Math Problem Statement

If you want to roll a ball down at rack in the shortest time possible, what shape should the track make? A straight line? A semi-circle? This is a famous physics questions with a surprising answer. The “curve of fastest decent” is called a brachistochrone, and is the same shape that a single point on a rolling wheel traces. It has this parametric equation: x(t) = t − sin(t), y(t) = cos(t) − 1. At what points does the brachistochrone have a horizontal tangent?

Solution

To find the points at which the brachistochrone has a horizontal tangent, we need to determine where the slope of the curve is zero. The slope of the curve is given by the derivative of y(t)y(t) with respect to x(t)x(t).

Given the parametric equations for the brachistochrone:

  • x(t)=tsin(t)x(t) = t - \sin(t)
  • y(t)=cos(t)1y(t) = \cos(t) - 1

We can find the derivative of yy with respect to xx using the chain rule:

dydx=dydtdxdt\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}

Step 1: Compute the derivatives of x(t)x(t) and y(t)y(t)

  1. dxdt=1cos(t)\frac{dx}{dt} = 1 - \cos(t)
  2. dydt=sin(t)\frac{dy}{dt} = -\sin(t)

Step 2: Set the slope equal to zero

To find where the tangent is horizontal, we need dydx=0\frac{dy}{dx} = 0, which means: dydt=0\frac{dy}{dt} = 0

Since dydx=sin(t)1cos(t)\frac{dy}{dx} = \frac{-\sin(t)}{1 - \cos(t)}, the horizontal tangent occurs when dydt=sin(t)=0\frac{dy}{dt} = -\sin(t) = 0, so:

sin(t)=0\sin(t) = 0

Step 3: Solve for tt

The sine function is zero at integer multiples of π\pi, so: t=nπforn=0,1,2,t = n\pi \quad \text{for} \quad n = 0, 1, 2, \dots

Step 4: Find the corresponding points

Substitute these values of tt back into the parametric equations to find the coordinates where the brachistochrone has a horizontal tangent.

  1. For t=0t = 0: x(0)=0sin(0)=0x(0) = 0 - \sin(0) = 0 y(0)=cos(0)1=11=0y(0) = \cos(0) - 1 = 1 - 1 = 0 So, the first point is (0,0)(0, 0).

  2. For t=πt = \pi: x(π)=πsin(π)=πx(\pi) = \pi - \sin(\pi) = \pi y(π)=cos(π)1=11=2y(\pi) = \cos(\pi) - 1 = -1 - 1 = -2 So, the second point is (π,2)(\pi, -2).

Thus, the brachistochrone has horizontal tangents at the points (0,0)(0, 0) and (π,2)(\pi, -2).

Would you like more details or have any questions about this solution?


Related questions:

  1. How is the brachistochrone problem related to the concept of the cycloid?
  2. What is the derivation of the parametric equations for a cycloid?
  3. How does the brachistochrone curve differ from a straight line in terms of time of descent?
  4. What is the history behind the brachistochrone problem in physics and mathematics?
  5. How can we use the calculus of variations to derive the brachistochrone curve?

Tip: The brachistochrone problem is an excellent example of how physics and mathematics intersect to solve real-world optimization problems.

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Math Problem Analysis

Mathematical Concepts

Parametric Equations
Derivatives
Cycloid
Tangent Line

Formulas

x(t) = t - sin(t)
y(t) = cos(t) - 1
dy/dx = (dy/dt) / (dx/dt)

Theorems

Chain Rule of Differentiation
Brachistochrone Problem

Suitable Grade Level

University Level (Calculus)