Explanation Given information: θ = θ M sin ( k t ) ...... ( 1 ) Calculation: Differentiate with respect to t, we get d θ d t = θ M cos ( k t ) × k d θ d t = θ M k cos ( k t ) Again, differentiate with respect to t, we get d 2 θ d t 2 = θ M k ( − sin ( k t ) × k ) <

7th Edition

ISBN: 9781524916817

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

Publisher: Kendall Hunt Publishing

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Chapter 3, Problem 63SP

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Toproof: d2θdt2+k2θ=0

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Chapter 3, Problem 62SP

Chapter 3, Problem 64SP

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2. A tank with a capacity of 650 gal. originally contains 200 gal of water with 100 lb. of salt in solution. Water containing 1 lb. of salt per gallon is entering at a rate of 4 gal/min, and the mixture is allowed to flow out of the tank at a rate of 3 gal/min. a. Find the amount of salt in the tank at any time prior to the instant when the tank begins to overflow (650 gallons). b. Find the concentration (in pounds per gallon) of salt in the tank when the tank hits 400 gallons. D.E. for mixture problems: dv dt=11-12 dA A(t) dt

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Chapter 3 Solutions

Calculus

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Toproof : d 2 θ d t 2 + k 2 θ = 0

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Toproof: d2θdt2+k2θ=0

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Chapter 3 Solutions