Question

Fastest descent time The cycloid is the curve traced by a point on the rim of a rolling wheel. Imagine a wire shaped like an inverted cycloid (see figure). A bead sliding down this wire without friction has some remarkable properties. Among all wire shapes, the cycloid is the shape that produces the fastest descent time (see the Guided Project The amazing cycloid for more about the brachistochrone property). It can be shown that the descent time between any two points \(0 \leq a < b \leq \pi\) on the curve is $$ \text { descent time }=\int_{a}^{b} \sqrt{\frac{1-\cos t}{g(\cos a-\cos t)}} d t $$ where \(g\) is the acceleration due to gravity, \(t=0\) corresponds to the top of the wire, and \(t=\pi\) corresponds to the lowest point or the wire. a. Find the descent time on the interval \([a, b]\) by making the substitution \(u=\cos t\) b. Show that when \(b=\pi\), the descent time is the same for all values of \(a\); that is, the descent time to the bottom of the wire is the same for all starting points.

Short answer

Question: Show that the descent time to the bottom of the wire is the same for all starting points when \(b = \pi\). Answer: This is true because the integral used to calculate the descent time has terms that are independent of \(a\), and only the lower limit of integration depends on \(a\). However, since we integrate from \([-1-\cos a]\) to 0, scaling the wire would only affect the amount of the curve traversed while not affecting the time dependent on \(a\). Therefore, for all values of \(a\), the descent time to the bottom of the wire will be the same.

Step-by-step solution

Perform the substitution \(u = \cos t\)

To find the descent time on the interval \([a, b]\), we'll make the substitution \(u = \cos t\). To do that, we first need to find \(du\) in terms of \(dt\). Since \(u = \cos t\), the derivative with respect to \(t\) is \(-\sin t dt = du\). Now, we need to change the limits of integration. When \(t = a\), \(u = \cos a\) and when \(t = b\), \(u = \cos b\). The integral now becomes: $$ \int_{\cos b}^{\cos a}\sqrt{\frac{1 - u}{g(u - \cos a)}} \frac{-1}{\sin t} du $$

Simplify the integral

Now, we need to simplify the integral. We have: $$-\int_{\cos b}^{\cos a}\frac{\sqrt{1-u}}{\sqrt{g(u-\cos a)}\sin{t}}du$$ Recall that we have \(\sin^2 t + \cos^2 t = 1\), which means that \(\sin^2 t = 1 - \cos^2 t = 1 - u^2\). Taking the square root of both sides, we have \(\sin t = \sqrt{1 - u^2}\). Substitute this into our integral: $$ \int_{\cos b}^{\cos a}\frac{\sqrt{1-u}}{\sqrt{g(u-\cos a)}\sqrt{1-u^2}}du $$

Find the descent time on the interval \([a, b]\)

Now we can find the descent time on the interval \([a, b]\) by evaluating the integral: $$ \text{descent time} = \int_{a}^{b} \sqrt{\frac{1-\cos t}{g(\cos a-\cos t)}} dt = \int_{\cos b}^{\cos a}\frac{\sqrt{1-u}}{\sqrt{g(u-\cos a)}\sqrt{1-u^2}}du $$

Show that the descent time is the same for all values of \(a\) when \(b = \pi\)

Now we need to show that when \(b = \pi\), the descent time is the same for all values of \(a\). When \(b = \pi\), \(\cos b = -1\). Plugging this into our descent time formula: $$ \text{descent time} = \int_{-1}^{\cos a}\frac{\sqrt{1-u}}{\sqrt{g(u-\cos a)}\sqrt{1-u^2}}du $$ Notice that the two terms \(\sqrt{1 - u^2}\) and \(\sqrt{1 - u}\) in the integrand don't depend on \(a\). Furthermore, when taking the antiderivative with respect to \(u\), only the term \(\sqrt{g(u - \cos a)}\) in the denominator will have any dependence on \(a\). Therefore, we need to show that the dependence on \(a\) cancels out in the final antiderivative to demonstrate that the descent time is the same for all values of \(a\). Without loss of generality, we can assume \(g = 1\) for this purpose. Then the integral becomes: $$ \text{descent time} = \int_{-1}^{\cos a}\frac{\sqrt{1-u}}{\sqrt{(u-\cos a)}\sqrt{1-u^2}}du $$ Let's make another substitution: \(v = u - \cos a\). Then, \(dv = du\) and when \(u = -1\), \(v = -1 - \cos a\), and when \(u = \cos a\), \(v = 0\). The integral now becomes: $$\text{descent time} = \int_{-1-\cos a}^{0}\frac{\sqrt{1-(v+\cos a)}}{\sqrt{v}\sqrt{1-(v+\cos a)^2}}dv$$ Now, notice that each term in the integrand does not depend on \(a\), and only the lower limit of integration depends on \(a\). But since we integrate from \([-1-\cos a]\) to 0, this means that the descent time would be the same, as scaling the wire would only affect the amount of the curve traversed while not affecting the time dependent on \(a\). Therefore, for all values of \(a\), the descent time to the bottom of the wire will be the same.

Key concepts

Brachistochrone Problem

The Brachistochrone Problem is a famous challenge in the field of physics and mathematics, particularly involving calculus of variations. This problem asks a seemingly simple question: What is the shape of the path or curve down which a bead will slide (in the absence of friction) from one point to another in the shortest amount of time? The answer to this puzzle is not a straight line or a simple arc but a Cycloid.

A Cycloid is the curve traced by a point on the rim of a rolling circle. Johann Bernoulli first postulated in the 17th century that this curvaceous path provides the quickest descent under the influence of gravity, and this has been a source of rich mathematical exploration ever since. When a bead slides along a cycloid-shaped wire, it exemplifies the fastest descent, regardless of the starting point on the curve—a surprising and counterintuitive revelation that may at first baffle the student.

Calculus of Variations

The calculus of variations is a field within mathematics that deals with optimizing functionals, which are mappings from a function to the real numbers. Unlike typical calculus, which seeks the maxima and minima of functions, calculus of variations is concerned with finding functions that optimize certain quantities.

In the context of the Brachistochrone Problem, calculus of variations is used to derive the equation of the cycloid curve by finding the function that minimizes the descent time of an object under the force of gravity. This involves complex equations that describe the potential and kinetic energy of the sliding object and leads to an understanding of how the shape of the wire affects the speed of descent. This field is profound as it extends beyond simple geometric paths and has implications for physics, engineering, and economics.

Integration by Substitution

Integration by substitution, also known as u-substitution, is a method used in calculus to simplify the evaluation of integrals. It is the counterpart to the chain rule for derivatives. In practice, the substitution is chosen so that the integral is transformed into a simpler one that is more straightforward to compute.

In the provided step-by-step solution for the cycloid's descent time, integration by substitution becomes an invaluable tool. By setting up a substitution, we can rewrite the difficult portion of the integral in terms of a new variable, usually denoted by 'u'. This new variable can help us eliminate complex relationships within the integral, thereby simplifying the integration process. Understanding this concept is critical for solving many calculus problems, including those involved in the Brachistochrone Problem.

Acceleration Due to Gravity

Acceleration due to gravity, denoted by 'g', is the acceleration of an object due to the gravitational force of Earth. This constant acceleration affects all objects near Earth’s surface and is approximately equal to 9.8 meters per second squared (m/s²).

In physics problems and those in calculus involving motion, 'g' is a crucial value that dictates how quickly an object will accelerate towards Earth when in freefall. The descent time of a bead along a cycloid curve as posed in the Brachistochrone Problem is contingent on this acceleration. Understanding that gravity is the only force acting on the bead in this frictionless scenario, and integrating 'g' within the calculus equations allows us to derive precise measures of time and motion for various descent paths.