- #1

Westlife

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## Homework Statement

Parameterize the part of the curve which allows an equilateral triangle, with the height 3R, to roll from one vertex to the next one, while its center travels at a constant height.

## Homework Equations

I will include some pictures to show what I'm doing

## The Attempt at a Solution

Hey I'm just checking soo far if I have the right idea.

let ##f(t)=(x(t),y(t))## be our curve

Here I defined the parameter ##t## to be the angel between the height and the point at which the triangle intersects the curve .

I said that the length of ##f## has to be equal to the length of the side from the vertex to the point of intersection and using the sine rule I got.

so the length of the curve is ##l=\frac{2R\sin(t)}{\sin(\frac{2\pi}{3}-t)}##

so from the arc length equation we know that ##\sqrt{x'^2+y'^2}=l##

the tangential unit vector of the curve is ##(x',y')/l## and the normal unit vector is ##(-y',x')/l##

Here i defined the curve of the center of the triangle to be ##S(t)=(m(t),n(t))## where

$$n(t)=y(t)+\lambda y'/l +Rx'/l=2R\quad \lambda=\frac{R\sin(60-t)}{sin(30+t)}$$

Where I got lambda from using the sine rule again.

And now I'm lost. I don't know how to continue from here

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