In the derivation of a cycloid - problem in Stewart

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In summary, the derivation of a cycloid states that arc PT is equal to r theta. This does not necessarily imply that arc PT and line segment OT are homeomorphisms, as homeomorphisms are functions, not sets of points. However, they are homeomorphic as topological spaces. The term "isomorphic" is often used to describe a relationship between algebraic structures, but in this case, it may not be applicable as there are no operations defined on the set. It is important to be careful with the use of words when discussing mathematical concepts.
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tronter
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In the derivation of a cycloid, it says (pg. 655 Stewart, 5th ed) [tex] |OT| = \text{arc} \ PT = r \theta [/tex].

I was just wondering if if this implies that [tex] \text{arc}\ PT [/tex] and the line segment [tex] OT [/tex] are homeomorphisms?

When I first learned about cycloids, I didn't know that homeomorphisms were. But now, having learned about them, it seems that they are in fact homeomorphisms.
 
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  • #2
i would go even further and say they are isomorphic, they're both just curves
 
  • #3
Careful - you've not said what you mean by isomorphic. Sure, as sets they are, as topological spaces, but not all curves are isomorphic in all categories (eg, rational curves with rational maps).
 
  • #4
I would say you need to be more careful with your use of words! No, arc PT and line segment OT are not homeomorphisms because hemeomorphisms are functions, not sets of points. It is true, of course, that they are homeomorphic.

My understanding of "isomorphism" is that it is a relationship between algebraic structures. In you do not have operations on the set, then "isomorphic" does not apply.
 
  • #5
I would say you need to be more careful with your use of words! No, arc PT and line segment OT are not homeomorphisms because hemeomorphisms are functions, not sets of points. It is true, of course, that they are homeomorphic.

My understanding of "isomorphism" is that it is a relationship between algebraic structures. In you do not have operations on the set, then "isomorphic" does not apply.
 

1. What is a cycloid?

A cycloid is a curve formed by tracing a point on the circumference of a circle as it rolls along a straight line. It is a specific type of trochoid, which is a curve traced by a fixed point on a circle as it rolls along a straight line.

2. Who is Stewart and what is his problem in the derivation of a cycloid?

James Stewart was a mathematician who published a problem in his book "Calculus: Early Transcendentals" about finding the area under a cycloid curve. The problem involves using integration to derive the equation for a cycloid.

3. What is the equation for a cycloid?

The parametric equations for a cycloid are x = r(θ - sinθ) and y = r(1 - cosθ), where r is the radius of the rolling circle and θ is the angle of rotation.

4. How is the equation for a cycloid derived?

The equation for a cycloid can be derived using integration, specifically the arc length formula. By finding the arc length of a quarter of a circle, the equations for x and y can be derived using trigonometric identities.

5. What are some real-life applications of the cycloid curve?

The cycloid curve has several real-life applications, including the design of gear teeth, the shape of bicycle wheels, and the path of a projectile launched at an angle. It is also used in engineering and physics to study rolling motion and optimize designs for efficiency.

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