiral Equation" You must log in or register to reply here.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top