Problem 14 - Calculus Of Variations, With Applications To Physics And Engineering

Question

(a) A brachistochrone is required to be constructed from a given curve \(h(x, y)=0\) to a second given curve \(g(x, y)-0 .\) What relationship must the two given curves bear to one another at the respective points of intersection with the brachistochrone? (b) A brachistochrone extends from the line \(y=x+4\) to the parabola \(y^{2}=x\). Show that the point of intersection of the brachistochrone with the parabola is \(\left(\frac{1}{3}, \frac{1}{3}\right)\)

Step-by-Step Solution

Answer

The curves should be tangent to the brachistochrone at the points of intersection, and the point of intersection between the line and parabola that the brachistochrone extends to is \(\left(\frac{1}{3}, \frac{1}{3}\right)\).

Step 1: Understand The Relationship Between Curves And A Brachistochrone

Let there be a curve \( h(x, y)=0 \) and \( g(x,y)=0 \). The brachistochrone intersects with the curves at these points. The path of quickest descent (or ascent) between these intersections is the brachistochrone. Hence, the curves must be tangent to the brachistochrone at the intersection points.

Step 2: Establish The Line And Parabola

We have a line \( y=x+4 \) and a parabola \( y^{2}=x \). The brachistochrone extends from the line to the parabola. We need to show that the point of intersection with the brachistochrone is at \(\left(\frac{1}{3}, \frac{1}{3}\right)\).

Step 3: Find the Intersection

To find the intersection between the line and the parabola, we must solve the equations \( y=x+4 \) and \( y^{2}=x \) simultaneously. By substituting the equation of the line into the equation of the parabola, we get: \( (x+4)^{2}= x \). Solving the equation results in \( x=\frac{1}{3} \). Substituting \( x=\frac{1}{3} \) into the line equation \( y=x+4 \), we get \( y=\frac{1}{3} + 4 \), but since both x and y should be equal we've got \( y=\frac{1}{3} \). Hence, the intersection point is \(\left(\frac{1}{3}, \frac{1}{3}\right)\).

Key Concepts

Brachistochrone Problem

The Brachistochrone problem is a classic problem in calculus of variations. It seeks the path of fastest descent between two points under gravity, ignoring friction.
Imagine a bead sliding down a curve from point A to B. Instead of just the shortest path (a straight line), we want the quickest path. The word "Brachistochrone" comes from Greek, meaning "shortest time."
It turns out, the quickest path on any curve is not a straight line but a cycloid.

• A cycloid is a curve traced by a point on the rim of a wheel as it rolls along a straight line.

This property, that a brachistochrone connects points via a cycloid, showcases the power of calculus in solving physical problems. To solve brachistochrone problems, one must usually consider the geometry and tangent requirements at both ends of the curves involved.

Tangent to Curves

When solving a brachistochrone problem, a key requirement is understanding tangency. If a curve from which the brachistochrone departs, or towards which it arrives, is to be smoothly connected, they must be tangent to each other.
In mathematical terms, when two curves are tangent at a point, they share a common line at that point, and hence their derivatives are equal.

• This means that the brachistochrone touches the other curve but doesn't cross it at the tangent point.
• This tangency requirement ensures a seamless transition of motion, without sharp turns.

Thus, one must carefully assess the curves’ properties at potential intersection points to ensure tangency.

Descent Path

A descent path is a path along which an object moves under the action of gravity. In the context of the brachistochrone problem, it refers to the path taken by an object descending from one curve to another quickly.
To determine the correct descent path, we leverage the principles of calculus of variations.

• We aim to minimize the time taken along the descent path.
• This involves considering factors like initial and final heights, as well as the shape of the descent path.

Typically, a cycloid serves as the ideal descent path in brachistochrone challenges because it minimizes travel time between points.