What is the simplest solution to the Brachistochrone problem and the Tautochrone problem involving calculus? (I know that the cycloid is the solution but I need a simple calculus proof as to why this is the case)

The Brachistochrone (shortest sliding time) and the Tautochrone (equal sliding time wherever release occurs) are the same curve. Both are cycloids - loci of the a point at the edge of a wheel as it rolls. The situation is discussed in G. B. Thomas' classic textbook "Calculus and Analytic Geometry", 3rd edition, published by Addison Wesley in 1960. There is a ninth edition with the same title with Thomas and Finney listed as authors, published in 1999. I don't know if the proof is there. For a complete proof, Thomas refers to the book "Calculus of Variations" by G.A. Bliss, published in 1925. The original proof was by James and John Bernoulli.

a can containing 54 in cubed of tuna and water is to be made in the form of a circular cylinder. What dimensions of the can will require the least amount of material?

To understand the simplest solution to the Brachistochrone problem and the Tautochrone problem using calculus, it's important to first understand the problems themselves.

The Brachistochrone problem seeks to find the curve along which a bead, subject only to gravity, will slide from one point to another in the shortest amount of time. The Tautochrone problem, on the other hand, aims to find the curve where the time of descent is the same for all points along the curve.

Now, let's focus on the Brachistochrone problem. To solve it using calculus, we need to minimize the time of descent. However, this is equivalent to minimizing the potential energy, as the total mechanical energy (kinetic energy + potential energy) is conserved during the bead's motion.

Let's assume the bead starts at the origin (0,0) and needs to reach a point (x, y) along the curve. We consider the curve as a function y = f(x) in the Cartesian coordinate system.

Using calculus, we can express the change in potential energy as the work done against gravity during the displacement of the bead. This can be written as:

dU = -mg(dy/dx)dx

Where:
dU = change in potential energy
m = mass of the bead
g = acceleration due to gravity
(dy/dx) = derivative of y with respect to x
dx = small change in x

We aim to minimize the potential energy, so we need to minimize the integral of dU over the curve.

∫dU = ∫-mg(dy/dx)dx

To perform this optimization, we can use the calculus of variations, specifically the Euler-Lagrange equation. By applying this equation, we will obtain the differential equation that the solution curve must satisfy.

The Euler-Lagrange equation for this problem is:

d/dx (∂f/∂(dy/dx)) - ∂f/∂y = 0

Here, f represents the integrand -mg(dy/dx) in our case.

By solving this differential equation for the curve y = f(x), we can find the solution to the Brachistochrone problem.

It can be shown that the cycloid curve satisfies this differential equation, and thus it is the solution to the Brachistochrone problem. The cycloid is the curve traced by a point on the circumference of a rolling circle.

Similarly, for the Tautochrone problem, it can be proven that the cycloid curve is the solution by considering the conservation of mechanical energy and minimizing the time of descent.

In summary, solving the Brachistochrone and Tautochrone problems involves formulating the problem in terms of potential energy and time optimization, then applying the calculus of variations to find the curves that minimize the objective. In both cases, it can be shown that the cycloid curve is the simplest solution to these problems through the derivation of the Euler-Lagrange equation and solving the resulting differential equation.