Explanation Given information: Using the intermediate value theorem, prove that the hands of a clockcoincide at least once between 1 P.M. and 1:15 P.M . Calculation: Let us assume the function, f(t)= α ( t ) − − − − − ( 1 ) Where, α is the angle between largest and smallest watch hands. This is a continuous function where t=0 at 12 : 00pm and it is given inminutes. For example at 12:00 pm both hands math pointing to the 12. Then f(t) is given by, f ( t = 0 ) = 0 ° − − − − − ( 2 ) Now at, 1:00 pm t=60 and f(t) is given by, f ( t = 60 ) = ( 360 12 ) = 30 ° − − − − ( 3 ) Because one hand is pointing to the 12 while the other hand is pointing to the 1.S that, the hands form an angle of 30 ° as can be observed. On the other hand, at1:15 pm, the largest hand points to the 3(15 min), while the other hand has travelled 1 4 of the angular distance between the 1 and the 2. This angular distance is given by, 1 4 ( 30 ° ) = 15 ° 2 − − − − ( 4 ) In this case the angle is α < 0 , since the angles are measured in counter clockwise direction

6th Edition

ISBN: 9781465208880

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 2.3, Problem 51PS

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To calculate: To prove that the hands of a clockcoincide at least once between 1 P.M. and 1:15 P.M.

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Chapter 2.3, Problem 50PS

Chapter 2.3, Problem 52PS

Chapter 2 Solutions

Calculus

Ch. 2.1 - Prob. 1PS Ch. 2.1 - Prob. 2PS Ch. 2.1 - Prob. 3PS Ch. 2.1 - Prob. 4PS Ch. 2.1 - Prob. 5PS Ch. 2.1 - Prob. 6PS Ch. 2.1 - Prob. 7PS Ch. 2.1 - Prob. 8PS Ch. 2.1 - Prob. 9PS Ch. 2.1 - Prob. 10PS

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To prove that the hands of a clockcoincide at least once between 1 P.M. and 1:15 P.M.

Solution Summary: The author explains how the intermediate value theorem proves that the hands of a clockcoincide at least once between 1 P.M.

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To calculate: To prove that the hands of a clockcoincide at least once between 1 P.M. and 1:15 P.M.

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Chapter 2 Solutions