Explanation It is given that the function for the intensity I of the light at an angle θ is I = I 0 [ sin ( k sin θ ) k sin θ ] 2 . Differentiate the function I = I 0 [ sin ( k sin θ ) k sin θ ] 2 with respect to θ . d I d θ = d d θ [ I 0 [ sin ( k sin θ ) k sin θ ] 2 ] = 2 I 0 [ sin ( k sin θ ) k sin θ ] [ k sin θ d d θ [ sin ( k sin θ ) ] − sin ( k sin θ ) d d θ [ k sin θ ] ( k sin θ ) 2 ] = 2 I 0 [ sin ( k sin θ ) k sin θ ] [ k sin θ cos ( k sin θ ) d d θ [ k sin θ ] − sin ( k sin θ ) [ k cos θ ] ( k sin θ ) 2 ] = 2 I 0 [ sin ( k sin θ ) sin θ ] [ sin θ cos ( k sin θ ) [ k cos θ ] − cos θ sin ( k sin θ ) ( k sin θ ) 2 ] Set d I d θ = 0 and obtain the value of θ where I has maximum and minimum

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ISBN: 9780134437705

Author: Washington

Publisher: PEARSON

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Chapter 27, Problem 104RE

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To show: The equation ksinθ=tan(ksinθ) represents the maximum and minimum values of I.

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Chapter 27, Problem 103RE

Chapter 27, Problem 105RE

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By considering appropriate series expansions, e². e²²/2. e²³/3. .... = = 1 + x + x² + · ... when |x| < 1. By expanding each individual exponential term on the left-hand side the coefficient of x- 19 has the form and multiplying out, 1/19!1/19+r/s, where 19 does not divide s. Deduce that 18! 1 (mod 19).

Proof: LN⎯⎯⎯⎯⎯LN¯ divides quadrilateral KLMN into two triangles. The sum of the angle measures in each triangle is ˚, so the sum of the angle measures for both triangles is ˚. So, m∠K+m∠L+m∠M+m∠N=m∠K+m∠L+m∠M+m∠N=˚. Because ∠K≅∠M∠K≅∠M and ∠N≅∠L, m∠K=m∠M∠N≅∠L, m∠K=m∠M and m∠N=m∠Lm∠N=m∠L by the definition of congruence. By the Substitution Property of Equality, m∠K+m∠L+m∠K+m∠L=m∠K+m∠L+m∠K+m∠L=°,°, so (m∠K)+ m∠K+ (m∠L)= m∠L= ˚. Dividing each side by gives m∠K+m∠L=m∠K+m∠L= °.°. The consecutive angles are supplementary, so KN⎯⎯⎯⎯⎯⎯∥LM⎯⎯⎯⎯⎯⎯KN¯∥LM¯ by the Converse of the Consecutive Interior Angles Theorem. Likewise, (m∠K)+m∠K+ (m∠N)=m∠N= ˚, or m∠K+m∠N=m∠K+m∠N= ˚. So these consecutive angles are supplementary and KL⎯⎯⎯⎯⎯∥NM⎯⎯⎯⎯⎯⎯KL¯∥NM¯ by the Converse of the Consecutive Interior Angles Theorem. Opposite sides are parallel, so quadrilateral KLMN is a parallelogram.

By considering appropriate series expansions, ex · ex²/2 . ¸²³/³ . . .. = = 1 + x + x² +…… when |x| < 1. By expanding each individual exponential term on the left-hand side and multiplying out, show that the coefficient of x 19 has the form 1/19!+1/19+r/s, where 19 does not divide s.

Chapter 27 Solutions

Basic Technical Mathematics

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Chapter 27 Solutions