My Math Forum  

Go Back   My Math Forum > College Math Forum > Topology

Topology Topology Math Forum


Reply
 
LinkBack Thread Tools Display Modes
May 26th, 2014, 04:52 AM   #1
Newbie
 
Joined: May 2014
From: de

Posts: 2
Thanks: 0

Find a ray intersection on a hyperbolic secant curve

Hallo

i need some help to fix a hyperbolic math problem.
the problem is to find a intersection point of a ray and a hyperbolic secant curve.


known is the ray origin and the direction/angle
searched is the intersection point on the hyperbolic secant curve, or the length of the ray hitting the curve.
the origin of the ray is somewhere on the y axis at the x0 point and the hyperbolic secant curve is on the x axis perpendicular to the ray origin.


if the solution is a differential equation, pleas explain the result in basic math too.
i have not studied mathematics and what i know i taught me self.
pleas explain a solution as close to a programming language or primitive code as possible.

thanks very much for any help!!

Code:
                     o RayOrigin
                     |\
                     | \
                     |  \ ? Lenght
                     |   \
                 ....1.... \
              ..     |     .x
           ...       |       ...
...........          |          ....... HypSec/SecH 
---------------0--------------------
Quader is offline  
 
Reply

  My Math Forum > College Math Forum > Topology

Tags
curve, find, hyperbolic, intersection, ray, secant



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Inverse hyperbolic secant function Tooperoo Calculus 1 November 27th, 2013 04:27 AM
Solving hyperbolic cosine equation (catenary curve) willmac Trigonometry 0 May 26th, 2011 12:55 PM
inverse hyperbolic secant, non-table integration oddlogic Calculus 6 March 22nd, 2011 06:17 PM
Curve of Intersection help mattnx Calculus 1 September 5th, 2009 04:59 PM
How find intersection inbetween curve sathyboopathiraja Algebra 3 March 2nd, 2009 03:08 PM





Copyright © 2019 My Math Forum. All rights reserved.