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Chapter 17: Problem 18

18\. Find the curve joining two points in the upper half-plane, which, when revolved around the \(x\) axis, generates a surface of minimal surface area. If \(y=f(x)\) is the equation of the curve, then the desired surface area is \(I=\) \(2 \pi \int y d s=2 \pi \int y \sqrt{1+y^{2}} d x .\) So use the Euler equation in the modified form \(F-y^{\prime}\left(\partial F / \partial y^{\prime}\right)=c\), where \(F=y \sqrt{1+y^{\prime 2}} .\) (Hint: Begin by multiplying the equation through by \(\sqrt{1+y^{\prime 2}}\).)

Short Answer

Expert verified
Question: Find the curve y = f(x) that, when revolved around the x-axis, generates a surface of minimal surface area, given the formula for the desired surface area as \(I = 2 \pi \int y ds = 2 \pi \int y \sqrt{1+y'^2} dx\). Answer: The explicit equation for the curve y = f(x) could not be obtained in a simpler form; however, we derived a relationship between y and its derivatives: \(\frac{(y')^3}{3} - \frac{y'}{2} = -\frac{c^2}{y} + k\). This relationship determines the shape of the desired curve generating a surface of minimal area when revolved around the x-axis.

Step by step solution

01

Write down the given formulas and notations

Let's list down the given formulas and notations: - The equation of the curve is y = f(x). - The desired surface area is \(I = 2 \pi \int y ds = 2 \pi \int y \sqrt{1+y'^2} dx\). - Modified Euler equation: \(F-y'\left(\dfrac{\partial F}{\partial y'}\right)=c\). - Function F: \(F = y\sqrt{1+y'^2}\). Now, we'll use the Euler equation to find the relationship between f(x) and its derivatives.
02

Compute the partial derivative of F with respect to y'

Let's find the partial derivative of F with respect to y': \(\frac{\partial F}{\partial y'} = \frac{\partial \left(y \sqrt{1+y'^2}\right)}{\partial y'}\). Using the chain rule, we get \(\frac{\partial F}{\partial y'} = y\frac{\partial \left(\sqrt{1+y'^2}\right)}{\partial y'}=y\frac{y'}{\sqrt{1+y'^{2}}}\).
03

Write down the Euler equation and substitute the partial derivative

Now, let's use the modified Euler equation and substitute the partial derivative we computed in the previous step: \(F - y'\left(\frac{\partial F}{\partial y'}\right) = y\sqrt{1+y'^2} - y'\left(y\frac{y'}{\sqrt{1+y'^{2}}}\right) = c\).
04

Simplify the Euler equation

We will simplify the Euler equation by multiplying both sides by \(\sqrt{1+y'^{2}}\) as hinted: \(y\sqrt{1+y'^2} - y'\left(y\frac{y'}{\sqrt{1+y'^{2}}}\right)\left(\sqrt{1+y'^{2}}\right)= c\left(\sqrt{1+y'^{2}}\right)\). This simplifies to \(y\sqrt{1+y'^2} - yy'^{2} = c\sqrt{1+y'^{2}}\). Let's further simplify the equation: \(y = c\frac{\sqrt{1+y'^{2}}}{1-y'^2}\).
05

Integrate the simplified equation to find the curve equation

Now, let's solve for y by integrating the equation we derived in the previous step: \(y'\left(1 - y'^2\right) = \frac{c^2}{y^2}\). Integrate both sides with respect to x: \(\int\frac{dy}{dx}\left(1 - \frac{dy}{dx}^2\right)dx = \int\frac{c^2}{y^2}dx\). To solve this equation, let \(u = y'\), then \(du = y''dx\). Thus, we get \(\int u(1-u^2) du = c^2 \int \frac{1}{y^2} dy\). Integrating, we obtain \(\frac{u^3}{3} - \frac{u}{2} = -\frac{c^2}{y} + k\). Substitute \(u = y'\): \(\frac{(y')^3}{3} - \frac{y'}{2} = -\frac{c^2}{y} + k\). This equation gives a relationship between y and its derivatives. We could not obtain an explicit equation for the curve y = f(x) in a simpler form; however, we got a relationship involving the curve's equation and its derivatives that determines the shape of the desired curve generating a surface of minimal area when revolved around the x-axis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Calculus of Variations
Calculus of Variations is a field of mathematical analysis that deals with maximizing or minimizing functional expressions. In simpler terms, it looks for the shape or the path a function should take to produce the highest or lowest value for a given integral. One common application of this calculus is in finding the curve of minimal length or surface area among all possible curves connecting two points.

For our exercise, we are applying the calculus of variations to determine the curve that, when revolved around the x-axis, produces a surface with the minimal possible surface area. This is a classic problem known as the minimal surface area problem, and finding a solution involves a deep understanding of how changes in a curve's shape affect the surface area it generates upon revolution.
Euler-Lagrange Equation
The Euler-Lagrange Equation is a fundamental equation in the calculus of variations. It provides a necessary condition for a function to be an extremum (maximum or minimum) of a functional. This equation comes from the idea that if a curve minimizes or maximizes a certain quantity, small changes to the curve will not change the value of that quantity significantly.

In our minimal surface area problem, we use the Euler-Lagrange Equation to determine the function y=f(x) that minimizes the surface area. The equation involves the function itself, its first derivative (represented as y'), and partial derivatives with respect to these quantities. The step-by-step solution involves calculating these partial derivatives as well as manipulating the equation to isolate y and its derivatives.
Surface of Revolution
A surface of revolution is created when a curve in two dimensions is revolved around an axis. The resulting three-dimensional shape's surface area can be calculated by integrating along the path of the curve. Many objects in engineering and nature can be modeled as surfaces of revolution, such as wine glasses, satellite dishes, and even the Earth's zones between latitudes.

For our example, the curve described by the equation y=f(x) is revolved around the x-axis. This rotation forms the shape whose surface area we're seeking to minimize. Understanding surfaces of revolution is crucial in visualizing the physical shape that derives from the theoretical curve we are studying.
Partial Derivatives
Partial derivatives are derivatives of multi-variable functions where only one variable is allowed to change while keeping other variables constant. They tell us how a function changes as each variable changes a little bit. This concept is central to finding extremums in multi-variable calculus and, by extension, the calculus of variations.

In our problem, we calculate partial derivatives of the function F, which represents the area of the surface of revolution, with respect to the derivative of y, denoted by y'. These partial derivatives are then used in the Euler-Lagrange Equation to find a condition that the curve must satisfy to have a minimal surface area when revolved. Understanding partial derivatives is key to working through the steps of this problem and deciphering the relationship between the curve and its resulting surface area.

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Most popular questions from this chapter

25\. Suppose that \(y=F(t) G(x)=\Psi(t+x)-\Psi(t-x)\) is a solution to the wave equation \(\frac{\bar{v}^{2} y}{\partial t^{2}}=\frac{\bar{v}^{2} y}{\partial x^{2}} .\) Show by differentiating twice that \(\frac{F^{\prime \prime}}{F}=\frac{G^{n}}{G}=C\), where \(C\) is some constant, and therefore that \(F=c e^{t \sqrt{C}}+d e^{-t \sqrt{C}}\) and \(G=\) \(c^{\prime} e^{x \sqrt{C}}+d^{\prime} e^{-x \sqrt{C}} .\) Apply the condition \(y(t, 0)=y(t, l)=\) 0 to show that \(C\) must be negative, and hence derive the solution \(F(t)=A \cos N t, G(x)=B \sin N x\) for the appropriate choice of \(A, B\), and \(N\) (d'Alembert).

38\. If \(y=\arctan x\), show that \(\sin y=x / \sqrt{1+x^{2}}\) and \(\cos y=\) \(1 / \sqrt{1+x^{2}}\). Then, if \(p=x / \sqrt{1+x^{2}}\), show that \(\sqrt{1-p^{2}}=\) \(1 / \sqrt{1+x^{2}}\). Since \(y=\arcsin p\), it follows that \(d y=\) \(d p / \sqrt{1-p^{2}}\) and \(d p=d x /\left(1+x^{2}\right)^{3 / 2}\). Conclude that $$ d y=\frac{d x}{1+x^{2}} \quad \text { (Euler) } $$

29\. Calculate the first four nonzero terms of the power series for \(y=\cos z\) using Maclaurin's technique without explicitly using the derivatives of the cosine or sine. Assume that the radius of the circle is 1 .

The modern way to derive Kepler's area law is to break the force into its radial and transverse components, rather than the tangential and normal components used by Hermann, and use polar coordinates whose origin is the center of force. Assume then that the center of force is at the origin of a polar coordinate system. Using vector notation, \(\operatorname{set} \mathbf{u}_{r}=\mathbf{i} \cos \theta+\mathbf{j} \sin \theta\) and \(\mathbf{u}_{\theta}=-\mathbf{i} \sin \theta+j \cos \theta .\) Show that \(d \mathbf{u}_{r} / d \theta=\mathbf{u}_{\theta}\) and \(d \mathbf{u}_{\theta} / d \theta=-\mathbf{u}_{r}\). Then show that if \(\mathbf{r}=r \mathbf{u}_{r}\), then the velocity \(\mathbf{v}\) is given by \(r(d \theta / d t) \mathbf{u}_{\theta}+\) \((d r / d t) \mathbf{u}_{r}\). Show next that the radial component \(a_{r}\) and the transverse component \(a_{\theta}\) of the acceleration are given by $$ a_{r}=\frac{d^{2} r}{d t^{2}}-r\left(\frac{d \theta}{d t}\right)^{2} \quad \text { and } \quad a_{\theta}=r \frac{d^{2} \theta}{d t^{2}}+2 \frac{d r}{d t} \frac{d \theta}{d t} $$ Since the force is central, \(a_{\theta}=0 .\) Multiply the differential equation expressing that fact by \(r\) and integrate to get \(r^{2} \frac{d \theta}{d t}=k\), where \(k\) is a constant. Show finally that \(r^{2} \frac{d \theta}{d t}=\) \(\frac{d A}{d t}\), where \(A\) is the area swept out by the radius vector. This proves Kepler's law of areas.

44\. Use Lagrange's technique to calculate the quantities \(p, q\), \(r\) for the function \(f(x)=\sqrt{x}\) and thus determine the first. three terms of its power series representation.
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