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Orthogonal Trajectories. A geometric problem occurring

Fundamentals of Differential Equations | 8th Edition | ISBN: 9780321747730 | Authors: R. Kent Nagle, Edward B. Saff, Arthur David Snider ISBN: 9780321747730 43

Solution for problem 32E Chapter 2.4

Fundamentals of Differential Equations | 8th Edition

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Fundamentals of Differential Equations | 8th Edition | ISBN: 9780321747730 | Authors: R. Kent Nagle, Edward B. Saff, Arthur David Snider

Fundamentals of Differential Equations | 8th Edition

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Problem 32E

Orthogonal Trajectories. A geometric problem occurring often in engineering is that of finding a family of curves (orthogonal trajectories) that intersects a given family of curves orthogonally at each point. For example, we may be given the lines of force of an electric field and want to find the equation for the equipotential curves. Consider the family of curves described by F (x,y) = k , where k is a parameter. Recall from the discussion of equation (2) that for each curve in the family, the slope is given by Recall that the slope of a curve that is orthogonal (perpendicular) to a given curve is just the negative reciprocal of the slope of the given curve. Using this fact, show that the curves orthogonal to the family F (x,y) = k satisfy the differential equation (b) Using the preceding differential equation, show that the orthogonal trajectories to the family ofcircles x2 +y2 = k are just straight lines through the origin (see Figure 2.10). (c) Show that the orthogonal trajectories to the family of hyperbolas are the hyperbolasx 2 = y2 = k (see Figure 2.11).

Step-by-Step Solution:

SolutionStep 1In this problem, we have to show that the curves orthogonal to the family F (x,y) = k satisfy the differential equation Step 2Here, we have to find the perpendicular slope of orthogonal curve If the curve is perpendicular then the slope is Rewrite the above equation where Hence, it is proved that the differential equation will be Step 2Let us assume that Now, differentiate with respect to y and Now, the differential equation will be Divide by 2 on both...

Step 2 of 3

Chapter 2.4, Problem 32E is Solved
Step 3 of 3

Textbook: Fundamentals of Differential Equations
Edition: 8
Author: R. Kent Nagle, Edward B. Saff, Arthur David Snider
ISBN: 9780321747730

This textbook survival guide was created for the textbook: Fundamentals of Differential Equations , edition: 8. Since the solution to 32E from 2.4 chapter was answered, more than 301 students have viewed the full step-by-step answer. Fundamentals of Differential Equations was written by and is associated to the ISBN: 9780321747730. The full step-by-step solution to problem: 32E from chapter: 2.4 was answered by , our top Calculus solution expert on 07/11/17, 04:37AM. This full solution covers the following key subjects: family, Orthogonal, given, curves, trajectories. This expansive textbook survival guide covers 67 chapters, and 2118 solutions. The answer to “Orthogonal Trajectories. A geometric problem occurring often in engineering is that of finding a family of curves (orthogonal trajectories) that intersects a given family of curves orthogonally at each point. For example, we may be given the lines of force of an electric field and want to find the equation for the equipotential curves. Consider the family of curves described by F (x,y) = k , where k is a parameter. Recall from the discussion of equation (2) that for each curve in the family, the slope is given by Recall that the slope of a curve that is orthogonal (perpendicular) to a given curve is just the negative reciprocal of the slope of the given curve. Using this fact, show that the curves orthogonal to the family F (x,y) = k satisfy the differential equation (b) Using the preceding differential equation, show that the orthogonal trajectories to the family ofcircles x2 +y2 = k are just straight lines through the origin (see Figure 2.10). (c) Show that the orthogonal trajectories to the family of hyperbolas are the hyperbolasx 2 = y2 = k (see Figure 2.11).” is broken down into a number of easy to follow steps, and 187 words.

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