Explanation Given: The given expression is, | 2 π − 6 | − π (1) Calculation: Since ( 2 π − 6 ) > 0 , | 2 π − 6 | = ( 2 π − 6 ) Substitute ( 2 π − 6 ) for | 2 π − 6 | in equation (1)

Applied Calculus for the Managerial, Life, and Social Sciences

10th Edition

ISBN: 9781305887831

Author: Tan

Publisher: CENGAGE CO

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Chapter 1, Problem 7RE

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To find: The value of expression |2π−6|−π .

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Chapter 1, Problem 6RE

Chapter 1, Problem 8RE

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1. Abel's Theorem. The goal in this problem is to prove Abel's theorem by following a series of steps (each step must be justified). Theorem 0.1 (Abel's Theorem). If y1 and y2 are solutions of the differential equation y" + p(t) y′ + q(t) y = 0, where p and q are continuous on an open interval, then the Wronskian is given by W (¥1, v2)(t) = c exp(− [p(t) dt), where C is a constant that does not depend on t. Moreover, either W (y1, y2)(t) = 0 for every t in I or W (y1, y2)(t) = 0 for every t in I. 1. (a) From the two equations (which follow from the hypotheses), show that y" + p(t) y₁ + q(t) y₁ = 0 and y½ + p(t) y2 + q(t) y2 = 0, 2. (b) Observe that Hence, conclude that (YY2 - Y1 y2) + P(t) (y₁ Y2 - Y1 Y2) = 0. W'(y1, y2)(t) = yY2 - Y1 y2- W' + p(t) W = 0. 3. (c) Use the result from the previous step to complete the proof of the theorem.

2. Observations on the Wronskian. Suppose the functions y₁ and y2 are solutions to the differential equation p(x)y" + q(x)y' + r(x) y = 0 on an open interval I. 1. (a) Prove that if y₁ and y2 both vanish at the same point in I, then y₁ and y2 cannot form a fundamental set of solutions. 2. (b) Prove that if y₁ and y2 both attain a maximum or minimum at the same point in I, then y₁ and Y2 cannot form a fundamental set of solutions. 3. (c) show that the functions & and t² are linearly independent on the interval (−1, 1). Verify that both are solutions to the differential equation t² y″ – 2ty' + 2y = 0. Then justify why this does not contradict Abel's theorem. 4. (d) What can you conclude about the possibility that t and t² are solutions to the differential equation y" + q(x) y′ + r(x)y = 0?

Question 4 Find an equation of (a) The plane through the point (2, 0, 1) and perpendicular to the line x = y=2-t, z=3+4t. 3t, (b) The plane through the point (3, −2, 8) and parallel to the plane z = x+y. (c) The plane that contains the line x = 1+t, y = 2 − t, z = 4 - 3t and is parallel to the plane 5x + 2y + z = 1. (d) The plane that passes through the point (1,2,3) and contains the line x = 3t, y = 1+t, and z = 2-t. (e) The plane that contains the lines L₁: x = 1 + t, y = 1 − t, z = 2t and L2 : x = 2 − s, y = s, z = 2.

Chapter 1 Solutions

Applied Calculus for the Managerial, Life, and Social Sciences

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The value of expression | 2 π − 6 | − π .

Solution Summary: The author explains the value of expression left|2pi -6right|-Pi in equation (1).

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To find: The value of expression |2π−6|−π .

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Chapter 1 Solutions