Explanation Given: k ( t ) = a b ( b 2 c o s 2 t + a 2 s i n 2 t ) 3 2 , a ≥ b Calculation: For a ≥ b . b 2 c o s 2 t + a 2 s i n 2 t ≥ b 2 c o s 2 t + b 2 s i n 2 t b 2 c o s 2 t + a 2 s i n 2 t ≥ b 2 ( c o s 2 t + s i n 2 t ) b 2 c o s 2 t + a 2 s i n 2 t ≥ b 2 ( 1 ) b 2 c o s 2 t + a 2 s i n 2 t ≥ b 2 Take square root on both sides. ( b 2 c o s 2 t + a 2 s i n 2 t ) 1 2 ≥ b Now, cube on both sides, ( b 2 c o s 2 t + a 2 s i n 2 t ) 3 2 ≥ b 3 1 ( b 2 c o s 2 t + a 2 s i n 2 t ) 3 2 ≤ 1 b 3 Multiplying ‘ ab’ on both sides, we get a b ( b 2 c o s 2 t + a 2 s i n 2 t ) 3 2 ≤ a b b 3 a b ( b 2 c o s 2 t + a 2 s i n 2 t ) 3 2 ≤ a b 2 This implies k ( t ) ≤ a b 2 ..... (i) Again, for a ≥ b

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Chapter 13.4, Problem 27E

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Chapter 13 Solutions

CALCULUS 4E (HC) W/ ACHIEVE ACCESS

Ch. 13.1 - Prob. 1PQ Ch. 13.1 - Prob. 2PQ Ch. 13.1 - Prob. 3PQ Ch. 13.1 - Prob. 4PQ Ch. 13.1 - Prob. 5PQ Ch. 13.1 - Prob. 6PQ Ch. 13.1 - Prob. 1E Ch. 13.1 - Prob. 2E Ch. 13.1 - Prob. 3E Ch. 13.1 - Prob. 4E

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