What is the rope's parametric equation?

  • Thread starter david90
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suppose u have an ellipse and u put a rope around it and at distance h from the original ellipse. Any point from the ellipse to the rope wrap around the ellipse is = to distance h. what is the rope's parametric equation? What shape is this rope in?
 

NateTG

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I don't understand your question. Unless the rope has non-zero thickness, the equation of the rope is the same as the equation for the elipse.
 
ellipse?

x^2/r + y^2/h = 0? or something like that?
 

Hurkyl

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If I understand you correctly, you are saying you have an ellipse, you have a loop of rope around the ellipse, and each point on the rope is a distance h from the ellipse. Correct?

Let P be a point, γ be a curve (such as an ellipse), and let Q be the point on γ closest to P.

By definition, the distance from P to γ is the length of the line segment PQ. You may recall that a necessary condition is that PQ be perpendicular to the tangent line to γ at Q.

This tells you everything you need to solve the problem!

Start with a parametric representation of your ellipse. Say (u(t), v(t)). Then, for each t you find the equation for the tangent line to the ellipse at (u(t), v(t)), then you find the equation for the line perpendicular to the tangent line at (u(t), v(t)), then you find the point on the perpendicular that is a distance h from (u(t), v(t)).

This answer will then be a parametric equation for the curve you seek.

(note: there will be two solutions; one for the rope inside the ellipse and one for the rope outside the ellipse)
 
303
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am having trouble finding the normal/perpendicular line to parametric equation acost,bsint
 

NateTG

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If you can find the slope of the tangent, then the slope of the perpendicular is the opposite reciprocal.

If that's not painfull enough, you can also normalize the tanget vector and take it's derivative. By normalizing the tangent, you make it travel in a circle, so the derivative will always be perpendicular or zero.
 

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