Figure 1.4.11 shows a bead sliding down a frictionless wire from point P to point Q

Chapter 3, Problem 68

(choose chapter or problem)

Figure 1.4.11 shows a bead sliding down a frictionless wire from point P to point Q. Thebrachistochrone problem asks what shape the wire should be in order to minimize the beads time of descent from P to Q. In June of 1696, John Bernoulli proposed this problem as a public challenge, with a 6-month deadline (later extended to Easter 1697 at George Leibnizs request). Isaac Newton, then retired from academic life and serving as Warden of the Mint in London, received Bernoullis challenge on January 29, 1697. The very next day he communicated his own solutionthe curve of minimal descent time is an arc of an inverted cycloidto the Royal Society of London. For a modern derivation of this result, suppose the bead starts from rest at the origin P and let y D y.x/ be the equation of the desired curve in a coordinate system with the y-axis pointing downward. Then a mechanical analogue of Snells law in optics implies that sin v Dconstant, (i) where denotes the angle of deection (from the vertical) of the tangent line to the curveso cot D y0.x/ (why?)and v Dp2gy is the beads velocity when it has descended a distance y vertically (from KE D 1 2mv2 Dmgy D!PE).(a) First derive from Eq. (i) the differential equationdy dx Ds2a!y y; (ii)where a is an appropriate positive constant. (b) Substitute y D 2asin2 t, dy D 4asint costdtin (ii) to derive the solution x D a.2t !sin2t/; y D a.1!cos2t/ (iii) for which t D y D 0 when x D 0. Finally, the substitution of & D 2t in (iii) yields the standard parametric equations x D a.& !sin&/, y D a.1!cos&/ of the cycloid that is generated by a point on the rim of a circular wheel of radius a as it rolls along the xaxis. [See Example 5 in Section 9.4 of Edwards and Penney, Calculus: Early Transcendentals, 7th edition (Upper Saddle River, NJ: Prentice Hall, 2008).]

Unfortunately, we don't have that question answered yet. But you can get it answered in just 5 hours by Logging in or Becoming a subscriber.

Becoming a subscriber
Or look for another answer

×

Login

Login or Sign up for access to all of our study tools and educational content!

Forgot password?
Register Now

×

Register

Sign up for access to all content on our site!

Or login if you already have an account

×

Reset password

If you have an active account we’ll send you an e-mail for password recovery

Or login if you have your password back